Chapter 11 Resource Masters - anderson1.k12.sc.us · PDF fileChapter 11 Resource Masters ... Mid-Chapter Test This one-page test provides an option to assess the first half of the

Chapter 11 Resource Masters

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Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such materials be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with the Glencoe Precalculus program. Any other reproduction, for sale or other use, is expressly prohibited.

Send all inquiries to:Glencoe/McGraw-Hill8787 Orion PlaceColumbus, OH 43240 - 4027

ISBN: 978-0-07-893812-2MHID: 0-07-893812-0

Printed in the United States of America.

2 3 4 5 6 7 8 9 10 079 18 17 16 15 14 13 12 11 10

StudentWorks PlusTM includes the entire Student Edition text along with the worksheets in this booklet.

TeacherWorks PlusTM includes all of the materials found in this booklet for viewing, printing, and editing.

Cover: Jason Reed/Photodisc/Getty Images

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Contents

Teacher’s Guide to Using the Chapter 11 Resource Masters ........................................... iv

Chapter ResourcesStudent-Built Glossary ....................................... 1Anticipation Guide (English) .............................. 3Anticipation Guide (Spanish) ............................. 4

Lesson 11-1Descriptive StatisticsStudy Guide and Intervention ............................ 5Practice .............................................................. 7Word Problem Practice ..................................... 8Enrichment ........................................................ 9TI-Nspire Activity ............................................. 10

Lesson 11-2Probability DistributionsStudy Guide and Intervention .......................... 11Practice ............................................................ 13Word Problem Practice ................................... 14Enrichment ...................................................... 15

Lesson 11-3The Normal DistributionStudy Guide and Intervention .......................... 16Practice ............................................................ 18Word Problem Practice ................................... 19Enrichment ...................................................... 20Graphing Calculator Activity ............................ 21

Lesson 11-4The Central Limit TheoremStudy Guide and Intervention .......................... 22Practice ............................................................ 24Word Problem Practice ................................... 25Enrichment ...................................................... 26

Lesson 11-5Confidence IntervalsStudy Guide and Intervention .......................... 27Practice ............................................................ 29Word Problem Practice ................................... 30Enrichment ...................................................... 31

Lesson 11-6Hypothesis TestingStudy Guide and Intervention .......................... 32Practice ............................................................ 34Word Problem Practice ................................... 35Enrichment ...................................................... 36

Lesson 11-7Correlation and Linear RegressionStudy Guide and Intervention .......................... 37Practice ............................................................ 39Word Problem Practice ................................... 40Enrichment ...................................................... 41

AssessmentChapter 11 Quizzes 1 and 2 ........................... 43Chapter 11 Quizzes 3 and 4 ........................... 44Chapter 11 Mid-Chapter Test .......................... 45Chapter 11 Vocabulary Test ........................... 46Chapter 11 Test, Form 1 ................................. 47Chapter 11 Test, Form 2A ............................... 49Chapter 11 Test, Form 2B ............................... 51Chapter 11 Test, Form 2C .............................. 53Chapter 11 Test, Form 2D .............................. 55Chapter 11 Test, Form 3 ................................. 57Chapter 11 Extended-Response Test ............. 59Standardized Test Practice ............................. 60

Answers ........................................... A1–A28

Chapter 11 iii Glencoe Precalculus

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Teacher’s Guide to Using theChapter 11 Resource Masters

The Chapter 11 Resource Masters includes the core materials needed for Chapter 11. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

Chapter ResourcesStudent-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 11-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson Resources Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Guided Practice exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply to the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI–Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

Chapter 11 iv Glencoe Precalculus

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Assessment OptionsThe assessment masters in the Chapter 11 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This one-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes two parts: multiple-choice questions with bubble-in answer format andshort-answer free-response questions.

Answers• The answers for the Anticipation Guide

and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

Chapter 11 v Glencoe Precalculus

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(continued on the next page)

This is an alphabetical list of key vocabulary terms you will learn in Chapter 11. As you study this chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Precalculus Study Notebook to review vocabulary at the end of the chapter.

Vocabulary TermFound

on PageDefinition/Description/Example

binomial distribution

Central Limit Theorem

confidence interval

correlation

correlation coefficient

critical values

empirical rule

hypothesis test

inferential statistics

level of significance

Student-Built Glossary11

Chapter 11 1 Glencoe Precalculus

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Vocabulary TermFound

on PageDefinition/Description/Example

normal distribution

percentiles

probability distribution

P-value

random variable

regression line

residual

standard error of the mean

t-distribution

z-value

Student-Built Glossary11



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11

Before you begin Chapter 11

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree ordisagree, write NS (Not Sure).

After you complete Chapter 11

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write anexample of why you disagree.

STEP 1 A, D, or NS

StatementSTEP 2 A or D

1. In a negatively skewed distribution, the mean is less than the median.

2. Percentiles divide a distribution into 100 equal groups.

3. A continuous random variable takes on a countable number of possible values.

4. In a probability distribution, the sum of P(X) must be 1.

5. In a normal distribution, the mean, median, and mode are equal and are located at the center of the distribution.

6. A z-value represents the number of standard deviations that a given data value is from the mean.

7. If the sample size is adequately large, the distribution of the sample means will a have a mean and standard deviation equal to the population mean and standard deviation.

8. As the probability of success increases to 0.5, the shape of the binomial distribution begins to resemble the normal distribution.

9. A confidence level gives the probability that the interval estimate will include a given parameter.

10. The null hypothesis states that there is a difference between the sample value and the population parameter.

11. Extrapolation uses an equation to make predictions over the range of the data.

12. A correlation coefficient of -0.95 shows a strong negative correlation between two variables.

Anticipation GuideInferential Statistics

Step 2

Step 1



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Ejercicios preparatoriosEstadística inferencial

11

Capítulo 11 4 Precálculo de Glencoe

Después de completar el Capítulo 11

• Relee cada enunciado y escribe A o D en la última columna.

• Compara la última columna con la primera. ¿Cambiaste de opinión sobre alguno de los enunciados?

• En los casos en que hayas estado en desacuerdo con el enunciado, escribe en una hoja aparte un ejemplo de por qué no estás de acuerdo.

Paso 2

Antes de comenzar el Capítulo 11

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a), escribe NS (no estoy seguro(a)).

Paso 1

PASO 1 A, D o NS

EnunciadoPASO 2 A o D

1. La media es menor que la mediana en una distribución asimétrica negativa.

2. Los percentiles dividen una distribución en 100 grupos iguales.

3. Se puede contar el número de valores posibles que puede tener una variable aleatoria continua.

4. La suma de P(X) debe ser igual a 1 en una distribución probabilística.

5. La media, la mediana y la moda de una distribución normal son iguales y están ubicadas en el centro de la distribución.

6. El valor de z representa el número de desviaciones estándar que separan la media de un valor dado de los datos.

7. Si la muestra es de un tamaño suficientemente grande, la distribución de las medias de la muestra tendrá una media y una desviación estándar igual a la media y la desviación estándar de la población.

8. A medida que la probabilidad de éxito se acerca a 0.5, la forma de la distribución binomial se empieza a parecer a la forma de la distribución normal.

9. Los niveles de confianza indican la probabilidad de que el intervalo del estimado incluya el parámetro dado.

10. La hipótesis nula establece que hay una diferencia entre el valor de la muestra y el parámetro de la población.

11. La extrapolación utiliza una ecuación para predecir valores más allá del rango de los datos.

12. Un coeficiente de correlación de -0.95 indica una fuerte correlación negativa entre dos variables.


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Study Guide and InterventionDescriptive Statistics

11-1

Describing Distributions When a data set has a symmetrical distribution, the data are evenly distributed on both sides of the mean. In a negatively skewed distribution, the left side of the distribution extends farther than the right and the mean is less than the median. In a positively skewed distribution, the right side extends farther than the left and the mean is greater than the median. If a distribution is reasonably symmetric, the mean and standard deviation can be used to describe it. Otherwise, the five-number summary is better.

The table shows the number of passengers who boarded planes at 36 airports in the United States in one year.

Number of Passengers

30,526 30,372 26,623 22,722 16,287 15,246 14,807 14,117 14,054

13,547 12,916 12,616 11,906 11,622 11,489 10,828 10,653 10,008

9703 9594 9463 9348 9125 8572 7300 6772 6549

6126 5907 5712 5287 4848 4832 4820 4750 4684

a. Construct a histogram and use it to describe the shape ofthe distribution.

Enter the data in L1, turn on Plot1, and select histogram.Press GRAPH . Press ZOOM and choose ZoomStat.Press TRACE to see how many data value are in each bar. The graph is positively skewed.

b. Summarize the center and spread of the data using eitherthe mean and standard deviation or the five-number summary. Justify your choice.Use the five-number summary since the data are skewed.

The number of passengers range from 4684 to 30,526, andthe median number of passengers is about 9856. Half of theairports had between 6338 and 13,800 passengers.

Exercise 1. A restaurant manager advertises a 20-ounce sirloin steak on his menu.

The weights of a sample of the steaks is shown.

Steak Weights (ounces)

20 20 20 18 20 18 18 19 19 19

21 20 17 20 19 21 18 19 20 20

a. Construct a histogram and use it to describe the shapeof the distribution.

b. Summarize the center and spread of the data usingeither the mean and standard deviation or the five-numbersummary. Justify your choice.

Example

[4000, 35,000] scl: 3000 by [0, 15] scl: 1

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11-1 Study Guide and Intervention (continued)

Descriptive Statistics

Measures of Position The quartiles given by the five-number summary specify the positions of data values within a distribution. For this reason, box plots are most useful for side-by-side comparisons of two or more distributions.

The ages of Oscar-winning actresses from two 20-year periods are shown. Construct side-by-side box plots of the data sets. Then use this display to compare the distributions.

1969–1988

61 35 34 34 26 37 42 41 35 31

41 33 30 74 33 49 38 61 21 41

1989–2008

26 80 42 29 33 36 45 49 39 34

26 25 33 35 35 28 30 29 61 32

Input the data into L1 and L2. Turn on Plot1 and Plot2 andchoose a box plot with outliers shown as the type of graph.

From the graph, we can see that the median age for the first20 years is just slightly higher than the median age for the second20 years. The range of ages for the middle 50% of the data setsis greater for the actresses from the second 20-year period.

Both data sets have outliers greater than the rest of the data. Ignoring the outliers, the distribution for the second 20-year period is more symmetric than the first and the range of data for the second is less than that of the first.

Exercise

The number of credit cards owned by students in a statisticsclass is shown. Construct side-by-side box plots of the data sets.Then use this display to compare the distributions.

Example

[15, 85] scl: 5 by [0, 0.5] scl: 0.125

Females

1 8 0 1 5

4 4 0 5 0

0 0 3 0 1

Males

0 4 0 0 2

0 5 7 0 0

3 5 0 0 1


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11-1 PracticeDescriptive Statistics

1. WEATHER The average wind speeds recorded at various weather stations in the United States are listed below.

StationSpeed

(mph)Station

Speed

(mph)Station

Speed

(mph)

Albuquerque 8.9 Anchorage* 7.1 Atlanta* 9.1

Baltimore* 9.1 Boston* 12.5 Chicago 10.4

Dallas– Ft. Worth 10.8 Honolulu* 11.3 Indianapolis 9.6

Kansas City 10.7 Las Vegas 9.3 Little Rock 7.8

Los Angeles* 6.2 Memphis 8.8 Miami* 9.2

Minneapolis– St. Paul 10.5 New Orleans 8.1 New York City* 9.4

Philadelphia* 9.5 Phoenix 6.2 Seattle* 9.0

Source: National Climatic Data Center

a. Construct a histogram and use it to describe the shapeof the distribution.

b. Summarize the center and spread of the data using eitherthe mean and standard deviation or the five-numbersummary. Justify your choice.

2. OCEANS The ten weather stations with an asterisk haverelatively close proximity to either the Atlantic Ocean orPacific Ocean. Construct side-by-side box plots of the datasets. Then use this display to compare the distributions.

3. SCORES The table gives the frequency distribution of thescores on a test.

a. Construct a percentilegraph of the data.

b. Estimate the percentile ranka score of 62 would have in thisdistribution. Interpret itsmeaning.

10

0

20

30

40

Cum

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ive

Perc

enta

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50

60

70

80

90

100

Score50 60 70 80 90 100

Class Boundaries f

55.5–60.5 3

60.5–65.5 8

65.5–70.5 12

70.5–75.5 5

75.5–80.5 9


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11-1 Word Problem PracticeDescriptive Statistics

1. SHOES A shoe store employee designs a display by placing shoe boxes in ten stacks. The number of boxes in each stack are 5, 7, 9, 11, 13, 10, 9, 8, 7, and 5.

a. Construct a box plot and use it to describe the shape of the distribution.

b. Summarize the center and spread of

the data using either the mean and standard deviation or the five-number summary. Justify your choice.

2. MEDICINE A histogram for the number of patients treated at 50 U.S. cancer centers in one year is shown.

Source: U.S. News Online

a. Which is greater, the mean or the median of the data set? Explain.

b. Sketch the general shape of a box plot that would represent the histogram.

3. EXAMS The table gives the frequencies of the final exam scores of 50 students in two precalculus classes.

Class BoundariesFrequency

f

42.5–52.5 4

52.5–62.5 10

62.5–72.5 15

72.5–82.5 13

82.5–92.5 7

92.5–102.5 1

a. Construct a percentile graph ofthe data.

10

0

20

30

40

Cum

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ive

Perc

enta

ge50

60

70

80

90

100

Score5040 60 70 80 90 100

b. Estimate the percentile rank a test score of 77 would have in this distribution. Interpret its meaning.

4. MOVIES The ages of movie patrons ina theater are 25, 47, 16, 45, 54, 17, 14, 16, 16, 39, 48, 48, 18, 12, 13, 62, 51, 46, and 18. Summarize the distribution of the data.

Patients in U.S. Cancer Centers

20

30

10

0Num

ber o

r Can

cer C

ente

rs 40

Number of Patients

3500500 1000 1500 2000 2500 3000


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11-1 Enrichment

Means

The average of a data set is known as the arithmetic mean. There are other means.

A trimmed mean is the arithmetic mean of a data set after the top 10% and bottom 10% of the data values are trimmed off. Therefore, if there are 100 data values, the least 10 values and greatest 10 values are removed.

A geometric mean is the nth root of the product of n data values. It is often used in economics to find an average rate of growth.

A harmonic mean H is the number of data values divided by the sum of the multiplicative inverses of all the data values. It is often used in averaging speeds. It cannot be used if 0 is a data value.

H = n −

∑

1 − x , where n is the number of data values and x is a data value.

Exercises 1. Find the trimmed mean of the data set.

32, 24, 56, 102, 54, 12, 27, 49, 35, 23, 44, 51, 66, 36, 52, 16, 63, 75, 21, 41

2. Find the arithmetic mean and median of the data set in Exercise 1. What is the benefit of using a trimmed mean instead of an arithmetic mean?

3. WATER The EPA recommends that quality freshwater be monitored for e.coli concentration. The geometric mean of the e.coli concentrations in 5 or more samples taken over 30 days should be less than 126 per 100 milliliters. Do the data {225, 181, 110, 118, 107} meet this criterion? Explain.

4. TRAVEL If you travel at one speed for half the distance of a trip and you travel at a different speed for the other half of the distance, then your average speed for the trip is the harmonic mean of the speeds. If Greg drove at 60 mph for half a trip and 70 mph when the speed limit increased for the second half, what was his average speed for the trip?


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11-1 TI-NspireTM Activity

Graphing DataYou can use a TI-Nspire to graph data sets and describe the shape of the distribution.

The lengths of fish caught on a one-dayfishing trip are shown in the table. Make a histogramand box plot of the data.

Step 1: Add a Lists & Spreadsheet page. List the datain column A and title the column lengths.

Step 2: With the cursor in column A, press b andchoose Data > Quick Graph. Press b andchoose Plot Type > Histogram. Press bagain and choose Plot Properties >Histogram Properties > Bin Settings to changethe width of the bars. Press b and chooseWindow/Zoom > Zoom-Data to automatically sizethe axes to see all the data.

You can change the vertical category to percentby pressing b and choosing Plot Properties >Histogram Properties > Histogram Scale >Percent.

Step 3: To change the graph to a box plot, press b andchoose Plot Type > Box Plot. Move the cursor overthe graph to view the five-number summary.

Exercises

Open a new Lists & Spreadsheet page. Title column A as data. Use it to complete the following.

1. Make a histogram of the data set: 44.5, 13.7, 29.4, 22.0, 32.5, 45.8, 38.6, 24.3,18.1, and 50.3. Use the graph to describe the shape of the distribution.

2. Change the histogram in Exercise 1 to a box plot. List the five-number summary.

3. Open a new Lists & Spreadsheet page. Title column A as data1 and column Bas data2. Enter the data shown below. Use Quick Graph in each column to makebox plots. Describe and compare the distributions.

Data1: 102, 116, 132, 111, 124, 103, 101, 108, 129, 103, 109, 115, 111 Data2: 115, 129, 101, 128, 125, 115, 101, 124, 124, 118, 121, 119, 120

Lengths of Fish (in.)

14 16 18 18 14

21 26 23 35 15

32 30 9 22 12

Example


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Probability Distribution A discrete random variable X can take on acountable number of values. A probability distribution links each possiblevalue of X with its probability of occurring.

Mean of probability distribution of X: μ = Σ [X � P(X)]

Standard deviation of probability distribution of X: σ = √

��

Σ [(X - μ)2 � P(X)

]

The number of books bought by each of 100 randomly selected bookstore customers during one week is shown.

a. Construct a probability distribution for X.

There are 100 customers, so P(0) = 0.45, P(1) = 0.30, P(2) = or 0.15, and P(3) = 0.10.

b. Find and interpret the mean in the context of the problem situation.Find the variance and standard deviation.

Organize your calculations in a table.

The mean of the distribution is 0.9. On average, each customer bought one book.The variance = 0.99, so the standard deviation is √

��

0.99 or about 0.995.

Exercise

1. The table shows the number of medical tests that15 randomly selected patients entering a particularhospital received one day.

a. Construct a probability distribution for X.

b. Find and interpret the mean in the context of the problem situation. Find thevariance and standard deviation.

11-2 Study Guide and InterventionProbability Distributions

Example

Books, X Frequency

0 45

1 30

2 15

3 10Books, X 0 1 2 3

Frequency 0.45 0.30 0.15 0.10

Books, X P(X) X � P(X) (X - μ)2 (X - μ)2 � P(X)

0 0.45 0 � 0.45 = 0 (0 - 0.9)2 = 0.81 0.81 � 0.45 = 0.3645

1 0.30 1 � 0.30 = 0.30 (1 - 0.9)2 = 0.01 0.01 � 0.30 = 0.003

2 0.15 2 � 0.15 = 0.30 (2 - 0.9)2 = 1.21 1.21 � 0.15 = 0.1815

3 0.10 3 � 0.10 = 0.30 (3 - 0.9)2 = 4.41 4.41 � 0.10 = 0.441

μ = 0.9 σ2 = 0.99

Tests, X Frequency

0 6

1 5

2 3

3 1


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Binomial Distribution In a binomial experiment, the outcomes are success or failure. There are a fixed number of independent trials n and the random variable X represents the number of successes. The probability of success p and the probability of failure q or 1 - p remain constant. The probability of X successes in n independent trials is

P(X) = nCx px qn - x = n! −

(n - x)!x! px qn - x .

A survey found that 20% of Americans have visited a doctor in the past six months. Five people will be selected at random and asked if they visited a doctor in the past six months. Construct and graph a binomial distribution for the random variable X, representing the number of people who say yes. Then find the probability that at least four of these people say yes.

For this binomial experiment, n = 5, p = 0.2, and q = 1 - 0.2 or 0.8. Compute each possible value of X using the Binomial Probability Formula.P(0) = 5C0 � 0.20 � 0.85 ≈ 0.328 P(3) = 5C3 � 0.23 � 0.82 ≈ 0.051P(1) = 5C1 � 0.21 � 0.84 ≈ 0.410 P(4) = 5C4 � 0.24 � 0.81 ≈ 0.006P(2) = 5C2 � 0.22 � 0.83 ≈ 0.205 P(5) = 5C5 � 0.25 � 0.80 ≈ 0.000

To find the probability that at least four people said yes, find the sum of P(4) and P(5).P(X ≥ 4) = P(4) + P(5) = 0.006 + 0.000 or 0.6%

Exercise

A survey found that 60% of American victims of health-care fraud were senior citizens. Six victims of health-care fraud will be chosen at random and their ages will be recorded. Construct and graph a binomial distribution for the random variable X, representing the number of senior citizens chosen. Then find the probability that at least three of the victims will be senior citizens.

Study Guide and Intervention (continued)

Probability Distributions

11-2

Example

X P(X)

0 0.328

1 0.410

2 0.205

3 0.051

4 0.006

5 0.000

00 1 2 3 4 5

0.1

0.2

0.3

0.4

0.5

People

Prob

abili

ty

P(x)

x


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Classify each random variable X as discrete or continuous. Explain your reasoning.

1. X represents the time it takes a randomly selected classroom to reach 68°Ffrom 60°F. .

2. X represents the number of photographs taken by a photographer at a randomlyselected wedding.

3. The table shows the number of cell phones owned by 100 randomly selected households. Construct and graph a probability distribution for X. Then find and interpret the mean in the context of the problem situation. Find the variance and standard deviation.

4. RACE A resort is planning a bicycle race. The cost of sponsoring the race is$8000. The resort expects to make $15,000 on the event. There is a 30% chanceof a hurricane arriving the day of the race. If this happens, the race will becancelled and will not be rescheduled. What is the resort’s expected profit?

5. COMMUTE In a recent poll, 45% of a town’s citizens said they use the bus toget to work. Five of these citizens will be randomly chosen and asked if theyuse the bus to get to work.

a. Construct a binomial distribution for the random variable X, representing thepeople who say yes.

b. Find the mean, variance, and standard deviation of this distribution. Interpretthe mean in the context of the problem situation.

11-2 PracticeProbability Distributions

Phones, X Frequency

0 2

1 30

2 48

3 13

4 7

005_042_PCCRMC11_893812.indd 13005_042_PCCRMC11_893812.indd 13 11/17/09 8:05:29 AM11/17/09 8:05:29 AM

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1. RETAIL A store manager made the probability distribution shown below. It shows the probability of selling Xswimsuits on a randomly selected dayin June.

Find the mean, variance, and standard deviation of the distribution.

2. INSURANCE An insurance company insures a painting worth $20,000 against theft for $300 per year. The company has assessed the probability of the painting being stolen in a given year as 0.002. What is the insurance company’s expected annual profit?

3. RESTAURANT A survey found that 25% of all parties at a restaurant were groups of five or larger. Eighteen parties are randomly selected.

a. Find the probability that exactlyfive parties are made up of five or more people.

b. Find the probability that 5, 6, or7 parties are made up of five ormore people.

4. PETS According to one poll, about 63% of American households include at least one pet. Six new homes are built and sold.

a. Construct a binomial distribution for the random variable X, representing the number of these homes that will have at least one pet.

b. Find the mean, variance, and standard deviation of this distribution.

c. Find the probability that at least half of the new homes have pets.

5. TESTING Mr. Hanlon distributed a 5-question multiple choice quiz to his students. There were 5 choices for each question. Ashley guesses the answer on each question.

a. What is Ashley’s probability of guessing exactly 3 questions correctly?

b. What would be the probability inpart a if there were 4 choices foreach question?

c. What would be the probability inpart a if the quiz contained onlytrue/false questions?

11-2 Word Problem PracticeProbability Distributions

Swimsuits, X 19 20 21 22 23

P(X) 0.20 0.20 0.30 0.20 0.10


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The Poisson distribution is a probability distribution for an event occurringx times over a given interval. The interval is usually a time interval, such as the number of people entering a store during one hour. The distribution is a discrete distribution, so x must be a whole number.

The probability of an event occurring x times over a given interval is λx · e-λ

−

x! ,

where λ is the average number of times the event occurs during the interval.

A bank manager determined that an average of 9.2 customers use a certain ATM every hour. You can create a Poisson distribution by using 9.2 for λ. For example, the probability that 5 customers use the ATM in a given hour

is given by 9.25 · e-9.2 −

5! . Use your calculator to verify P(5) ≈ 0.055 or

about 5.5%.

On a graphing calculator, you can also find the probability by using the poissonpdf( command found by pressing [DISTR]. In the parentheses, enter λ, x.

1. The bank manager determined that an average of 1.8 customers inquire about opening a new account at a certain branch every hour. Complete the table.

2. Notice that unlike a binomial distribution, x has no upper limit. The table in Exercise 2 stopped at x = 5, but it is possible for 6 or more people to inquire about opening a new account. How can you find P(x > 2)?

3. On average, the bank receives 4.2 online loan applications per day. Find each probability.

a. The bank receives 3 online loan applications one day.

b. The bank receives 4 online loan applications one day.

c. The bank receives 5 online loan applications one day.

d. The bank receives 6 online loan applications one day.

e. The bank receives 7 online loan applications one day.

4. Make a conjecture about the general shape of any Poisson distribution.

11-2 EnrichmentThe Poisson Distribution

X 0 1 2 3 4 5

P(X)


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The Normal Distribution A normal distribution is a continuous probability distribution. The graph of a normal distribution is symmetric and bell-shaped. It approaches but never touches the x-axis. It includes 100% of the data, so the area under the curve is 1. The z-value represents the number of standard deviations that a given data value is from the mean.

z = X - μ

−

σ , where X is the data value, μ is the mean, and σ is the standard deviation.

The standard normal distribution has a mean of 0 and a standard deviation of 1.

On his last 20 airline trips, an employee had an average layover of 82 minutes with a standard deviation of7.5 minutes. Find the number of layovers that were less than75 minutes.

First, find the z-value.

z = X - μ

−

σ Formula for z -values

= 75 - 82 −

7.5 or about -0.93 X = 75, μ = 82, and σ = 7.5

Use a graphing calculator to find the area under the curve that is to the left of 75. Press 2nd [DISTR] and choose normalcdf(. Enter the lower value (you can use -4 instead of negative infinity) and the upper value as -0.93. The resulting area is 0.176. This means that about 17.6% of the data values are less than-0.93 standard deviations from the mean.

Because there are 20 flights, about 20 � 0.176 or about 4 flights had layover times that were less than 75 minutes.

Exercises

1. At a restaurant, the average time between when an order is placed and when the entree is served is 12.5 minutes with a standard deviation of1.2 minutes. Out of 100 randomly selected customers, how many will be served their entrees within 14 minutes of ordering?

2. Mrs. Quan, a full professor at a community college, earns a salary of $48,600. The average salary for a full professor at the college is $52,000 with a standard deviation of $3600. How many of the 45 full professors earn less than Mrs. Quan?

3. During one October, the average water temperature of a pond was 53.2° with a standard deviation of 2.3°. How many days was the temperature greater than 50°?

Study Guide and InterventionThe Normal Distribution

11-3

Example


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Probability and the Normal Distribution The area under the normal curve corresponds to the probability of data values falling within a given interval.

The average monthly temperatures for a city for one year were normally distributed with μ = 65° and σ = 5. FindP(50 < X < 70). Use a graphing calculator to sketch the corresponding area under the curve.

Find the z-values for X = 50 and X = 70.

z = X - μ

−

σ z =

X - μ −

σ

= 50 - 65 −

5 = -3 = 70 - 65 −

5 = 1

Find the area between z = -3 and z = 1. On your calculator, press 2nd [DISTR] and choose ShadeNorm under the DRAW menu. Enter the lower and upper z-values and press enter.

P(50 < X < 70) ≈ 84%

Therefore, approximately 84% of the temperatures were between 50° and 70°.

Exercises 1. The average age of the swimmers on a master swim team is normally

distributed with μ = 56 and σ = 4. Find each probability. Use a graphing calculator to sketch the corresponding area under the curve.

a. P(53 < X < 59)

b. P(X < 53)

2. Students who score in the bottom 5% of a physical education test will be enrolled in a supplemental physical education program. The scores of all of the students who took the test are normally distributed with μ = 122.6 and σ = 18. What is the greatest score that a student who enrolled in the supplemental program could have received?


The Normal Distribution

Example

[–4, 4] scl: 1 by [0, 0.5] scl: 0.125


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1. TREES The heights of 200 trees in a nursery are normally distributed with a mean of 120 inches and a standard deviation of 16 inches.

a. Approximately how many trees are more than 136 inches tall?

b. What percent of the trees are between 88 inches and 104 inches tall?

Find each of the following.

2. z if X = 65, μ = 50, and σ = 10 3. X if z = -0.4, μ = 40, and σ = 5

Find the interval of z-values associated with each area.

4. the middle 60% of the data

5. the outside 30% of the data

6. DOGS The weights of the 42 full-grown German shepherds at a kennel are normally distributed. The mean weight is 86 pounds and the standard deviation is 3 pounds.

a. Determine the number of German shepherds that weigh more than 82 pounds.

b. How many German shepherds weigh less than 88 pounds?

7. HOTELS The prices of rooms at hotels around an airport are normally distributed with μ = $120 and σ = $20. Find each probability.

a. The cost of a room is greater than $150.

b. The cost of a room is between $110 and $130.

c. The cost of a room is between $90 and $100.

d. If only the most expensive 10% of the rooms are available, what is the least amount you will pay for a room?

8. EXAMS A student scored 65 on a biology exam with μ = 50 and σ = 10. She scored 30 on a literature exam with μ = 25 and σ = 5. Compare her scores on each test. Assume that both sets of scores were normally distributed.

11-3 PracticeThe Normal Distribution


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1. REAL ESTATE The average price of a one-bedroom condominium listed by a realtor is $145,500 with a standard deviation of $1500. The prices are normally distributed. Determine the probability that a randomly selected condominium costs between $143,580 and $147,420.

2. COMMUTING The average times spent commuting to work in a certain city are normally distributed with a mean of25.5 minutes and a standard deviationof 6.1 minutes. What is the probability that a randomly selected commute to work takes longer than a half hour?

3. SIGHTSEEING The times people spend viewing certain ancient ruins are normally distributed with a mean of 96 minutes with a standard deviation of 17 minutes.

a. Find the probability that a sightseer will spend at least two hours at the ruins.

b. Find the probability that a sightseer will spend at most 80 minutes at the ruins.

c. If a tour bus drops off a group of sightseers at 9 A.M., what time should the bus pick up the sightseers? Explain.

4. TRAINING To qualify for a security position, candidates must take a physical fitness test. The scores on the test are normally distributed with a mean of 400 and a standard deviation of 100.

a. Candidates scoring in the top 3% are later recruited as trainers in the program. What is the minimum score a candidate needs in order to be recruited later as a trainer?

b. Candidates scoring in the bottom 1.5% must retake the physical training program. What is the minimum score a candidate would need to avoid retaking the training program?

5. JUICE The amount of juice pouredinto bottles in a factory is normally distributed with a mean of 16 ouncesand a standard deviation of 0.3 ounce. A shipment contains 280 bottles.

a. How many bottles are expected to contain more than 16.5 ounces of juice?

b. How many bottles are expected to contain less than 15.75 ounces of juice?

11-3 Word Problem PracticeThe Normal Distribution


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Before graphing calculators, students had to find the area under a normal distribution curve by using a table of values. You can still find these tables in many statistics books.It is good to know how to use them, in case you find yourself in a situation without a graphing calculator.

Here is part of a standard normal probability table. It gives the area that is between the mean, 0, and the z-score. To find the area between a z-score of 0 and a z-score of 1.35, find 1.3 in the vertical column and 0.05 in the top row. It is 0.4115.

To find the area that is to the left of a certain positive z-score, remember to add the areathat is to the left of the mean, 0.5.

Because a normal distribution is symmetric about the mean, you can use the samechart to find the area between a negative z-score and the mean. For example, thearea between a z-score of -1.35 and 0 is the same as between 0 and 1.35: 0.4115.

Use the table to answer each question.

1. What is the area between a z-score of 0 and a z-score of 1.02?

2. What is the area between a z-score of 1.18 and a z-score of 1.43? Explain howyou found it.

3. Find the area to the right of a z-score of 1.17. Explain how you found it.

4. What is the area to the left of a z-score of 1.41?

5. What is the area between a z-score of -1.25 and a z-score of 0?

6. What is the area to the left of a z-score of -1.33?

7. What is the area between a z-score of -1.05 and a z-score of 1.38?

11-3 EnrichmentFinding Areas by Using a Table

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621

1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830

1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015

1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177

1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319


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The graph shown at the right is known as the standard normal

curve. The standard normal curve is the graph of f(x) = 1 −

√

��

2π e -

x2 −

2 .

You can use a graphing calculator to investigate properties of this function and its graph. Enter the function for the normal curve in the Y = list of a graphing calculator.

1. The standard normal curve models a probability distribution. As a result, probabilities for intervals of x-values are equal to areas of regions bounded by the curve, the x-axis, and the vertical lines through the endpoints of the intervals. The calculator can approximate the areas of such regions. To find the area of the region bounded by the curve, the x-axis, and the vertical lines x = -1 and x = 1, go to the CALC menu and select 7: ∫f(x) dx. Move the cursor to

the point where x = -1. Press ENTER . Then move the cursor to the

point where x = 1 and press ENTER . The calculator will shade the region and display its approximate area. What number does the calculator display for the area of the shaded region?

2. Enter 2nd [DRAW] 1. This causes the calculator to clear the shading and redisplay the graph. Find the area of the region bounded by the curve, the x-axis, and the vertical lines x = -2 and x = 2.

3. Find the area of the region bounded by the curve, the x-axis, and the vertical lines x = -3 and x = 3.

4. Without using a calculator, estimate the area of the region bounded by the curve, the x-axis, and the vertical lines x = -4 and x = 4 to four decimal places. Verify your conjecture.

11-3 Graphing Calculator ActivityThe Standard Normal Curve

[-4.7, 4.7] scl: 1 by [-0.2, 0.5] scl: 0.1


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11-4 Study Guide and InterventionThe Central Limit Theorem

The Central Limit Theorem The Central Limit Theorem states that as the sampling size n increases:

• the shape of the distribution of the sample means of a population with mean μand standard deviation σ will approach a normal distribution and

• the corresponding distribution will have a mean μ and standard deviation σ

−

x = σ −

√

�

n .

The z-value for a sample mean in a population is given by z = − x - μ

− σ − x , where − x is the

sample mean, μ is the mean of the population, and σ

−

x is the standard error.

A study was done in which parents reported the number of hours per day that their children, between the ages of 2.0 and 5.9, watched television or videos. The mean age of the children was 3.3 years with a standard deviation of 0.9 year. Assume that the variable is normally distributed. If a random sample of 30 children in this study is selected, find the probability that the mean age is less than 3 years.

The distribution of sample means will be approximately normal with μ = 3.3 and

σ

−

x = 0.9 −

√

��

30 or about 0.164. Find the z-value.

z = − x - μ

− σ − x z-value for a sample mean

z = 3 - 3.3 −

0.164 or about -1.83 − x = 3, μ = 3.3, σ − x = 0.164

The area to the left of a z-value of -1.83 is 0.03359.

The probability that the mean age of the sampleis less than 3 years or P( − x < 3) is about 3.36%.

Exercises 1. The mean pitch of the expert slopes at the ski resorts in a certain region

is 25˚ with a standard deviation of 3˚. Assume that the variable is normally distributed. If 20 expert slopes are chosen at random, what is the probability that the mean of the slopes will be greater than 26.3˚?

2. A veterinarian reports that the average age of the cats that she treats is 96 months with a standard deviation of 16 months. If a random sample of 36 of her cat patients is selected, find the probability that the mean age is between 90 and 100 months.

Example


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The Central Limit Theorem

The Normal ApproximationThe normal distribution can be used to approximate a binomial distribution if:

• the original variable is normally distributed or n ≥ 30 and

• np ≥ 5 and nq ≥ 5,where n is the number of trials, p is the probability of success, and q is theprobability of failure.When using the normal distribution to approximate a binomial distribution, thecontinuity correction factor must be used. This involves adding 0.5 unit to or subtracting 0.5 unit from a given discrete boundary.

Ten percent of the members of a golf league are youngerthan 30. If 200 members are randomly selected, find the probability thatmore than 10 will be younger than 30.

Step 1 Find the mean μ and standard deviation σ of the binomial distribution.

μ = np σ = √

��

npq

= 200 · 0.10 or 20 = √

��

200 · 0.10 · 0.90 or about 4.24

Since np = 20 and nq = 180, the normal distribution can be used.

Step 2 Write the problem in probability notation: P (X > 10) .

Step 3 Rewrite the problem with the continuity factor included. Since the question is asking for the probability that more than 10 members will be younger than 30, add 0.5 to 10.

P (X > 10) = P (X > 10 + 0.5) or P (X > 10.5)

Step 4 Find the corresponding z-value for X = 10.5.

z = X - μ

−

σ

= 10.5 - 20 −

4.24 ≈ -2.24

Step 5 Find the corresponding area above z = -2.24. It isabout 0.987. The probability is about 98.7%.

Exercises 1. A study found that 40% of all of a town’s citizens approve of a new train station.

If 50 citizens are randomly selected, find the probability that fewer than 20 citizens approve of a new train station.

2. A baseball player gets a hit 32% of the time. Find the probability that theplayer will get at least 26 hits in his next 100 times at bat.

Example


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11-4 PracticeThe Central Limit Theorem

1. NURSING The mean salary for nurses in a city is $52,129 with a standard deviationof $1800. What is the probability that the mean salary for a randomly selectedgroup of 50 nurses in the city is greater than $52,500?

2. UTILITIES The average electric bill in a residential area in June is $72. Assumethis variable is normally distributed with a standard deviation of $6. Find theprobability that the mean electric bill for a randomly selected group of 15 residentsis less than $75.

3. SHOES The prices of shoes in a store are normally distributed with a mean of$93 and a standard deviation of $18. If nine pairs of shoes are randomly selected,find the probability that the mean cost is between $100 and $110.

4. COLLEGE Of the total population at a small college, 20% are from the Mid-Atlantic states. If 200 students are randomly selected, find the probability that at least50 are from the Mid-Atlantic states.

5. WEATHER Kyle has researched the average annual precipitation in his city forseveral years and calculated the mean as 19.32 inches. Assume the averageprecipitation is normally distributed and the standard deviation is 2.44 inches.

a. If one year from the time period that Kyle researched is randomly selected,what is the probability that the precipitation is more than 18 inches?

b. If five years from the time period that Kyle researched are randomly selected,what is the probability that the mean precipitation is more than 18 inches?

6. GROCERY Eighty-five percent of the shoppers at a grocery store have afrequent-buyer card. Thirty-five shoppers are randomly selected for a tastetest. What is the probability that at least 25 and at most 30 of the tastetesters have a frequent-buyer card?

Find the minimum sample size needed for each probability so that the normal distribution can be used to approximate the binomial distribution.

7. p = 0.6 8. p = 0.15



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Chapter 11 25 Glencoe PrecalculusChapter 11 25 Glencoe Precalculus

Word Problem PracticeThe Central Limit Theorem

11-4

1. HEATING Workers at a public utilities company surveyed 200 of their customers and found that the mean temperature at which the customers’ thermostats were set in January was 70˚ with a standard deviation of 1.8˚. If 30 customers are randomly selected for a follow-upsurvey, find the probability that the mean temperature at which they set their thermostats in January is lessthan 69.5˚.

2. MILK A food study in a city found that the price of a gallon of whole milk in its stores was $3.72. This variable is normally distributed with a standard deviation of $0.08.

a. If a gallon of whole milk is randomly selected from one store, find the probability that it costs less than $3.70.

b. If a gallon of milk is randomly selected from each of 40 different stores, find the probability that the mean cost of the sample is less than $3.70.

3. BICYCLES Thirty-seven percent of the residents in a neighborhood own bicycles. If a group of 40 residents are randomly selected, what is the probability that at least half own bicycles?

4. TESTING The average time it takes a group of students to complete a reading test is 46.2 minutes with a standard deviation of 8 minutes. The times are normally distributed.

a. A group of 10 students is randomly selected. Find the probability that the mean time to complete the test is more than 45 minutes.

b. How does your answer to part achange if the group consists of15 students?

5. CAFETERIA College freshmen were asked if they ate breakfast on Sunday morning in the cafeteria. The graph shows the percent of males and females who said yes.

FemaleMale

40

60

20

0

Perc

ent

80

a. Out of 30 randomly selected female freshmen, what is the approximate probability that at least 25 of them ate breakfast in the cafeteria on Sunday morning?

b. Out of 30 randomly selected male freshmen, what is the approximate probability that at least 25 of them ate breakfast in the cafeteria on Sunday morning?


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11-4

Experience the Central Limit TheoremThe Central Limit Theorem is best appreciated when experienced. This activitywill allow you this privilege. You may wish to work with a partner. Simulate spinning a spinner with equal sections labeled 1–10two times by selecting the randInt( command from thePRB menu after pressing MATH . The screen shown indicatesthe spinner landed on 7 on the first spin and 1 on the second spin.The mean of these spins is 4.

1. Complete 25 simulations and list the means of the spins.

2. Find the mean and standard deviation of the means in Exercise 1.

Now simulate spinning the same spinner four times.



Now simulate spinning the same spinner six times.



7. What do you notice about the means in Exercises 2, 4, and 6?

8. What do you notice about the standard deviations in Exercises 2, 4, and 6?

9. Find σ −

√

�

n for n = 2, 4, and 6 given σ = 2.87. What do you notice?

10. If n = 1, that is the spinner is spun once, what is the distribution of the samples?How does it change as n increases?

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Normal Distribution The confidence level c is the probability that an interval estimate contains a given parameter, such as a population mean. The maximum error of estimate E is the maximum difference between the point estimate and the actual value of the parameter. A confidence interval CI is a specific interval estimate of a parameter. For a population mean:

CI = − x ± E or − x ± z · σ −

√

�

n .

z is the critical value that corresponds to a certain confidence level. The most common are shown below.

Confidence Level 90% 95% 99%

z-Value 1.645 1.960 2.576

When σ is unknown, the sample standard deviation s can be substituted for σ, provided the sample size n is greater than or equal to 30.

In a sample of 50 people who buy magazines, a researcher finds the mean amount spent per month to be $12. Assume a standard deviation of $4.50. Find the 95% confidence interval for the mean amount spent for magazines each month.CI = − x ± z · σ

−

√

�

n Confidence Interval for the Mean

= 12 ± 1.96 · 4.50 −

√

��

50 − x = 12, z = 1.96, σ = 4.50, and n = 50

≈ 12 ± 1.25 Simplify.

Add and subtract the margin of error.

Left Boundary: 12 - 1.25 = 10.75 Right Boundary: 12 + 1.25 = 13.25

The 95% confidence interval is 10.75 < μ < 13.25. Therefore, with 95% confidence, the mean amount spent on magazines per month is between $10.75 and $13.25.

Exercises

The number of days with temperatures above freezing for a sample of 35 cities had a mean of 190.7 days and a sample standard deviation of 54.2 days.

1. Find the 95% confidence interval for the mean number of days with temperatures above freezing.

2. Find the 90% confidence interval for the mean number of days with temperatures above freezing.

Study Guide and InterventionConfidence Intervals

Example

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t-Distribution To find a confidence interval when σ is unknown and the sample size n is less than 30, use the t-distribution, provided the variable is approximately normally distributed. The formula for using the t-distribution to construct a confidence interval is

CI = − x ± t · s −

√

�

n ,

where − x is the sample mean, t is a critical value with n - 1 degrees of freedom, s is the sample standard deviation, and n is the sample size.

The mean price of 9 randomly selected televisions at an electronics store is $783 with a standard deviation of $116. Find the 90% confidence interval of the mean price of all of the televisions at the store. Assume that the variable is normally distributed.

Because the population standard deviation is unknown and n < 30, use the t-distribution. Because n = 9, there are 9 - 1 or 8 degrees of freedom.

To find the t-value, determine the area in the lower tail of the distribution. Since 100% - 90% or 10% is in the tails, then 5% is in each tail. Press 2nd DISTR and choose invT(. Type thearea to the left of each critical value, as a decimal, followed by the degrees of freedom.

CI = − x ± t · s −

√

�

n Confidence interval using t-distribution

≈ 783 ± 1.860 · 116 −

√

�

9 − x = 783, t ≈ 1.860, s = 116, and n = 9

≈ 783 ± 71.92 Simplify.

Therefore, the 90% confidence interval is $711.08 < μ < $854.92.

Exercises

1. At a manufacturing plant, the threads on ten randomly selected screws have a mean depth of 0.32 inch and a standard deviation of 0.03 inch. Find the 95% confidence interval of the mean depth of all the screws, assuming that the variable is normally distributed.

2. An environmental study involves counting the number of light bulbs in randomly selected rooms of a building. A building manager counts the bulbs in 16 rooms and finds the mean number of bulbs to be 21 with a standard deviation of 4.8. Find the 99% confidence interval of the mean number of bulbs in all the rooms of the building, assuming the variable is normally distributed.


Confidence Intervals

Example

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1. CHILD CARE A random sample of 50 parents of young children found that they spend a mean of $7648 each year for child care. The standard deviation of the sample was $630. Find the 90% confidence interval of the mean annual cost of child care.

2. FITNESS Twenty-eight people who enrolled in a fitness program lost a mean of 14.3 pounds with a standard deviation of 2 pounds. Find the 95% confidence interval of the mean weight loss in pounds of all of the members enrolled in the program.

3. COMMUTE The average number of minutes it takes 8 people to commute to and from work during rush hour is shown. Assume that the times are normally distributed.

Commuting Time (minutes)

55 70 60 55 60 56 55 60

a. Decide the type of distribution that can be used to estimate the commuting time mean. Explain your reasoning.

b. Calculate the mean and standard deviation to the nearest hundredth.

c. Construct a 90% confidence interval for the average commuting time in minutes for a commuter from this city.

4. TRAINING In a survey of 442 employees at a call center, the mean time that employees felt was needed for adequate training for their jobs was 7 days. The sample standard deviation was 1.5 days. Construct a 98% confidence interval for the amount of training time that employees felt was adequate to begin their jobs.

Determine whether the normal distribution or t-distribution should be used for each question. Then find each confidence interval given the following information.

5. 95%; − x = 115, s = 6, n = 6

6. 96%; − x = 18.5, s = 1.2, n = 40

7. 99%; − x = 236, σ = 8, n = 45

8. FOOD The owners of a sandwich shop want to find the 95% confidence interval of the true mean cost of a hamburger in their city. How large should their sample be if they want to be accurate to within 15 cents? In an earlier survey, the standard deviation of the price was 26 cents.

9. SPORTS A teacher wants to estimate the average number of hours per week that her students are at practice or at games for a sports team. The standard deviation from a previous year was 6.2 hours. How large of a sample must she select if she wants to be 99% confident of finding the average amount of time students spent participating in sports within 1.5 hours?

PracticeConfidence Intervals

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1. READING A sample of normally distributed reading scores of forty eighth-grade students has a mean of 82 and a standard deviation of 15. Find the 95% confidence interval for the mean of all of the reading scores.

2. CHOLESTEROL The serum cholesterol level was collected for a group of525 college women. The mean of the sample was 191.7

mg −

100 mL with a

standard deviation of 41.0.

a. Construct a 90% confidence level for the mean serum cholesterol level.

b. Construct a 95% confidence level for the mean serum cholesterol level.

c. Suppose you hear a claim that the mean serum cholesterol level for women in college is 200. What would be your reaction based on your answers to parts a and b? Why?

3. CAR POLLUTION Seven cars were tested for nitrogen-oxide emissions. The results are shown in the table.

Emissions (grams per mile)

0.05 0.12 0.16 0.15 0.14 0.19 0.14

a. Find the mean of the data.

b. Find the standard deviation of the data.

c. Find the 99% confidence interval of the mean nitrogen-oxide emissions.

4. FUEL CONSUMPTION The mean and standard deviation for city and highway fuel consumption in miles per gallon for 33 randomly selected pre-owned cars on a dealer’s lot is shown. Assume the variables are normally distributed.

− x s

City 21.35 4.13

Highway 29.65 3.65

a. Find the 98% confidence interval for the mean fuel consumption in the city.

b. Find the 98% confidence interval for the mean fuel consumption on the highway.

c. Compare the confidence intervals in parts a and b.

5. INTELLIGENCE QUOTIENT Suppose managers of a corporation want to estimate the IQ score for their employees. How many employees must be randomly selected for IQ tests if the managers want to be 95% confident that the mean is within 2 IQ points of the population mean? They know from previous studies that the standard deviation is 15 points.

Word Problem PracticeConfidence Intervals

11-5


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The Population Correction FactorThe formula for the maximum error of estimate, E = z · σ

−

√

�

n , assumes a very large

population or sampling with replacement. If the population size is N and the sample number is n, the formula should be modified with a correction factor if n > 0.05N.

The population correction factor is √

��

N - n −

N - 1 . Therefore, if n > 0.05N, the maximum

error of estimate is E = z · σ −

√

�

n ·

√

��

N - n −

N - 1 .

Determine if the population correction factor would be used when determininga confidence interval. Justify your answer.

1. 29 of the 140 employees in a company are surveyed

2. 75 of the town’s residents are surveyed and the town population is 2998

3. 10 of the 80 pages printed by a new printer are examined

4. There are 250 students in a school. A sample of 35 randomly selected students finds that their mean daily study time is 52 minutes with a standard deviationof 3.3 minutes. Find the 95% confidence interval for the mean daily study timeof all of the students. Round E to the nearest hundredth.

5. What would be the confidence interval in Exercise 4 if there were 750 students in the school?

6. Forty of the 160 potatoes in a bin are randomly selected for weight. The meanweight of the sample is 150 grams with a standard deviation of 6.5 grams. Findthe 90% confidence interval for the mean weight of all of the potatoes. Round Eto the nearest hundredth. Assume the variable is normally distributed.

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Hypotheses A hypothesis test allows you to evaluate a claim about a population.

The null hypothesis or H0 states that there is not a significant difference between the sample value and the population parameter and it always contains a statement of equality.The alternative hypothesis or Ha states that there is a significant difference betweenthe sample value and the population parameter and it always contains a statement of inequality.

For each statement, write the null and alternative hypotheses. State which hypothesis represents the claim.

a. A doctor claims that the pulse rate of a patient changes from 82 beats per minute after taking a new medication.

This claim becomes μ ≠ 82 and is the alternative hypothesis since it includes an inequality symbol. The complement is μ = 82.

H0: μ = 82 Ha: μ ≠ 82 (claim)

b. A track member claims that he can run at least 10 miles that day.

This claim becomes μ ≥ 10 and is the null hypothesis since it includes an equality symbol. The complement is μ < 10.H0: μ ≥ 10 (claim) Ha: μ < 10

c. A contractor claims that installing a particular kind of insulation willlower the average July cooling bill of $68 in her area.

This claim becomes μ < $68 and is the alternative hypothesis since it includesan inequality symbol. The complement is μ ≥ $68.H0: μ ≥ $68 Ha: μ < $68 (claim)

Exercises

Write the null and alternative hypotheses for each statement. State which hypothesis represents the claim.

1. On a scale of 1 to 10, patients describe their anxiety levels as 8 during dentalprocedures. A dental assistant thinks this level is lower when soft music isplayed during the procedures.

2. A real estate agent claims that the average home price in an area is $250,000.

3. A hiker claims that the average trail length in a park is at most 10 miles.

4. A restaurant owner claims that the average age of diners in a certain area is greater than 40.

Study Guide and InterventionHypothesis Testing

Example

11-6


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Significance and Tests To validate a claim, the null hypothesis is always tested at a chosen level of significance, α, which defines the area of the critical region. If the test statistic z- or t-value is in the critical region, then H0 should be rejected. Otherwise, H0 should not be rejected.

An employee at a sporting goods store claims that the average price of a pair of baseball cleats is less than $80. Another employee randomly selects a sample of 36 pairs and finds − x = 75 and s = 19.2. Determine if theresults are statistically significant at α = 0.10.Step 1 State the null and alternative hypotheses and identify the claim. H0: μ ≥ $80 Ha: μ < 80 (claim)

Step 2 Determine the critical value and region.Because n ≥ 30, use the z-value. The test is left-tailed since μ < 80. All of the critical region, with an area of 0.10, is to the left of the critical value. Use 2nd DISTR invNorm to see that the criticalvalue is -1.28. The critical region is the area to the left of z = -1.28.

Step 3 Calculate the test statistic. Use σ − x = 19.2 −

√

��

36 = 3.2.

z = − x - μ

− σ − x Formula for z-statistic

= 75 - 80 −

3.2 or -1.5625 − x = 75, μ = 80, and σ

−

x = 3.2

Step 4 Accept or reject the null hypothesis. H0 is rejected since the test statistic of -1.56 falls within the critical region. There is enough evidence to support the claim that the average cost of baseball cleats is less than $80.

Exercises

1. Managers of a large department store claim that the average salary for theirpart-time employees is $24,000. A sample of 10 part-time employees has a meansalary of $23,450 and a standard deviation of $1400. Determine whether thereis enough evidence to support the claim at α = 0.05.

2. In Exercise 1, suppose that the mean of $23,450 was taken from a sample of50 part-time employees. Would there be enough evidence to support the claimat α = 0.05?


Hypothesis Testing

Example

-1.28

Critical region


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Write the null and alternative hypotheses for each statement. State which hypothesis represents the claim.

1. A florist claims that a certain type of flower has a vase life of at least 7 days.

2. A brand of cereal claims that a serving contains less than 2 grams of sugar.

3. Robert claims that he swims 100 laps each week.

For each claim k, use the specified information to calculate the test statistic and determine whether there is enough evidence to reject the null hypothesis. Then make a statement regarding the original claim.

4. k: μ ≥ 60, α = 0.10, − x = 58.88, s = 5.08, n = 8

5. k: μ = 8, α = 0.05, − x = 8.2, s = 0.6, n = 32

6. MAIL A mail carrier claims that the average number of pieces of mail received daily by households in a certain neighborhood is 7. Sample data for 15 households is collected. The average number of mail pieces was 6.5 with a standard deviation of 0.8.

a. Is there enough evidence to reject the mail carrier’s claim at α = 0.05?

b. Is there enough evidence to reject the mail carrier’s claim at α = 0.01?

7. WEDDINGS A wedding planner wants to test the claim that the average wedding has 125 guests. In a random sample of 35 weddings, he found the mean to be 110 guests with a standard deviation of 30 guests. Find the P-value and determine whether there is enough evidence to support the claim at α = 0.01.

PracticeHypothesis Testing

11-6


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1. BOTTLED WATER The average volume in ounces of a random sample of 36 bottles of water at a packaging plant was found to be 12.19 ounces with a standard deviation of 0.11 ounce. The floor supervisor made the claim that the mean volume was greater than 12 ounces. Test her claim at α = 0.01.

a. Write the null and alternative hypotheses and state which hypothesis represents the claim.

b. Is there enough evidence to reject the null hypothesis? Why?

c. Make a statement regarding the original claim.

2. TEMPERATURE It is a long-established claim that the mean body temperature for humans is 98.6°F. In a random sample of 10 people, the mean body temperature was 98.3°F with a standard deviation of 0.23°F. Test the claim atα = 0.02.

a. What are the critical values?

b. What is the test statistic?

c. Is this evidence enough to supportthe claim that the mean body temperature for humans is 98.6°?

3. PULSE RATES An aerobics instructor had a pulse rate of 110 beats per minute after a warm-up routine with her class. Students recorded their pulse rates at the same time. Their rates are recorded in the table below. The instructor claims that their average pulse rate is lower than hers. Test her claim at α = 0.05.

80 70 90 75 110

105 120 110 85 115

95 95 105 90 70

105 95 100 105 90

a. Write the null and alternative hypotheses and state which hypothesis represents the claim.

b. What are the critical values and test statistic?

c. Does the data support the instructor’s claims?

4. MOUNTAIN CLIMBING A mountain climbing instructor claims that his students take longer than 5 minutes to pack their backpacks. In a random sample of 80 students, the average time it took to pack a backpack was 5.3 minutes with a standard deviationof 1.7 minutes. Find the P-value and determine whether there is enough evidence to support the claim atα = 0.05.

11-6 Word Problem PracticeHypothesis Testing


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Hypothesis TestingYou have tested hypotheses by finding a test statistic and seeing if it falls within the critical region. Alternatively, you can test a hypothesis by finding a confidence interval and seeing if the claim falls within that interval.

A condominium superintendent buys 5-pound bags of salt for melting ice.He weighs the salt in 50 randomly selected bags and finds the mean to be4.6 pounds with a standard deviation of 0.7 pound. Use a confidence levelto test the superintendent’s claim that the mean weight is not 5 pounds at α = 0.05.

Write the hypotheses: H0: μ = 5 Ha: μ ≠ 5

Use the normal distribution since n > 30. Since α = 0.05, the confidence level is 1 - α = 0.05, or 0.95. The z-value associated with a 95% confidence level is 1.96.

Find the 95% confidence interval for the mean weight of the salt in the bags.CI = − x ± z · σ

−

√

�

n Confidence interval for the mean

= 4.6 ± 0.194 − x = 4.6, and z · σ

−

√

�

n = 1.96 · 0.7

−

√

��

50 or about 0.194

So, the 95% confidence interval is 4.406 < μ < 4.794.

The confidence interval for the population does not contain the claim of 5, so we reject the null hypothesis. Evidence supports the claim that the mean is not 5 pounds.

Exercises

1. SOCCER A soccer coach claims that the mean weight of the players on the opposing teams is 200 pounds. A random sample of 10 players from opposing teams has a mean of 198.2 pounds and a standard deviation of 3.3 pounds. At α = 0.05, can the claim be supported? Support your answer with a confidence interval.

2. CALORIES A teacher walked by a vending machine and claimed that the average number of Calories in the snacks was 250. Students who overheard her wanted to prove her incorrect. They bought a random sample of 8 snacks and found the mean number of Calories of those snacks to be 210 with a standard deviation of 24. At α = 0.10, does the data support the teacher or the students? Support your answer with a confidence interval.

3. HOMEWORK A teacher claims her students spend an average of 45 minutes on her homework each night. To test the claim, she asks 35 students how much time, on average, they spend on her homework each night and the results show a mean of 40 minutes with a standard deviation of 9 minutes. At α = 0.05, does the data support her claim? Support your answer with a confidence interval.

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11-7

Correlation To determine if the correlation coefficient r is significant,

perform a hypothesis test with H0 : ρ = 0, and t as the test statistic, t = r √

��

n - 2 −

1 - r2 ,

where the degrees of freedom is n - 2.

The table shows the number of absences of 7 students from a psychology class and their final grades.

Absences 9 10 15 2 8 2 6

Grades 70 58 43 90 78 87 79

a. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.

Enter the data into L1 and L2. Turn on Plot1 and choose a scatter plot. The data appears to have a negative linear correlation.

Press STAT and select LinReg(ax + b) under the Calc menu. The correlation coefficient r is about -0.9609. Because r is close to -1, this suggests a strong negative correlation.

b. Test the significance of the correlation coefficient from part a at the 5% level.

Step 1 State the hypotheses. H0: ρ = 0 Ha: ρ ≠ 0

Step 2 Determine the critical values using n - 2 or 5 degrees of freedom. The critical values are ±2.6.

Step 3 Calculate the test statistic.

t = -0.9609 √

��

5 −

1 - (-0.9609)2 ≈ -7.7597

Step 4 Since t < -2.6, reject the null hypothesis. The evidence supports a significant correlation between the number of absences and a student’s grade.

Exercises

The table shows data about geysers at Yellowstone National Park.

Duration (min) 3.25 3 5 7.5 17.5 3.5 6 35 10 6.5 10 1 20 4

Height (ft) 184 25 150 20 75 75 10 100 160 60 5 25 75 5

Source: National Park Service

1. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.

2. Test the significance of the correlation coefficient from Exercise 1 at the 5% level.

Study Guide and InterventionCorrelation and Linear Regression

Example

[1, 17] scl: 1 by [35, 100] scl: 5


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Linear Regression If the correlation between two variables is significant, you can determine the least-squares regression line, which is a line of best fit. Then you can use the equation of the line to make predictions over the range of data.

The table shows the data from the previous page, which showed a significant correlation. Find the equation of the regression line for the data. Interpret the slope and intercept in context. Use the equation to predict the expected grade for a student with 4 absences and state whether this prediction is reasonable. Explain.

Absences 9 10 15 2 8 2 6

Grades 70 58 43 90 78 87 79

Press STAT and select LinReg(ax + b) under the Calc menu. The equation is approximately y = -3.48x + 97.993.

The slope indicates that for every additional absence a student has, his or her grade will drop by about 3.5 points. The y-intercept indicates that a student with no absences will have a grade of about 98.

Evaluate the regression equation for x = 4 and calculate y.y = -3.48x + 97.993 Regression equation

= -3.48(4) + 97.993 x = 4

= 84.073 Simplify.

We expect that a student with 4 days of absences would have a final grade of about 84. This is reasonable because 4 is in the range of the given x-values and 84 is in the range of the given y-values.

Exercises

A sports writer has already determined that there is a negative correlation between the batting average and the number of home runs for third basemen whose statistics are shown in the table.

Average 0.307 0.328 0.305 0.294 0.306 0.311 0.271 0.267 0.267 0.320

Home Runs 96 118 317 65 78 103 512 268 548 58

1. Find the equation of the regression line for the data.

2. Use the equation to predict the number of home runs for a third baseman with a batting average of 0.350. State whether this prediction is reasonable. Explain.


Correlation and Linear Regression

11-7

Example


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A supervisor of a cleaning business has the data in the table that shows the age of her workers and the number of sick days they take each year. She wonders if there is a significant linear relationship between the age of an employee and the number of sick days he or she takes each year.

Age 38 25 60 18 45 54 19 22 43 34

Days 9 12 2 16 4 5 15 17 3 1

1. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.

2. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain your reasoning.

3. If the correlation is significant at the 10% level, find the least-squares regression equation and interpret the slope and y-intercept in context.

4. Graph and analyze the residual plot.

5. Identify any influential outliers. Describe the effect the outlier has on the strength of the correlation and on the slope and intercept of the original regression line.

6. If any data were removed, reassess the significance of the correlation at the 10% level and, if still appropriate, recalculate the regression equation.

7. If appropriate, predict the number of sick days for employees who are 30, 50, and 70 years old. Interpret your results and state whether the predictions are reasonable. Explain your reasoning.

PracticeCorrelation and Linear Regression

11-7


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1. COLLEGE The data in the table represent the American College Test (ACT) composite scores and grade point averages (GPA) of 20 randomly selected students after their first semester in college. A college counselor wants to determine if there is a correlation between ACT scores and first semester GPAs.

ACT 27 18 17 15

GPA 3.9 2.9 3.3 3.0

ACT 22 20 17 21

GPA 3.6 2.7 2.9 3.4

ACT 25 17 25 18

GPA 3.5 3.1 4.0 3.0

ACT 23 19 20 29

GPA 3.6 2.6 3.0 3.4

ACT 23 28 22 20

GPA 1.8 4.0 3.0 4.0

a. Calculate and interpret the correlation coefficient.

b. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels.

c. Graph and analyze the residual plot.

2. SALES A sales associate wants to know if there is a relationship between the average number of times his coworkers contact clients each month and the average monthly sales volume in thousands of dollars. He collected the data shown in the table.

Clients 21 23 48 50 46

Sales 30 30 95 110 80

Clients 12 55 14 50 16

Sales 15 130 25 90 30

a. Make a scatter plot of the data and identify the relationship. Then calculate and interpret the correlation coefficient.

b. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain your reasoning.

c. If the correlation is significant at the 10% level, find the least-squares regression equation and interpret the slope and y-intercept in context.

11-7 Word Problem PracticeCorrelation and Linear Regression


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Multiple RegressionIn multiple regression, there are two or more independent variables and one dependent variable. The multiple regression equation is y = a + b1x1 + b2x2 + … + bkxk, where x1, x2, …, xk are the independent variables.

The strength of the relationship between the independent variables and the dependent variable is measured by the multiple correlation coefficient R. This value can range from 0 to 1, where values closer to 1 indicate a stronger relationship than those closer to 0. The formula for a multiple correlation coefficient with two independent variables is

R =

√

��

(

ry x 1 ) 2 +

(

r yx2 )

2 - 2 r yx1 (

r yx2 )

(

r x1x2 )

−−

√

��

1 - (

r x1x2 )

2 ,

where ry x 1 is the correlation coefficient for y and x1; ry x 2

is the correlation coefficient for y and x2 ; and r x1x2

is the correlation coefficient for x1 and x2.

Hospital administrators wish to see if a nursing applicant’s GPA and age are related to the nurse’s score on the state nursing board exam. The table shows these statistics for five nurses.

1. Find the values of ry x 1 , ry x 2

, and r x1x2 .

2. Find and interpret R.

3. A researcher is conducting a study to see if a person’s age x1 and cholesterol level x2 are related to his or her systolic blood pressure y. Find and interpret R if ry x 1

= 0.681,

ry x 2 = 0.872, and r x1x2

= 0.746.

A track coach wants to see if the 5k times for the first two meets of the season are related to the 5k times at the state meet. The table shows these statistics, in minutes, for five runners.

4. Find the values of r yx1 , r yx2

, and r x1x2 .

5. Find and interpret R.

Enrichment11-7

GPA

x1

Age

x2

State Board

Exam y

3.5 28 675

2.7 28 570

3.2 22 560

2.2 23 490

2.4 24 550

Meet 1

x1

Meet 2

x2

State Meet

y

17.4 16.5 16.5

15.4 16.1 15.8

18 18.2 16.9

16.5 16.3 16.5

17.2 17.5 16.1


Pdf Pass005_042_PCCRMC11_893812.indd 42005_042_PCCRMC11_893812.indd 42 3/25/09 12:39:40 PM3/25/09 12:39:40 PM

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11 Chapter 11 Quiz 1(Lessons 11-1 and 11-2)

11 Chapter 11 Quiz 2(Lessons 11-3 and 11-4)

1. MULTIPLE CHOICE Find X if z = 2.88, μ = 43, and σ = 5.2.

A -10.6 B -7.7 C 43.6 D 58.0

2. Find the interval of z-values associated with the middle 60%.

Wait-listed students at a certain college had a mean ACT reading score of 21.3 with a standard deviation of 6.

3. If a sample of 50 students is selected, find the probability that the mean score of the sample is less than 20.

4. If a sample of 20 students is selected, find the probability that the mean score is between 19 and 21.

5. For a normal approximation of a binomial distribution, how would you rewrite P(X < 16) to account for the continuity correction factor?

1.

2.

3.

4.

5.

X 1 2 3 4

P(X) 0.18 0.34 0.21 0.27

The exam scores of 19 students are shown.

78 92 76 66 88

52 76 86 84 74

74 72 78 82 82

62 96 76 68

1. Construct a histogram.

2. Describe the shape of the distribution.

3. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.

4. What is the mean of the probability distribution?

5. MULTIPLE CHOICE One of the competitors in a dart competition hits the center 90% of the time. Find the probability that he will hit the center on at least 4 of his next 5 throws.

A 0.328 B 0.590 C 0.919 D 0.950

1.

2.

3.

4.

5.

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11

11

A random sample of 60 college students showed a mean commute of 10.22 miles, with a standard deviation of 2.4 miles. 1. Find the maximum error of estimate given a

95% confidence level. 2. Write the 95% confidence interval for the mean

commuting time. 3. A doctor wants to estimate the mean weight of all

7-year old boys to within 1 pound with 95% confidence. How large should her sample size be if a previous study showed a standard deviation of 3 pounds?

Kurt claims that the mean car rental rate in his city is at least$60 per day. A sample of rates from 10 local companies gives a mean of $63.50 and a standard deviation of $2.75. 4. MULTIPLE CHOICE Determine the critical value to test

the claim at α = 0.05.A 1.36 C 2.26B 1.83 D 2.31

5. Is there sufficient evidence to reject Kurt’s claim? Explain.

Chapter 11 Quiz 3(Lessons 11-5 and 11-6)

Students who participated in a bully awareness project were given tolerance surveys before and after the project. The pre- and post-scores are shown.

Pre-score 18 14 11 23 19 21 21 11 22

Post-score 17 17 10 25 20 15 24 10 24

1. Calculate and interpret the correlation coefficient.

2. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain your reasoning.

3. If the correlation is significant at the 10% level, find the least-squares regression equation.

4. If appropriate, use the regression equation to predict a student’s post-score if their pre-score is 15.

5. MULTIPLE CHOICE Which data point is influential?

A (14, 17) C (19, 20)

B (11, 10) D (21, 15)

1.

2.

3.

4.

5.

Chapter 11 Quiz 4(Lesson 11-7)

1.

2.

3.

4.

5.


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11

Write the letter for the correct answer in the blank at the right of each question.

1. Which random variable X is a discrete variable?

A the number of degrees a liquid rises after being heated

B the number of female babies in the birth of twins

C the thickness of a bolt

D the height of a tomato plant 20 days after germination

2. BREAD The mean weight of a loaf of Italian bread at a bakery is 482 grams with a standard deviation of 18 grams. In a random sample of 40 loaves, what is the probability that the mean of the sample will be less than 478 grams?

F 0.08 G 0.16 H 0.41 J 0.92

3. MEDICINE Seventy-five percent of those surveyed reported that a certain medicine was effective in reducing arthritis pain. If 7 respondents are randomly selected, what is the probability that 3 thought the medicine was effective?

A 0.058 B 0.071 C 0.093 D 0.173

Part II

DEXTERITY The data show the time, in minutes, it takes people with carpal tunnel syndrome to complete a manual task. Half of the group are taking medication and half are taking a placebo.

Placebo

2.9 4.1 2.7 3.5 5.6 4.0 13.3 5.2

5.6 6.5 5.2 4.8 4.8 4.0 5.7 5.5

Medicine

1.4 3.9 2.9 3.3 3.4 3.2 4.0 4.4

3.1 2.6 5.3 2.6 2.3 2.8 3.4 2.9

4. Construct side-by-side box plots of the data sets.

5. Use the box plots to compare the center and spread of the distributions.

6. Find the interval of z-values associated with the outside 35% of the data.

7. TOOLS Weights of hammers are normally distributed withμ = 2.2 pounds and σ = 0.3 pound. If the heaviest 15% of hammers are on sale, what is the lightest weight a hammer can be and still be on sale?

Mid-Chapter Test(Lessons 11-1 through 11-4)

Part I

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7.


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alternative hypothesis

binomial distribution

Central Limit Theorem

confidence interval

continuous random variable

correlation

correlation coefficient

critical values

discrete random variable

empirical rule

explanatory variable

extrapolation

hypothesis test


interpolation

least squares regression line

negatively skewed distribution

normal distribution

null hypothesis

percentiles

positively skewed distribution

probability distribution

random variable

regression line

sampling distribution

sampling error

standard normal distribution

symmetrical distribution

t-distribution

z-value

Choose a term from the vocabulary list above to complete each sentence.

1. The graph of a(n) _______________ is a bell-shaped curve that is symmetric about the mean of the data.

2. A(n) _______________ represents a numerical value assigned to an outcome of a probability experiment.

3. A(n) _______________ assesses evidence provided by data about a claim concerning a population parameter.

4. A(n) _______________ can be found when the maximum error of the estimate is added to and subtracted from the sample mean.

5. A(n) _______________ is used when the population standard deviation is unknown and n < 30.

6. In _______________, a sample of data is analyzed and conclusions are made about the entire population.

7. The _______________ represents the number of standard deviations that a given data value is from the mean in a normal distribution.

8. A(n) _______________ gives the type and strength of a linear relationship.

Define each term in your own words. 9. Central Limit Theorem

10. probability distribution

Pdf Pass


11 Chapter 11 Vocabulary Test

1.

2.

3.

4.

5.

6.

7.

8.


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CARS The table shows the number of cars sold by 20 salespeople in one week.

Number of Cars

10 7 6 9 7 3 5 6 8 4

8 2 7 5 7 9 11 5 7 10

1. Which best describes the shape of the distribution?

A fairly symmetric C positively skewed B bimodal D negatively skewed

2. What is the median of the data? F 6.8 G 6 H 7 J 7.5

3. What is the mean of the data?

A 6.5 B 6.8 C 7 D 7.2

For Questions 4 and 5, use the probability distribution.

4. What is the mean of the distribution? F 1.58 G 3 H 3.34 J 13.66

5. What is the standard deviation of the distribution? A 1.58 B 2.50 C 3 D 3.34

6. INVESTMENT If an investment of $10,000 is successful, the investor makes $50,000. Otherwise, he or she loses everything. Which is the expected value if the probability of success is 40%?

F $4000 G $6000 H $10,000 J $14,000

7. What is z if X = 237, μ = 220, and σ = 12.3? A -4.55 B -1.38 C 1.38 D 4.55

8. LICENSES In a certain region, the ages of licensed drivers is normally distributed with a mean of 44.5 years and a standard deviation of 9.1 years. Find the probability that a randomly selected driver is younger than 25.

F 1.6% G 2.7% H 4.9% J 88.4%

9. If there are 42,000 drivers in the region described in Question 8, what is the approximate number of drivers younger than 25?

A 672 B 1134 C 2058 D 2625

10. In a normal distribution with μ = 120 and σ = 4, a random sample of 35 values is chosen. Find the probability that the sample mean is between 119 and 121.

F 11.2% G 40.1% H 50% J 86.1%

11. In a binomial distribution, n = 50, p = 0.20, and q = 0.80. Find P(X < 12).

A 70.2% B 76% C 79.8% D 81.2%

X 1 2 3 4 5 6

P(X) 0.15 0.20 0.18 0.22 0.13 0.12

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

Pdf Pass


11 Chapter 11 Test, Form 1


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EXAMS A random sample of 1000 exams resulted in an average score of 500 points. Assume that the standard deviation is 80 points.

12. Find the maximum error of estimate for a 99% confidence level.

F 4.96 G 6.52 H 30.99 J 40.7

13. Find a 99% confidence interval for the mean population score.

A 499–501 B 495–505 C 479–520 D 493–507

14. For a maximum error of ±4 points and a 90% confidence level, what is the minimum number of samples that should be selected?

F 1008 G 1083 H 1537 J 2655

RAIN Ryan believes that the mean monthly rainfall in his town is more than 4 inches. He randomly selects monthly rainfall amounts, as shown below, to test his claim at α = 0.05. Use this data for Questions 15–17.

Monthly Average Rainfall (inches)

8.8 6.6 5.9 3.0 2.5 0.9 0.1 0.0 0.6 3.4 5.9 9.0

15. Identify the null and alternative hypotheses and the claim.

A H0: μ > 4 (claim); Ha: μ ≤ 4 C H0: μ < 4; Ha: μ ≥ 4 (claim)

B H0: μ ≥ 4 (claim); Ha: μ < 4 D H0: μ ≤ 4; Ha: μ > 4 (claim)

16. What is the value of the test statistic?

F -0.381 G -0.034 H -0.116 J -0.11

17. What is the critical value?

A -2.2 B -1.8 C -1.6 D -1.4

HEALTH For Questions 18–20, use the data given in the table.

18. What is the correlation coefficient?

F -0.983 G -0.923 H 0.923 J 0.983

19. At which level is the correlation coefficient significant for H0: ρ = 0?

A 1% B 5% C 10% D all of these

20. Find the least-squares regression line for the data.

F ^y = -0.81x + 440.03 H ^y = 0.81x + 440.03 G ^y = -0.81x - 440.03 J ^y = 0.81x - 440.03

Bonus In a binomial distribution, n = 15 and q = 0.55. Find P(X > 1).

NAME DATE PERIOD

Calories Sodium (mg) Calories Sodium (mg) Calories Sodium (mg)

130 320 120 315 180 300

160 340 128 335 45 410

60 395 90 370 150 310

145 322 80 385 100 355

12.

13.

14.

15.

16.

17.

18.

B:

19.

20.

Pdf Pass


11 Chapter 11 Test, Form 1 (continued)


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MOVIES The playing times for 20 movies are recorded in the table. Use the table for Questions 1–3.

Playing Time of Movies (minutes)

102 128 123 132 104 95 109 121 108 124

92 140 117 102 124 115 113 89 111 108


A symmetric B bimodal C skewed D none of these

2. What is the standard deviation of the data?

F 5.7 G 10.8 H 13.5 J 20

3. Which is not true?

A IQR = 20.5 B P50 = 112 C Q1 = 89 D Q3 = 123.5


4. What is the mean of the distribution?

F 1.92 G 2.45 H 3.5 J 4.5

5. What is the standard deviation of the distribution? A 1.28 B 1.65 C 2.45 D 7.65

6. RAFFLE You buy a raffle ticket for $2. If your ticket is selected from the500 sold, you will win $750. What is the expected value of your net gain?

F -$0.75 G -$0.50 H -$0.25 J $0.50

7. What is X if z = 2.88, μ = 43, and σ = 5.2?

A -42 B 28 C 44 D 58

8. STRESS Recruits for a security position must take a stress test. The scores are normally distributed with a mean of 400 and a standard deviation of 100. What percent of the recruits scored higher than a recruit who scored 644?

F 0.73% G 7.3% H 49.3% J 99.3%

9. Out of 500 randomly selected recruits in Question 8, about how many scored higher than a recruit who scored 644?

A 4 B 37 C 247 D 497

10. In a normal distribution with μ = 88 and σ = 1.8, a random sample of 40 values is selected. Find the probability that the sample mean is between 87 and 87.5.

F 1.8% G 3.9% H 10.1% J 28.9%

11. In a binomial distribution, n = 35 and p = 0.20. Find P(X < 5).

A 14.5% B 20.0% C 26.3% D 32.8%

X 1 2 3 4 5 6

P(X) 0.21 0.46 0.13 0.10 0.07 0.03

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

Pdf Pass


11 Chapter 11 Test, Form 2A


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TIRES A random sample of 30 similar tires found an average life span of 36,200 miles. Assume that the standard deviation is 3800 miles.


F 248.27 G 326.29 H 1359.81 J 1787.18

13. Find a 99% confidence interval for the mean population lifespan. A 34,413-37,987 C 35,506-36,894 B 34,840-37,560 D 35,952-36,448

14. For a maximum error of ±1500 miles and a 90% confidence level, what is the minimum number of samples that should be chosen?

F 10 G 15 H 18 J 26

WEATHER A travel agent believes that the mean monthly temperature of a resort town is at least 70°. She randomly chose the mean monthly temperatures as shown to test her claim at α = 0.05.

Mean Monthly Temperatures

75.0 54.5 61.7 68.7 50.7 81.3 84.0 52.7 79.0 70.0 59.9 84.0

15. Identify the null and alternative hypotheses and the claim. A H0: μ > 70 (claim); Ha: μ ≤ 70 C H0: μ < 70; Ha: μ ≥ 70 (claim)


16. What is the value of the test statistic? F -5.7 G -1.54 H -0.43 J -0.13

17. What is the critical value? A -2.2 B -1.8 C -1.6 D -1.4

FOOD For Questions 18–20, use the data given in the table.

Calories Sugar (g) Calories Sugar (g) Calories Sugar (g)

80 1 120 2.2 150 5

160 7 145 5 100 2

128 2.5 90 2 85 1.2

130 3 180 10 60 0

18. What is the correlation coefficient? F -0.927 G -0.860 H 0.860 J 0.927

19. At which level is the correlation coefficient significant for H0: ρ = 0? A 1% B 5% C 10% D all of these


F y^ = -0.07x + 5.26 H y^ = 0.07x + 5.26

G y^ = -0.07x - 5.26 J y^ = 0.07x - 5.26

Bonus In a standard normal distribution, if P(z > a) = 0.0865, what is a?

NAME DATE PERIOD

12.

13.

14.

15.

16.

17.

18.

B:

19.

20.

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11 Chapter 11 Test, Form 2A (continued)


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Write the letter for the correct answer in the blank at the right of each problem.

MUSIC The playing times of 20 songs on a radio station are recorded in the table. Use the table for Questions 1–3.

Playing Time (seconds)

212 204 174 187 292 225 159 189 214 206

215 257 243 232 310 299 187 240 187 176


A fairly symmetric C positively skewed B bimodal D negatively skewed

2. What is the standard deviation of the data?

F 6.5 G 11 H 36 J 42.7

3. Which is not true?

A Q1 = 159 B P50 = 213 C IQR = 54.5 D Q3 = 241.5

For Questions 4 and 5, use the probability distribution below.

4. What is the mean of the distribution?

F 1.48 G 2.48 H 3.19 J 3.5

5. What is the standard deviation of the distribution?

A 1.34 B 1.80 C 2.24 D 3.25

6. RAFFLE You sold your friend a raffle ticket for $1. If his ticket is selected from the 1000 sold, he will win $500. What is the expected value of his net gain?

F -$0.99 G -$0.50 H -$0.10 J $0.05

7. What is X if z = 1.38, μ = 220, and σ = 12.3?

A -203 B -211 C 229 D 237

8. WRITING Applicants for an editorial position must take a grammar test. The scores are normally distributed with a mean of 150 and a standard deviation of 8. What percent of the applicants scored higher than an applicant who scored 163?

F 2.3% G 5.2% H 94.8% J 97.7%

9. Out of 200 randomly selected applicants in Question 8, about how manyscored higher than an applicant who scored 163?

A 10 B 14 C 20 D 24

10. In a normal distribution with μ = 112 and σ = 5.3, a random sample of 50 values is chosen. Find the probability that the sample mean is between 110 and 111.

F 0.4% G 8.8% H 9.1% J 40.9%

11. In a binomial distribution, n = 30 and p = 0.60. Find P(X > 16).

A 68.1% B 71.2% C 77.2% D 82.4%

X 1 2 3 4 5 6

P(X) 0.25 0.18 0.16 0.11 0.14 0.16

Pdf Pass


11 Chapter 11 Test, Form 2B

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8.

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DRINKS A random sample of 200 drink machines found the average amount dispensed to be 8.1 ounces. Assume that the standard deviation is 0.75 ounce. 12. Find the maximum error of estimate for a 99% confidence level.

F 0.087 G 0.104 H 0.137 J 1.23

13. Find a 99% confidence interval for the population mean. A 7.35-8.85 B 7.96-8.24 C 8.00-8.20 D 8.10-8.11

14. For a maximum error of ±0.15 ounce and a 90% confidence level, what is the minimum number of samples that should be chosen?

F 68 G 97 H 166 J 178

CARS Sasha believes that the mean number of cars sold per week by her staff is greater than 7. She randomly selected the sales for one week, as shown, to test her claim at α = 0.05.

Number of Cars

10 7 6 9 7 3 5 6 8 4

8 2 7 5 7 9 11 5 7 10

15. Identify the null and alternative hypotheses and the claim. A H0: μ > 7 (claim); Ha: μ ≤ 7 C H0: μ < 7; Ha: μ ≥ 7 (claim)


16. What is the value of the test statistic? F -3.77 G -0.52 H -0.38 J -0.08

17. What is the critical value? A -2.3 B -2.1 C -1.9 D -1.7

MOVIES For Questions 18–20, use the data given in the table.

Weekly Income vs. Number of Movie Rentals

Income ($) Movies Income ($) Movies Income ($) Movies

500 9 900 7 1225 3

1150 4 650 8 850 8

400 10 950 7 1650 2

825 9 1800 1 1000 5

18. What is the correlation coefficient? F -0.950 G -0.903 H 0.903 J 0.950

19. At which level is the correlation coefficient significant for H0: ρ = 0? A 1% B 5% C 10% D all of these

20. Find the least-squares regression line for the data. F y^ = -0.007x + 12.8 H y^ = 0.007x + 12.8 G y^ = -0.007x - 12.8 J y^ = 0.007x - 12.8


NAME DATE PERIOD

B:

18.

19.

20.

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11 Chapter 11 Test, Form 2B (continued)

12.

13.

14.

15.

16.

17.


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11

MOVIES Fifty randomly selected people going to a science-fiction movie were asked their ages. The results are recorded in the table. Use the table for Questions 1–3.

Ages of Science-Fiction Moviegoers (years)

17 42 21 78 16 21 31 29 29 16

49 19 81 16 69 69 18 31 22 14

21 75 42 78 18 41 22 16 18 80

42 42 42 16 16 21 19 18 44 18

22 14 49 17 16 18 18 18 17 23

1. Construct a histogram of the data.

2. Describe the shape of the distribution of the data.

3. Find the mean and median of the distribution.

For Exercises 4 and 5, use the probability distribution.

X P(X) X P(X)

1 0.02 6 0.02

2 0.08 7 0.30

3 0.16 8 0.24

4 0.03 9 0.07

5 0.05 10 0.03

4. Find the mean of the distribution.

5. Find the standard deviation of the distribution.


7. Find X if z = 2.15, μ = 185, and σ = 7.5.

8. TRANSPORTATION The average time it takes a trolley to travel between two particular stops is 7.2 minutes with a standard deviation of 1.5 minutes. The times are normally distributed. What is the probability that on any given trip the trolley takes more than 8 minutes to travel between the two stops?

9. From a sample of 200 trolley trips described in Question 8, about how many trips will be less than 7 minutes long?

10. In a normal distribution with μ = 58 and σ = 3, a random sample of 30 values is selected. Find the probability that the sample mean is between 58.5 and 59.

Chapter 11 Test, Form 2C

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


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11

11. In a binomial distribution, n = 35 and p = 0.45. Find P(X < 15).

CLOCKS For Questions 12–14, a random sample of 225 homes found an average of 5.2 clocks per home. Assume from past studies the standard deviation is 0.8. .


13. Find a 99% confidence interval for the mean number of clocks in all the homes.

14. For a maximum error of ±0.25 clock and a 90% confidence level, what is the minimum number of samples that should be taken?

HOCKEY Jabir believes that the mean save percentages for NHL goalies is greater than 0.910. He randomly selected the save percentages for some goalies in a recent season, as shown below to test his claim at α = 0.05. Use these data for Questions 15–17.

0.940 0.937 0.930 0.929 0.926 0.925 0.925 0.923 0.923 0.923

0.921 0.920 0.919 0.918 0.916 0.914 0.911 0.911 0.910 0.910

0.909 0.907 0.907 0.904 0.903 0.903 0.902 0.900 0.898 0.895

15. Identify the null and alternative hypotheses and the claim for Jabir’s hypothesis test.

16. What is the critical value and the test statistic?

17. Is there sufficient evidence to reject the null hypothesis? Explain.

FABRIC For Questions 18–20, use the data in the table. It gives the amount of each fabric sold at a craft store and its price.

Sq. Yd Sold Price/Sq. Yd ($) Sq. Yd Sold Price/Sq. Yd ($)

900 8.99 965 32.99

1000 15.99 855 38.99

975 21.99 750 40.99

950 25.99 740 42.99

950 28.99 732 45.99

960 30.99 600 55.99

18. Calculate the correlation coefficient.

19. Is the correlation coefficient significant at α = 0.05 for ρ = 0? Explain.



Chapter 11 Test, Form 2C (continued)

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:


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11

TELEVISION Fifty students recorded the number of hours that the television was on during one week in their homes. The results are in the table. Use the table for Questions 1–3.

Weekly Television Hours (to the nearest hour)

54 28 9 15 3 54 35 32 0 34

72 57 62 33 58 23 57 53 24 27

36 63 3 58 53 13 12 75 66 57

18 53 53 46 77 26 32 42 43 88

44 71 22 57 45 73 44 11 45 34

1. Construct a histogram of the data.

2. Describe the shape of the distribution of the data.

3. Find the mean and median of the distribution.


X P(X) X P(X)

1 0.08 6 0.10

2 0.10 7 0.06

3 0.30 8 0.02

4 0.20 9 0.01

5 0.12 10 0.01


5. Find the standard deviation of the distribution.


7. Find X if z = 1.65, μ = 125, and σ = 5.5.

8. BIRD FEEDER The average number of birds that visit a bird feeder per day is normally distributed with a mean of 26.8 and a standard deviation of 4.7. What is the probability that on any given day the number of birds is greater than 30?

9. From a sample of 80 randomly selected days described in Question 8, about how many days will the number of birds be less than 20?

10. In a normal distribution with μ = 425 and σ = 24, a random sample of 40 values is chosen. Find the probability that the sample mean is between 420 and 428.

Chapter 11 Test, Form 2D

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


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11

11. In a binomial distribution, n = 42 and p = 0.72. Find P(X > 28).

FOOD For Questions 12–14, a random sample of256 people found that they ate fast food an average of2.6 times per week. Assume from past studies the standard deviation is 0.4.


13. Find a 99% confidence interval for the mean number of times people eat fast food each week.

14. For a maximum error of ±0.1 times and a 90% confidence level, what is the minimum sample that should be taken?

WORK A supervisor believes that the mean pay increase for her employees is 4.5%. She randomly selected the increases for 13 employees, as shown below, to test her claim at α = 0.05. Use these data for Questions 15–17.

Percent Pay Raise

3.2% 4.4% 4.1% 3.8% 1.5% 2.4% 4.6%

3.3% 1.7% 9.2% 4.5% 4.2% 5.1%

15. Identify the null and alternative hypotheses and the claim for her hypothesis test.

16. What is the critical value and the test statistic?

17. Is there sufficient evidence to reject the null hypothesis? Explain.

BUDGET For Questions 18–20, use the data in the table.

Cost of Gas vs. Number of Meals Out of House

Cost of gas

($ per gal)

Number of

Meals Out (wk)

Cost of gas

($ per gal)

Number of

Meals Out (wk)

1.79 6 2.45 3

1.84 5 2.51 2

2.20 6 3.15 2

2.28 5 3.89 1

2.34 4 4.05 1

2.39 3 4.10 0

18. Calculate the correlation coefficient.

19. Is the correlation coefficient significant at α = 0.05for ρ = 0? Explain.



Chapter 11 Test, Form 2D (continued)

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:


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11 Chapter 11 Test, Form 3

SPEEDS The speeds of 50 randomly selected cars crossing a state border are recorded in the tables. Use the tables for Questions 1–3.

Massachusetts Speeds

(miles per hour)

New Hampshire Speeds

(miles per hour)

66 69 70 67 74 88 71 75 68 73

69 72 67 69 68 78 68 58 69 77

72 65 72 59 63 73 72 65 76 77

58 63 73 66 77 76 82 72 73 81

64 71 72 67 65 73 64 71 65 84

1. Construct side-by-side box plots of the data.

2. Compare the distributions.

3. Describe the shape of the Massachusetts distribution.


X P(X) X P(X)

5 0.06 55 0.24

15 0.22 65 0.23

25 0.26 75 0.18

35 0.24 85 0.23

45 0.22 95 0.12


5. Find the variance and standard deviation of the distribution.

6. FIRE INSURANCE A company sells a one-year home fire insurance policy to a customer for $765. The probability that no claim for fire will be made is 0.996. If there is a fire, assume a pay-out to the customer for $165,000. What is the company’s expected profit?

7. Find X if z = -2.15, μ = 16.9, and σ = 1.85.

8. FURNITURE A decorator finds that retail costs of sofas are normally distributed with a mean of $950 and a standard deviation of $215. What is the probability that a randomly selected sofa costs between $700 and $800?

9. From a sample of 50 randomly selected sofas described in Question 8, how many will cost more than $1250?

10. In a normal distribution with μ = 102 and σ = 8.6, a random sample of 42 values is selected. Find the probability that the sample mean is between 101.5 and 103.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


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11

11. In a binomial distribution, n = 22 and q = 0.178. Find P(X ≥ 19).

BATTERIES For Questions 12–14, a random sample of 85 batteries found a mean battery life of 450 minutes. Assume from past studies the standard deviation is 18.4 minutes.


13. Find the 99% confidence interval for the mean battery life of all the batteries.

14. For a maximum error of ±6 minutes and a 90% confidence level, what is the minimum number of samples to be taken?

LAW ENFORCEMENT Myra believes that the average number of traffic tickets issued per day in her town is no more than 10. She finds the number of traffic tickets issued on 20 random days, as shown below, to test her claim at a = 0.05. Use these data for Questions 15–17.

7 12 10 8 7 12 15 10 10 7

14 6 10 12 9 8 8 2 9 12

15. Identify the null and alternative hypotheses and the claim for Myra’s hypothesis test.

16. State the critical value, test statistic, and P-value.

17. Is there sufficient evidence to reject the null hypothesis? Explain.For Questions 18–20, use the data in the table.

Amount of Garbage Produced by Selected Households

Garbage

(lb)

Number in

HouseholdGarbage (lb)

Number in

Household

35.6 4 38.1 6

49.0 5 21.8 1

27.6 3 28.0 4

21.9 2 33.3 6

10.8 2 20.0 3

18. Calculate and interpret the correlation coefficient.

19. Determine whether the correlation coefficient is significant at the 1%, 5%, and 10% levels. Explain.

20. Use the least-squares regression line for the data to predict the size of a household that produces 60 pounds of garbage.

Bonus In a standard normal distribution, if P(-a < z < a) = 0.4775, what is a?

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 11 Test, Form 3 (continued)


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11 Extended-Response Test

Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solutions in more than one way or investigate beyond the requirements of the problem.

1. The table shows the weights of 37 dogs in a park’s large-dog play area.

Weights of Large Dogs (lb)

78 76 88 68 87 78 84 83 71 8285 91 86 92 80 93 98 73 93 93

96 80 94 64 97 91 84 100 89 81

86 74 87 97 90 91 101

a. Construct a histogram and use it to describe the shape of the distribution.

b. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.

2. The combined test scores for all of the advanced mathematics classes in a school are normally distributed. The mean score is 85 and the standard deviation is 6. There are 200 students in the classes.

a. Those who had scores above 94 were given a grade of A. How many students received an A? Explain your reasoning.

b. Students who score in the top 30% are eligible to participate in a regional competition. What is the minimum score required for eligibility? Explain.

c. If a random sample of 35 students are chosen, what is the probability that the mean of the sample is between 83 and 84?

3. Company executives report that their cereal boxes contain 14 ounces of cereal. A journalist weighs a random sample of 15 boxes and finds the mean weight to be 13.9 ounces with a standard deviation of 0.17 ounce. Use a 5% level of significance to test the executives’ claim.

4. The table shows the volume and cost of dried backpacking food.

Volume (oz) 6 3.75 21 7.13 1.4 3.5 4.8 4.6 2.1 5 3

Cost ($) 5.5 4.5 14.99 6.5 3 3.25 5.99 5.99 3.99 7.5 3.99

a. Calculate and interpret the correlation coefficient.

b. Determine if the correlation coefficient is significant at the 1%, 5%, and 10% levels.

c. Find the least-squares regression equation. Interpret the slope and intercept in context.


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11

1. Which best describes the distribution of the data in the table?

790 750 840 617 687 700

599 545 527 840 905 813

A fairly symmetric C positively skewed

B bimodal D negatively skewed 1.

2. The cost of renting a washer and dryer is $30 for one month, $55 for two months, and $20 per month for more than two months, up to one year. Which function represents the situation, where m represents the number of months?

F c(m) =

⎧

⎨

⎩

30 if 0 < m ≤ 1

55 if 1 < m ≤ 2

20m if 2 < m ≤ 12

H c(m) =

⎧

⎨

⎩

30 if 0 < m < 1

55 if 1 ≤ m < 2

20m if 2 ≤ m

G c(m) = 20m + 30 J c(m) = 30(m + 1) + 55(m + 2) + 20(m + 3) 2.

3. Find the distance between (

2, π

−

6 )

and (

5, 2π

−

3 )

.

A 0 B 3.39 C 7.68 D 9.43 3.

4. Use a half-angle identity to find the exact value of sin 67.5°.

F √

2 + √

2 −

2 G

√

2 - √

2 −

2 H -

√

2 + √

2 −

2 J -

√

2 - √

2 −

2 4.

5. Which is not a possible rational root of 2x3 - 9x2 - 11x + 8 = 0?

A -2 B -

1 −

2 C 1 −

4 D 8 5.

6. Find the partial fraction decomposition of 6x - 14 −

x2 -2x - 15 .

F 2 −

x + 3 + 4 −

x - 5 G 2 −

x + 3 + -4 −

x - 5 H 4 −

x + 3 + 2 −

x - 5 J 4 −

x + 3 + -2 −

x - 5 6.

7. What is -4 + 3i in polar form?

A 5(cos 53.13 - i sin 53.13) C 25(cos 53.13 + i sin 53.13)

B 25(cos 36.89 + i sin 36.89) D 5(-cos 36.89 + i sin 36.89) 7.

8. What is the coefficient of the third term in the expansion of (2x - y)5?

F -80 G 32 H 40 J 80 8.

9. Solve 19e4x = 95. A 0.19 B 0.25 C 0.40 D 1.16 9.

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.

Standardized Test Practice(Chapters 1–11)

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D


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10. Which statement is an identity?

F tan θ −

4

= tan

θ −

4

H sin 2θ = 2 sin θ cos θ

G csc (

θ - π −

2 )

= sec θ J 3 tan2 θ - 1 = sec2 θ 10.

11. Find the angle between u = ⟨4, -3⟩ and v = ⟨1, 2⟩.

A 78.7° B 79.7° C 100.3° D 101.3° 11.

12. Which is the inverse of f(x) = x2 - 3? What are the restrictions on the domain?

F f -1(x) = ± √

�

x + 3, x ≠ 0 H f -1(x) = ± √

�

x - 3, x ≠ 0

G f -1(x) = ± √

��

x - 3 , x ≥ 3 J f -1(x) = ± √ ��

x + 3 , x ≥ -3 12.

13. Expand log 5x8y-2.

A log 5 + 8 log x - 2 log y C 5 log x + 8 log x - 2 log y

B log 1 −

5 + log 8 x - log 2 y D 1 −

5 log x + log 8x - log 2y 13.

14. Given f(x) = 2x2 and g(x) = 3x, find (g ◦ f)x.

F (g ◦ f)x = 6x2 G (g ◦ f)x = 9x2 H (g ◦ f)x = 18x2 J (g ◦ f)x = 8x4 14.

15. Find the exact value of cos (-210°).

A -

√

�

3 −

3 B -

√

�

3 −

2 C

√

�

3 −

2 D

√

�

3 −

3 15.

16. A car salesman is testing the gas mileage of cars in his lot. He knows from previous tests that the standard deviation is 4 miles per gallon. If he wants results that are accurate to within ±3 miles per gallon, with a 95% confidence level, what is the minimum number of cars he must test?

F 5 G 7 H 12 J 15 16.

17. What is the mean of the probability distribution?

X 1 2 3 4

P(X) 0.4 0.25 0.15 0.2

A 2.15 B 2.5 C 2.75 D 2.85 17.

18. What is the sum of the infinite geometric series 18 + 6 + 2 + … ?

F 27 G 36 H 45 J 54 18.

Standardized Test Practice (continued)

(Chapters 1–11)

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J


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19. Find three arithmetic means between 16 and 52.

20. State the amplitude, period, frequency, phase shift,

and vertical shift of y = -sin (

x + π −

2 )

+ 2.

21. Find the dot product of u = ⟨4, -5⟩ and v = ⟨10, 8 ⟩. Then determine if u and v are orthogonal.

22. Identify, the vertex, focus, axis of symmetry, and directrix of the graph of y2 - 12x + 2y = -37.

23. Find all solutions to tan2 θ + tan θ = 0 on the interval [0, 2π].

24. Find the inverse matrix required

to solve ⎡

⎢

⎣

12 5 7 3 ⎤

⎦

·

⎡

⎢

⎣

x1 x2

y1 y2

⎤

⎦

= ⎡

⎢

⎣

2 3

-1

2

⎤

⎦

.

25. An engineer has developed a new microprocessing chip that he believes has a longer usable life than his company’s existing chip. The existing chip has a usable life of 20,000 hours with a standard deviation of 3200 hours. The engineer has found that in a randomly selected sample of 1000 new chips, the mean usable life is 20,500 hours. He wishes to test his claim at α = 0.01.

a. Identify the hypotheses and the claim.

b. Find the critical value.

c. Find the test statistic.

d. Is there sufficient evidence to show the new chip has a longer life? Justify your answer.

Part 2: Short Response

Instructions: Write your answers in the space provided.

19.

20.

21.

22.

23.

24.

25a.

25b.

25c.

25d.

Standardized Test Practice (continued)

(Chapters 1–11)


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Chapter 11 A1 Glencoe Precalculus

An

swer

s

Answers (Anticipation Guide and Lesson 11-1)

Pdf Pass

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

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umm

ariz

e th

e ce

nter

and

spr

ead

of t

he d

ata

usin

g ei

ther

the

mea

n an

d st

anda

rd d

evia

tion

or

the

five

-num

ber

sum

mar

y. J

usti

fy y

our

choi

ce.

Use

the

five

-num

ber

sum

mar

y si

nce

the

data

are

ske

wed

.

The

num

ber

of p

asse

nger

s ra

nge

from

468

4 to

30,

526,

and

the

med

ian

num

ber

of p

asse

nger

s is

abo

ut 9

856.

Hal

f of t

heai

rpor

ts h

ad b

etw

een

6338

and

13,

800

pass

enge

rs.

Exer

cise

1. A

res

taur

ant

man

ager

adv

erti

ses

a 20

-oun

ce s

irlo

in s

teak

on

his

men

u.Th

e w

eigh

ts o

f a s

ampl

e of

the

ste

aks

is s

how

n.

S

tea

k W

eig

hts

(o

un

ce

s)

20

20

20

18

20

18

18

19

19

19

21

20

17

20

19

21

18

19

20

20

a. C

onst

ruct

a h

isto

gram

and

use

it t

o de

scri

be t

he s

hape

of t

he d

istr

ibut

ion.

Th

e g

rap

h i

s n

eg

ati

ve

ly s

ke

we

d.

b. S

umm

ariz

e th

e ce

nter

and

spr

ead

of t

he d

ata

usin

gei

ther

the

mea

n an

d st

anda

rd d

evia

tion

or

the

five-

num

ber

sum

mar

y. J

usti

fy y

our

choi

ce.

Be

ca

us

e t

he

dis

trib

uti

on

is

sk

ew

ed

, u

se

th

e f

ive

-nu

mb

er

su

mm

ary

. T

he

we

igh

ts r

an

ge

fro

m 1

7 o

un

ce

s t

o

21

ou

nc

es

an

d t

he

me

dia

n w

eig

ht

is 1

9.5

ou

nc

es

. H

alf

th

e w

eig

hts

a

re b

etw

ee

n 1

8.5

ou

nc

es

an

d 2

0 o

un

ce

s.

Exam

ple

[ 17,

22]

scl:

1 b

y [ 0

, 10]

scl:

1

[ 400

0, 3

5,00

0] s

cl: 3

000

by [ 0

, 15]

scl:

1

005_

042_

PC

CR

MC

11_8

9381

2.in

dd5

3/25

/09

12:3

6:30

PM


NA

ME

DA

TE

PE

RIO

D

Chapter Resources

A A A A A D DD A A D A

11

B

efor

e yo

u be

gin

Cha

pter

11

•

Rea

d ea

ch s

tate

men

t.

•

Dec

ide

whe

ther

you

Agr

ee (A

) or

Dis

agre

e (D

) wit

h th

e st

atem

ent.

•

Wri

te A

or

D in

the

firs

t co

lum

n O

R if

you

are

not

sur

e w

heth

er y

ou a

gree

or

disa

gree

, wri

te N

S (N

ot S

ure)

.

A

fter

you

com

plet

e C

hapt

er 1

1

•

Rer

ead

each

sta

tem

ent

and

com

plet

e th

e la

st c

olum

n by

ent

erin

g an

A o

r a

D.

•

Did

any

of y

our

opin

ions

abo

ut t

he s

tate

men

ts c

hang

e fr

om t

he fi

rst

colu

mn?

•

For

thos

e st

atem

ents

tha

t yo

u m

ark

wit

h a

D, u

se a

pie

ce o

f pap

er t

o w

rite

an

exam

ple

of w

hy y

ou d

isag

ree.

ST

EP

1

A,

D,

or

NS

Sta

tem

en

tS

TE

P 2

A

or

D

1.

In a

neg

ativ

ely

skew

ed d

istr

ibut

ion,

the

mea

n is

less

tha

n th

e m

edia

n. 2

. Pe

rcen

tile

s di

vide

a d

istr

ibut

ion

into

100

equ

al g

roup

s.

3.

A c

onti

nuou

s ra

ndom

var

iabl

e ta

kes

on a

cou

ntab

le n

umbe

r of

pos

sibl

e va

lues

. 4

. In

a p

roba

bilit

y di

stri

buti

on, t

he s

um o

f P(X

) mus

t be

1.

5.

In a

nor

mal

dis

trib

utio

n, t

he m

ean,

med

ian,

and

mod

e ar

e eq

ual a

nd a

re lo

cate

d at

the

cen

ter

of t

he d

istr

ibut

ion.

6.

A z

-val

ue r

epre

sent

s th

e nu

mbe

r of

sta

ndar

d de

viat

ions

tha

t a

give

n da

ta v

alue

is fr

om t

he m

ean.

7.

If t

he s

ampl

e si

ze is

ade

quat

ely

larg

e, t

he d

istr

ibut

ion

of t

he

sam

ple

mea

ns w

ill a

hav

e a

mea

n an

d st

anda

rd d

evia

tion

eq

ual t

o th

e po

pula

tion

mea

n an

d st

anda

rd d

evia

tion

. 8

. A

s th

e pr

obab

ility

of s

ucce

ss in

crea

ses

to 0

.5, t

he s

hape

of t

he

bino

mia

l dis

trib

utio

n be

gins

to

rese

mbl

e th

e no

rmal

di

stri

buti

on.

9.

A c

onfid

ence

leve

l giv

es t

he p

roba

bilit

y th

at t

he in

terv

al

esti

mat

e w

ill in

clud

e a

give

n pa

ram

eter

.10

. Th

e nu

ll hy

poth

esis

sta

tes

that

the

re is

a d

iffer

ence

bet

wee

n th

e sa

mpl

e va

lue

and

the

popu

lati

on p

aram

eter

. 11

. E

xtra

pola

tion

use

s an

equ

atio

n to

mak

e pr

edic

tion

s ov

er t

he

rang

e of

the

dat

a.12

. A

cor

rela

tion

coe

ffici

ent

of -

0.95

sho

ws

a st

rong

neg

ativ

e co

rrel

atio

n be

twee

n tw

o va

riab

les.

Antic

ipat

ion

Guid

eIn

fere

nti

al

Sta

tist

ics

Step

2

Step

1

Ch

ap

ter

11

3

Gle

ncoe

Pre

calc

ulus

0ii_

004_

PC

CR

MC

11_8

9381

2.in

ddS

ec1:

33/

25/0

912

:40:

05P

M

A01_A19_PCCRMC11_893812.indd 1A01_A19_PCCRMC11_893812.indd 1 11/17/09 7:53:13 AM11/17/09 7:53:13 AM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.


Answers (Lesson 11-1)

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

6

Gle

ncoe

Pre

calc

ulus

11-1

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Desc

rip

tive

Sta

tist

ics

Mea

sure

s of

Pos

itio

n Th

e qu

arti

les

give

n by

the

five

-num

ber

sum

mar

y sp

ecify

the

pos

itio

ns o

f dat

a va

lues

wit

hin

a di

stri

buti

on. F

or t

his

reas

on, b

ox p

lots

are

mos

t us

eful

for

side

-by-

side

com

pari

sons

of t

wo

or m

ore

dist

ribu

tion

s.

T

he a

ges

of O

scar

-win

ning

act

ress

es f

rom

tw

o 20

-yea

r pe

riod

s ar

e sh

own.

Con

stru

ct s

ide-

by-s

ide

box

plot

s of

the

dat

a se

ts.

The

n us

e th

is d

ispl

ay t

o co

mpa

re t

he d

istr

ibut

ions

.

19

69

–1

98

8

61

35

34

34

26

37

42

41

35

31

41

33

30

74

33

49

38

61

21

41

19

89

–2

00

8

26

80

42

29

33

36

45

49

39

34

26

25

33

35

35

28

30

29

61

32

Inpu

t th

e da

ta in

to L

1 an

d L

2. T

urn

on P

lot1

and

Plo

t2 a

ndch

oose

a b

ox p

lot

wit

h ou

tlie

rs s

how

n as

the

typ

e of

gra

ph.

From

the

gra

ph, w

e ca

n se

e th

at t

he m

edia

n ag

e fo

r th

e fir

st20

yea

rs is

just

slig

htly

hig

her

than

the

med

ian

age

for

the

seco

nd20

yea

rs. T

he r

ange

of a

ges

for

the

mid

dle

50%

of t

he d

ata

sets

is g

reat

er fo

r th

e ac

tres

ses

from

the

sec

ond

20-y

ear

peri

od.

Bot

h da

ta s

ets

have

out

liers

gre

ater

tha

n th

e re

st o

f the

dat

a. I

gnor

ing

the

outl

iers

, the

dis

trib

utio

n fo

r th

e se

cond

20-

year

per

iod

is m

ore

sym

met

ric

than

the

firs

t an

d th

e ra

nge

of d

ata

for

the

seco

nd is

less

tha

n th

at o

f the

fir

st.

Exer

cise

The

num

ber

of c

redi

t ca

rds

owne

d by

stu

dent

s in

a s

tati

stic

scl

ass

is s

how

n. C

onst

ruct

sid

e-by

-sid

e bo

x pl

ots

of t

he d

ata

sets

.Th

en u

se t

his

disp

lay

to c

ompa

re t

he d

istr

ibut

ions

.

Mo

re t

ha

n o

ne

-fo

urt

h o

f th

e s

tud

en

ts i

n e

ac

h g

rou

p h

av

e n

o c

red

it c

ard

s.

Th

e m

ed

ian

nu

mb

er

of

cre

dit

ca

rds

fo

r th

e f

em

ale

s i

s h

igh

er

tha

n f

or

ma

les

. T

he

ra

ng

e a

nd

ma

xim

um

nu

mb

er

of

cre

dit

ca

rds

is

sli

gh

tly

hig

he

r fo

r th

e

fem

ale

s t

ha

n t

he

ma

les

.

Exam

ple

[ 15,

85]

scl:

5 b

y [ 0

, 0.5

] scl:

0.1

25

[-1,

10]

scl:

1 b

y [ 0

, 1] s

cl: 0

.5

Fe

ma

les

18

01

5

44

05

0

00

30

1

Ma

les

04

00

2

05

70

0

35

00

1

005_

042_

PC

CR

MC

11_8

9381

2.in

dd6

11/1

4/09

3:54

:12

PM


NA

ME

DA

TE

PE

RIO

D

Lesson 11-1

Ch

ap

ter

11

7

Gle

ncoe

Pre

calc

ulus

11-1

Prac

tice

Desc

rip

tive

Sta

tist

ics

1. W

EATH

ER T

he a

vera

ge w

ind

spee

ds r

ecor

ded

at v

ario

us w

eath

er s

tati

ons

in t

he U

nite

d St

ates

are

list

ed b

elow

.

Sta

tio

nS

pe

ed

(mp

h)

Sta

tio

nS

pe

ed

(mp

h)

Sta

tio

nS

pe

ed

(mp

h)

Alb

uq

ue

rqu

e8

.9A

nch

ora

ge

*7

.1A

tlan

ta*

9.1

Ba

ltim

ore

*9

.1B

ost

on

*1

2.5

Ch

ica

go

10

.4

Da

llas–

Ft.

Wo

rth

10

.8H

on

olu

lu*

11

.3In

dia

na

po

lis9

.6

Ka

nsa

s C

ity1

0.7

La

s V

eg

as

9.3

Litt

le R

ock

7.8

Lo

s A

ng

ele

s*6

.2M

em

ph

is8

.8M

iam

i*9

.2

Min

ne

ap

olis

– S

t. P

au

l1

0.5

Ne

w O

rle

an

s8

.1N

ew

Yo

rk C

ity*

9.4

Ph

ilad

elp

hia

*9

.5P

ho

en

ix6

.2S

ea

ttle

*9

.0

Sour

ce: N

atio

nal C

limat

ic D

ata

Cent

er

a. C

onst

ruct

a h

isto

gram

and

use

it t

o de

scri

be t

he s

hape

of t

he d

istr

ibut

ion.

Th

e d

istr

ibu

tio

n i

s s

ing

le-p

ea

ke

d a

nd

ro

ug

hly

sy

mm

etr

ic.

b. S

umm

ariz

e th

e ce

nter

and

spr

ead

of t

he d

ata

usin

g ei

ther

the

mea

n an

d st

anda

rd d

evia

tion

or

the

five-

num

ber

sum

mar

y. J

usti

fy y

our

choi

ce.

Be

ca

us

e t

he

da

ta a

re s

ym

me

tric

, u

se

th

e

me

an

an

d s

tan

da

rd d

ev

iati

on

. T

he

me

an

win

d s

pe

ed

is

9.2

1 m

ph

an

d

the

sta

nd

ard

de

via

tio

n i

s 1

.53

mp

h.

2. O

CEA

NS

The

ten

wea

ther

sta

tion

s w

ith

an a

ster

isk

have

rela

tive

ly c

lose

pro

xim

ity

to e

ithe

r th

e A

tlan

tic

Oce

an o

rPa

cific

Oce

an. C

onst

ruct

sid

e-by

-sid

e bo

x pl

ots

of t

he d

ata

sets

. The

n us

e th

is d

ispl

ay t

o co

mpa

re t

he d

istr

ibut

ions

.T

he

me

dia

n w

ind

sp

ee

d f

or

sta

tio

ns

ne

ar

an

oc

ea

n i

s a

bo

ut

the

sa

me

as

th

e o

the

r s

tati

on

s.

Th

e m

idd

le h

alf

of

the

sp

ee

ds

fo

r th

os

e n

ea

r a

n o

ce

an

va

rie

s l

es

s t

ha

n

the

mid

dle

ha

lf o

f th

e s

pe

ed

s a

t th

e o

the

r s

tati

on

s.

3. S

CORE

S Th

e ta

ble

give

s th

e fr

eque

ncy

dist

ribu

tion

of t

hesc

ores

on

a te

st.

a. C

onst

ruct

a p

erce

ntile

grap

h of

the

dat

a.

b. E

stim

ate

the

perc

enti

le r

ank

a sc

ore

of 6

2 w

ould

hav

e in

thi

sdi

stri

buti

on. I

nter

pret

its

mea

ning

. 2

5th

pe

rce

nti

le;

A s

tud

en

t w

ith

a s

co

reo

f 6

2 s

co

red

be

tte

r th

an

ab

ou

t 2

5%

of

the

stu

de

nts

wh

o t

oo

k t

he

te

st.

[ 6, 1

4] s

cl: 1

by

[ 0, 1

0] s

cl: 1

[ 5, 1

4] s

cl: 1

by

[ 0, 1

] scl:

0.5

10 0203040

Cumulative Percentage

5060708090100

Scor

e50

6070

8090

100

Cla

ss

Bo

un

da

rie

sf

55

.5–

60

.53

60

.5–

65

.58

65

.5–

70

.51

2

70

.5–

75

.55

75

.5–

80

.59

005_

042_

PC

CR

MC

11_8

9381

2.in

dd7

11/1

9/09

7:10

:51

PM

A01_A19_PCCRMC11_893812.indd 2A01_A19_PCCRMC11_893812.indd 2 11/19/09 8:00:21 PM11/19/09 8:00:21 PM

Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.



An

swer

s

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

8

Gle

ncoe

Pre

calc

ulus

11-1

Wor

d Pr

oble

m P

ract

ice

Desc

rip

tive

Sta

tist

ics

1. S

HO

ES A

sho

e st

ore

empl

oyee

des

igns

a

disp

lay

by p

laci

ng s

hoe

boxe

s in

ten

st

acks

. The

num

ber

of b

oxes

in e

ach

stac

k ar

e 5,

7, 9

, 11,

13,

10,

9, 8

, 7, a

nd 5

.

a. C

onst

ruct

a b

ox p

lot

and

use

it t

o de

scri

be t

he s

hape

of t

he d

istr

ibut

ion.

[ 4

, 14]

scl:

1 b

y [ 0

, 1] s

cl: 0

.5

b. S

umm

ariz

e th

e ce

nter

and

spr

ead

of

the

data

usi

ng e

ithe

r th

e m

ean

and

stan

dard

dev

iati

on o

r th

e fiv

e-nu

mbe

r su

mm

ary.

Jus

tify

you

r ch

oice

.

th

e m

ea

n a

nd

sta

nd

ard

d

ev

iati

on

; m

ea

n:

8.4

, s

tan

da

rd

de

via

tio

n:

ab

ou

t 2

.5

2. M

EDIC

INE

A h

isto

gram

for

the

num

ber

of p

atie

nts

trea

ted

at 5

0 U

.S. c

ance

r ce

nter

s in

one

yea

r is

sho

wn.

Sour

ce: U

.S. N

ews

Onl

ine

a. W

hich

is g

reat

er, t

he m

ean

or t

he

med

ian

of t

he d

ata

set?

Exp

lain

.m

ean

; T

he

dat

a a

re p

osi

tive

ly

skew

ed.

b. S

ketc

h th

e ge

nera

l sha

pe o

f a b

ox p

lot

that

wou

ld r

epre

sent

the

his

togr

am.

Sa

mp

le a

ns

we

r:

3. E

XA

MS

The

tabl

e gi

ves

the

freq

uenc

ies

of t

he fi

nal e

xam

sco

res

of 5

0 st

uden

ts in

tw

o pr

ecal

culu

s cl

asse

s.

Cla

ss

Bo

un

da

rie

sF

req

ue

nc

y

f

42

.5–

52

.54

52

.5–

62

.51

0

62

.5–

72

.51

5

72

.5–

82

.51

3

82

.5–

92

.57

92

.5–

10

2.5

1

a. C

onst

ruct

a p

erce

ntile

gra

ph o

fth

e da

ta. 10 0203040 Cumulative Percentage 506070809010

0

Scor

e50

4060

7080

9010

0

b. E

stim

ate

the

perc

enti

le r

ank

a te

st

scor

e of

77

wou

ld h

ave

in t

his

dist

ribu

tion

. Int

erpr

et it

s m

eani

ng.

90

th p

erc

en

tile

; A

stu

de

nt

wit

h

a s

co

re o

f 7

7 h

as

a b

ett

er

sc

ore

th

an

ab

ou

t 9

0%

of

the

s

tud

en

ts t

ak

ing

th

e e

xa

m.

4. M

OV

IES

The

ages

of m

ovie

pat

rons

ina

thea

ter

are

25, 4

7, 1

6, 4

5, 5

4, 1

7, 1

4,

16, 1

6, 3

9, 4

8, 4

8, 1

8, 1

2, 1

3, 6

2, 5

1, 4

6,

and

18. S

umm

ariz

e th

e di

stri

buti

on o

f th

e da

ta.

bim

od

al;

tw

o c

lus

ters

th

at

are

e

ac

h r

ou

gh

ly s

ym

me

tric

Patie

nts i

n U.

S. C

ance

r Cen

ters

2030 10 0

Number or Cancer Centers

40

Num

ber o

f Pat

ient

s

3500

500

1000

1500

2000

2500

3000

Th

e d

ata

are

ro

ug

hly

s

ym

me

tric

.

500

1000

1500

2000

2500

3000

3500

005_

042_

PC

CR

MC

11_8

9381

2.in

dd8

11/1

6/09

12:0

1:18

PM


NA

ME

DA

TE

PE

RIO

D

Lesson 11-1

Ch

ap

ter

11

9

Gle

ncoe

Pre

calc

ulus

11-1

Enri

chm

ent

Mean

s

The

aver

age

of a

dat

a se

t is

kno

wn

as t

he a

rith

met

ic m

ean.

The

re a

re o

ther

m

eans

.

A t

rim

med

mea

n is

the

ari

thm

etic

mea

n of

a d

ata

set

afte

r th

e to

p 10

% a

nd

bott

om 1

0% o

f the

dat

a va

lues

are

tri

mm

ed o

ff. T

here

fore

, if t

here

are

100

dat

a va

lues

, the

leas

t 10

val

ues

and

grea

test

10

valu

es a

re r

emov

ed.

A g

eom

etri

c m

ean

is t

he n

th r

oot

of t

he p

rodu

ct o

f n d

ata

valu

es. I

t is

oft

en

used

in e

cono

mic

s to

find

an

aver

age

rate

of g

row

th.

A h

arm

onic

mea

n H

is t

he n

umbe

r of

dat

a va

lues

div

ided

by

the

sum

of t

he

mul

tipl

icat

ive

inve

rses

of a

ll th

e da

ta v

alue

s. I

t is

oft

en u

sed

in a

vera

ging

spe

eds.

It

can

not

be u

sed

if 0

is a

dat

a va

lue.

H

=

n −

∑

1 −

x , whe

re n

is t

he n

umbe

r of

dat

a va

lues

and

x is

a d

ata

valu

e.

Exer

cise

s 1.

Fin

d th

e tr

imm

ed m

ean

of t

he d

ata

set.

32

, 24,

56,

102

, 54,

12,

27,

49,

35,

23,

44,

51,

66,

36,

52,

16,

63,

75,

21,

41

4

2.1

25

2. F

ind

the

arit

hmet

ic m

ean

and

med

ian

of t

he d

ata

set

in E

xerc

ise

1. W

hat

is t

he b

enef

it o

f usi

ng a

tri

mm

ed m

ean

inst

ead

of a

n ar

ithm

etic

mea

n?

a

rith

me

tic

me

an

: 4

3.9

5,

me

dia

n:

42

.5;

Th

e t

rim

me

d m

ea

n i

s

les

s s

en

sit

ive

to

ou

tlie

rs.

3. W

ATE

R Th

e E

PA r

ecom

men

ds t

hat

qual

ity

fres

hwat

er b

e m

onit

ored

for

e.co

li co

ncen

trat

ion.

The

geo

met

ric

mea

n of

the

e.c

oli c

once

ntra

tion

s in

5

or m

ore

sam

ples

tak

en o

ver

30 d

ays

shou

ld b

e le

ss t

han

126

per

100

mill

ilite

rs. D

o th

e da

ta {2

25, 1

81, 1

10, 1

18, 1

07} m

eet

this

cri

teri

on?

Exp

lain

. N

o,

the

ge

om

etr

ic m

ea

n i

s a

bo

ut

14

1.4

pe

r m

L.

4. T

RAV

EL I

f you

tra

vel a

t on

e sp

eed

for

half

the

dist

ance

of a

tri

p an

d yo

u tr

avel

at

a di

ffere

nt s

peed

for

the

othe

r ha

lf of

the

dis

tanc

e, t

hen

your

av

erag

e sp

eed

for

the

trip

is t

he h

arm

onic

mea

n of

the

spe

eds.

If G

reg

drov

e at

60

mph

for

half

a tr

ip a

nd 7

0 m

ph w

hen

the

spee

d lim

it in

crea

sed

for

the

seco

nd h

alf,

wha

t w

as h

is a

vera

ge s

peed

for

the

trip

? a

bo

ut

64

.6 m

ph

005_

042_

PC

CR

MC

11_8

9381

2.in

dd9

3/25

/09

12:3

6:57

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.


Answers (Lesson 11-1 and Lesson 11-2)

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

10

Gle

ncoe

Pre

calc

ulus

11-1

TI-N

spir

eTM A

ctiv

ity

Gra

ph

ing

Data

You

can

use

a T

I-N

spir

e to

gra

ph d

ata

sets

and

des

crib

e th

e sh

ape

of t

he d

istr

ibut

ion.

T

he le

ngth

s of

fis

h ca

ught

on

a on

e-da

yfi

shin

g tr

ip a

re s

how

n in

the

tab

le. M

ake

a hi

stog

ram

and

box

plot

of

the

data

.

Step

1:

Add

a L

ists

& S

prea

dshe

et p

age.

Lis

t th

e da

tain

col

umn

A a

nd t

itle

the

col

umn

leng

ths.

Step

2:

Wit

h th

e cu

rsor

in c

olum

n A

, pre

ss b

and

choo

se D

ata

> Q

uick

Gra

ph. P

ress

b a

ndch

oose

Plo

t T

ype

> H

isto

gram

. Pre

ss b

agai

n an

d ch

oose

Plo

t P

rope

rtie

s >

His

togr

am P

rope

rtie

s >

Bin

Set

ting

s to

cha

nge

the

wid

th o

f the

bar

s. P

ress

b a

nd c

hoos

eW

indo

w/Z

oom

> Z

oom

-Dat

a to

aut

omat

ical

ly s

ize

the

axes

to

see

all t

he d

ata.

You

can

cha

nge

the

vert

ical

cat

egor

y to

per

cent

by p

ress

ing b

and

cho

osin

g P

lot

Pro

pert

ies

>H

isto

gram

Pro

pert

ies

> H

isto

gram

Sca

le >

Per

cent

.

Step

3:

To c

hang

e th

e gr

aph

to a

box

plo

t, pr

ess b

and

choo

se P

lot

Typ

e >

Box

Plo

t. M

ove

the

curs

or o

ver

the

grap

h to

vie

w t

he fi

ve-n

umbe

r su

mm

ary.

Exer

cise

s

Ope

n a

new

Lis

ts &

Spr

eads

heet

pag

e. T

itle

col

umn

A a

s da

ta. U

se it

to

com

plet

e th

e fo

llow

ing.

1. M

ake

a hi

stog

ram

of t

he d

ata

set:

44.5

, 13.

7, 2

9.4,

22.

0, 3

2.5,

45.

8, 3

8.6,

24.

3,18

.1, a

nd 5

0.3.

Use

the

gra

ph t

o de

scri

be t

he s

hape

of t

he d

istr

ibut

ion.

s

ym

me

tric

2. C

hang

e th

e hi

stog

ram

in E

xerc

ise

1 to

a b

ox p

lot.

List

the

five

-num

ber

sum

mar

y.1

3.7

, 2

2,

30

.95

, 4

4.5

, 5

0.3

3. O

pen

a ne

w L

ists

& S

prea

dshe

et p

age.

Tit

le c

olum

n A

as

data

1 an

d co

lum

n B

as d

ata2

. Ent

er t

he d

ata

show

n be

low

. Use

Qui

ck G

raph

in e

ach

colu

mn

to m

ake

box

plot

s. D

escr

ibe

and

com

pare

the

dis

trib

utio

ns.

D

ata1

: 102

, 116

, 132

, 111

, 124

, 103

, 101

, 108

, 129

, 103

, 109

, 115

, 111

D

ata2

: 115

, 129

, 101

, 128

, 125

, 115

, 101

, 124

, 124

, 118

, 121

, 119

, 120

T

he

ra

ng

es

are

alm

os

t th

e s

am

e.

Ab

ou

t 7

5%

of

the

da

ta i

n d

ata

2 a

re

gre

ate

r th

an

th

e m

ed

ian

of

da

ta1.

Da

ta1 i

s p

os

itiv

ely

sk

ew

ed

an

d d

ata

2 i

s

ne

ga

tiv

ely

sk

ew

ed

.

Le

ng

ths

of

Fis

h (

in.)

14

16

18

18

14

21

26

23

35

15

32

30

92

21

2

Exam

ple

005_

042_

PC

CR

MC

11_8

9381

2.in

dd10

3/25

/09

12:3

7:00

PM


Lesson X-2

NA

ME

DA

TE

PE

RIO

D

Lesson 11-2

Ch

ap

ter

11

11

Gle

ncoe

Pre

calc

ulus

Prob

abili

ty D

istr

ibut

ion

A d

iscr

ete

rand

om v

aria

ble

X c

an t

ake

on a

coun

tabl

e nu

mbe

r of

val

ues.

A p

roba

bili

ty d

istr

ibut

ion

links

eac

h po

ssib

leva

lue

of X

wit

h it

s pr

obab

ility

of o

ccur

ring

.

Mea

n of

pro

babi

lity

dist

ribu

tion

of X

: μ =

Σ

[ X �

P(X

)]

Stan

dard

dev

iati

on o

f pro

babi

lity

dist

ribu

tion

of X

: σ =

√

��

��

��

�

Σ [(X

- μ

)2 � P

(X) ]

T

he n

umbe

r of

boo

ks b

ough

t by

eac

h of

100

ran

dom

ly

sele

cted

boo

ksto

re c

usto

mer

s du

ring

one

wee

k is

sho

wn.

a. C

onst

ruct

a p

roba

bili

ty d

istr

ibut

ion

for

X.

Th

ere

are

100

cust

omer

s, s

o P(

0) =

0.4

5, P

(1) =

0.3

0,

P(2)

= o

r 0.

15, a

nd P

(3) =

0.1

0.

b. F

ind

and

inte

rpre

t th

e m

ean

in t

he c

onte

xt o

f th

e pr

oble

m s

itua

tion

.F

ind

the

vari

ance

and

sta

ndar

d de

viat

ion.

O

rgan

ize

your

cal

cula

tion

s in

a t

able

.

Th

e m

ean

of t

he d

istr

ibut

ion

is 0

.9. O

n av

erag

e, e

ach

cust

omer

bou

ght

one

book

.Th

e va

rian

ce =

0.9

9, s

o th

e st

anda

rd d

evia

tion

is √

��

0.

99 o

r ab

out

0.99

5.

Exer

cise

1. T

he t

able

sho

ws

the

num

ber

of m

edic

al t

ests

tha

t15

ran

dom

ly s

elec

ted

pati

ents

ent

erin

g a

part

icul

arho

spit

al r

ecei

ved

one

day.

a. C

onst

ruct

a p

roba

bilit

y di

stri

buti

on fo

r X

.

b. F

ind

and

inte

rpre

t th

e m

ean

in t

he c

onte

xt o

f the

pro

blem

sit

uati

on. F

ind

the

vari

ance

and

sta

ndar

d de

viat

ion.

T

he

me

an

is

ab

ou

t 0

.93

, s

o,

on

av

era

ge

, e

ac

h p

ati

en

t re

ce

ive

d o

ne

te

st;

va

ria

nc

e ≈

0.8

6;

sta

nd

ard

de

via

tio

n

≈0

.93

11-2

Stud

y Gu

ide

and

Inte

rven

tion

Pro

bab

ilit

y D

istr

ibu

tio

ns

Exam

ple

Bo

ok

s,

XF

req

ue

nc

y

04

5

13

0

21

5

31

0B

oo

ks

, X

01

23

Fre

qu

en

cy

0.4

50

.30

0.1

50

.10

Bo

ok

s,

XP

(X)

X �

P(X

)(X

- μ

)2(X

- μ

)2 �

P(X

)

00

.45

0 �

0.4

5 =

0(0

- 0

.9)2

= 0

.81

0.8

1 �

0.4

5 =

0.3

64

5

10

.30

1 �

0.3

0 =

0.3

0(1

- 0

.9)2

= 0

.01

0.0

1 �

0.3

0 =

0.0

03

20

.15

2 �

0.1

5 =

0.3

0(2

- 0

.9)2

= 1

.21

1.2

1 �

0.1

5 =

0.1

81

5

30

.10

3 �

0.1

0 =

0.3

0(3

- 0

.9)2

= 4

.41

4.4

1 �

0.1

0 =

0.4

41

μ =

0.9

σ2 =

0.9

9

Te

sts

, X

Fre

qu

en

cy

06

15

23

31

Te

sts

, X

01

23

Fre

qu

en

cy

2

−

5

1

−

3

1

−

5

1

−

15

005_

042_

PC

CR

MC

11_8

9381

2.in

dd11

3/25

/09

12:3

7:05

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

12

Gle

ncoe

Pre

calc

ulus

Bino

mia

l Dis

trib

utio

n In

a b

inom

ial e

xper

imen

t, t

he o

utco

mes

are

suc

cess

or

failu

re. T

here

are

a fi

xed

num

ber

of in

depe

nden

t tr

ials

n a

nd t

he r

ando

m v

aria

ble

X

repr

esen

ts t

he n

umbe

r of

suc

cess

es. T

he p

roba

bilit

y of

suc

cess

p a

nd t

he p

roba

bilit

y of

fa

ilure

q o

r 1

- p

rem

ain

cons

tant

. The

pro

babi

lity

of X

suc

cess

es in

n in

depe

nden

t tr

ials

is

P(X

) = nC

x px q

n -

x =

n!

−

(n -

x)!x

! px q

n -

x .

A

sur

vey

foun

d th

at 2

0% o

f A

mer

ican

s ha

ve v

isit

ed a

doc

tor

in t

he

past

six

mon

ths.

Fiv

e pe

ople

wil

l be

sele

cted

at

rand

om a

nd a

sked

if t

hey

visi

ted

a do

ctor

in t

he p

ast

six

mon

ths.

Con

stru

ct a

nd g

raph

a b

inom

ial d

istr

ibut

ion

for

the

rand

om v

aria

ble

X, r

epre

sent

ing

the

num

ber

of p

eopl

e w

ho s

ay y

es. T

hen

find

the

pr

obab

ilit

y th

at a

t le

ast

four

of

thes

e pe

ople

say

yes

.

For

this

bin

omia

l exp

erim

ent,

n =

5, p

= 0

.2, a

nd q

= 1

- 0

.2 o

r 0.

8. C

ompu

te e

ach

poss

ible

va

lue

of X

usi

ng t

he B

inom

ial P

roba

bilit

y Fo

rmul

a.P(

0) =

5C0 �

0.2

0 � 0

.85 ≈

0.3

28

P(3)

= 5C

3 � 0

.23 �

0.8

2 ≈ 0

.051

P(1)

= 5C

1 � 0

.21 �

0.8

4 ≈ 0

.410

P(

4) =

5C4 �

0.2

4 � 0

.81 ≈

0.0

06P(

2) =

5C2 �

0.2

2 � 0

.83 ≈

0.2

05

P(5)

= 5C

5 � 0

.25 �

0.8

0 ≈ 0

.000

To fi

nd t

he p

roba

bilit

y th

at a

t le

ast

four

peo

ple

said

yes

, fin

d th

e su

m o

f P(4

) and

P(5

).P(

X ≥

4) =

P(4

) + P

(5) =

0.0

06 +

0.0

00 o

r 0.

6%

Exer

cise

A s

urve

y fo

und

that

60%

of A

mer

ican

vic

tim

s of

hea

lth-

care

frau

d w

ere

seni

or c

itiz

ens.

Six

vi

ctim

s of

hea

lth-

care

frau

d w

ill b

e ch

osen

at

rand

om a

nd t

heir

age

s w

ill b

e re

cord

ed.

Con

stru

ct a

nd g

raph

a b

inom

ial d

istr

ibut

ion

for

the

rand

om v

aria

ble

X, r

epre

sent

ing

the

num

ber

of s

enio

r ci

tize

ns c

hose

n. T

hen

find

the

prob

abili

ty t

hat

at le

ast

thre

e of

the

vic

tim

s w

ill b

e se

nior

cit

izen

s.

P(X

≥ 3

) =

0.8

21

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Pro

bab

ilit

y D

istr

ibu

tio

ns

11-2

Exam

ple

XP

(X)

00

.32

8

10

.41

0

20

.20

5

30

.05

1

40

.00

6

50

.00

0

00

12

34

5

0.1

0.2

0.3

0.4

0.5

Peop

le

Probability

P(x)

x

XP

(X)

XP

(X)

00

.00

44

0.3

11

10

.03

75

0.1

87

20

.13

86

0.0

47

30

.27

6

01

23

45

6

0.050.1

0.150.2

0.250.3

0.35

Seni

or C

itize

ns

Probability

0

P(x)

x

005_

042_

PC

CR

MC

11_8

9381

2.in

dd12

3/27

/09

7:14

:30

PM


Lesson X-2

NA

ME

DA

TE

PE

RIO

D

Lesson 11-2

Ch

ap

ter

11

13

Gle

ncoe

Pre

calc

ulus

Cla

ssif

y ea

ch r

ando

m v

aria

ble

X a

s di

scre

te o

r co

ntin

uous

. Exp

lain

you

r re

ason

ing.

1. X

rep

rese

nts

the

tim

e it

tak

es a

ran

dom

ly s

elec

ted

clas

sroo

m t

o re

ach

68°F

from

60°

F.

co

nti

nu

ou

s;

Th

e t

ime

ca

n b

e a

ny

nu

mb

er.

2. X

rep

rese

nts

the

num

ber

of p

hoto

grap

hs t

aken

by

a ph

otog

raph

er a

t a

rand

omly

sele

cted

wed

ding

. d

isc

rete

; T

he

nu

mb

er

of

ph

oto

gra

ph

s i

s c

ou

nta

ble

.

3. T

he t

able

sho

ws

the

num

ber

of c

ell p

hone

s ow

ned

by

100

rand

omly

sel

ecte

d ho

useh

olds

. Con

stru

ct a

nd g

raph

a

prob

abili

ty d

istr

ibut

ion

for

X. T

hen

find

and

inte

rpre

t th

e m

ean

in t

he c

onte

xt o

f the

pro

blem

sit

uati

on. F

ind

the

vari

ance

and

sta

ndar

d de

viat

ion.

4. R

ACE

A r

esor

t is

pla

nnin

g a

bicy

cle

race

. The

cos

t of

spo

nsor

ing

the

race

is$8

000.

The

res

ort

expe

cts

to m

ake

$15,

000

on t

he e

vent

. The

re is

a 3

0% c

hanc

eof

a h

urri

cane

arr

ivin

g th

e da

y of

the

rac

e. I

f thi

s ha

ppen

s, t

he r

ace

will

be

canc

elle

d an

d w

ill n

ot b

e re

sche

dule

d. W

hat

is t

he r

esor

t’s e

xpec

ted

prof

it?

$8

10

0

5. C

OM

MU

TE I

n a

rece

nt p

oll,

45%

of a

tow

n’s

citi

zens

sai

d th

ey u

se t

he b

us t

oge

t to

wor

k. F

ive

of t

hese

cit

izen

s w

ill b

e ra

ndom

ly c

hose

n an

d as

ked

if th

eyus

e th

e bu

s to

get

to

wor

k.

a. C

onst

ruct

a b

inom

ial d

istr

ibut

ion

for

the

rand

om v

aria

ble

X, r

epre

sent

ing

the

peop

le w

ho s

ay y

es.

b. F

ind

the

mea

n, v

aria

nce,

and

sta

ndar

d de

viat

ion

of t

his

dist

ribu

tion

. Int

erpr

etth

e m

ean

in t

he c

onte

xt o

f the

pro

blem

sit

uati

on.

m

ea

n:

2.2

5,

va

ria

nc

e:

1.2

37

5,

std

de

v:

ab

ou

t 1

.11

1;

On

av

era

ge

, 2

of

ev

ery

5 r

an

do

mly

se

lec

ted

pe

op

le i

n t

he

to

wn

wo

uld

sa

y t

he

y t

ak

e t

he

b

us

to

ge

t to

wo

rk.

11-2

Prac

tice

Pro

bab

ilit

y D

istr

ibu

tio

ns

Th

e m

ea

n i

s 1

.93

, s

o e

ac

h

ho

us

eh

old

ha

s a

bo

ut

two

c

ell

ph

on

es

; v

ari

an

ce

≈ 0

.79

; s

tan

da

rd d

ev

iati

on

≈ 0

.88

6

Ph

on

es

, X

01

23

4

Fre

qu

en

cy

0.0

20

.30

0.4

80

.13

0.0

7

Ph

on

es

, X

Fre

qu

en

cy

02

13

0

24

8

31

3

47

01

23

4

Cell

Phon

es

Probability

00.1

0.20.3

0.4

0.5

P(x)

x

X0

12

34

5

P(X

)0

.05

00

.20

60

.33

70

.27

60

.11

30

.01

8

005_

042_

PC

CR

MC

11_8

9381

2.in

dd13

11/1

4/09

4:53

:59

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

14

Gle

ncoe

Pre

calc

ulus

1. R

ETA

IL A

sto

re m

anag

er m

ade

the

prob

abili

ty d

istr

ibut

ion

show

n be

low

. It

sho

ws

the

prob

abili

ty o

f sel

ling

Xsw

imsu

its

on a

ran

dom

ly s

elec

ted

day

in J

une.

Find

the

mea

n, v

aria

nce,

and

sta

ndar

d de

viat

ion

of t

he d

istr

ibut

ion.

2

0.8

; 1

.6;

1.2

2. I

NSU

RAN

CE A

n in

sura

nce

com

pany

in

sure

s a

pain

ting

wor

th $

20,0

00 a

gain

st

thef

t fo

r $3

00 p

er y

ear.

The

com

pany

has

as

sess

ed t

he p

roba

bilit

y of

the

pai

ntin

g be

ing

stol

en in

a g

iven

yea

r as

0.0

02.

Wha

t is

the

insu

ranc

e co

mpa

ny’s

expe

cted

ann

ual p

rofit

?

$

25

9.4

0

3. R

ESTA

URA

NT

A s

urve

y fo

und

that

25%

of

all

part

ies

at a

res

taur

ant

wer

e gr

oups

of

five

or

larg

er. E

ight

een

part

ies

are

rand

omly

sel

ecte

d.

a. F

ind

the

prob

abili

ty t

hat

exac

tly

five

part

ies

are

mad

e up

of f

ive

or

mor

e pe

ople

.

19

.9%

b. F

ind

the

prob

abili

ty t

hat

5, 6

, or

7 pa

rtie

s ar

e m

ade

up o

f fiv

e or

mor

e pe

ople

.

4

2.4

%

4. P

ETS

Acc

ordi

ng t

o on

e po

ll, a

bout

63%

of

Am

eric

an h

ouse

hold

s in

clud

e at

leas

t on

e pe

t. Si

x ne

w h

omes

are

bui

lt a

nd s

old.

a. C

onst

ruct

a b

inom

ial d

istr

ibut

ion

for

the

rand

om v

aria

ble

X, r

epre

sent

ing

the

num

ber

of t

hese

hom

es t

hat

will

ha

ve a

t le

ast

one

pet.

b. F

ind

the

mea

n, v

aria

nce,

and

st

anda

rd d

evia

tion

of t

his

dist

ribu

tion

. 3

.78

; 1

.39

86

; a

bo

ut

1.1

8

c. F

ind

the

prob

abili

ty t

hat

at le

ast

half

of t

he n

ew h

omes

hav

e pe

ts.

Ab

ou

t 8

6%

5. T

ESTI

NG

Mr.

Han

lon

dist

ribu

ted

a 5-

ques

tion

mul

tipl

e ch

oice

qui

z to

his

st

uden

ts. T

here

wer

e 5

choi

ces

for

each

qu

esti

on. A

shle

y gu

esse

s th

e an

swer

on

each

que

stio

n.

a. W

hat

is A

shle

y’s

prob

abili

ty o

f gu

essi

ng e

xact

ly 3

que

stio

ns

corr

ectl

y?

5.1

2%

b. W

hat

wou

ld b

e th

e pr

obab

ility

inpa

rt a

if t

here

wer

e 4

choi

ces

for

each

que

stio

n?

8.8

%

c.

Wha

t w

ould

be

the

prob

abili

ty in

part

a if

the

qui

z co

ntai

ned

only

true

/fals

e qu

esti

ons?

3

1.2

5%

11-2

Wor

d Pr

oble

m P

ract

ice

Pro

bab

ilit

y D

istr

ibu

tio

ns

Sw

ims

uit

s,

X1

92

02

12

22

3

P(X

)0

.20

0.2

00

.30

0.2

00

.10

X0

12

3

P(X

)0

.00

30

.02

60

.11

20

.25

3

X4

56

P(X

)0

.32

30

.22

00

.06

3

005_

042_

PC

CR

MC

11_8

9381

2.in

dd14

11/1

4/09

4:56

:17

PM


Lesson X-2

NA

ME

DA

TE

PE

RIO

D

Lesson 11-2

Ch

ap

ter

11

15

Gle

ncoe

Pre

calc

ulus

The

Pois

son

dist

ribu

tion

is a

pro

babi

lity

dist

ribu

tion

for

an e

vent

occ

urri

ngx

tim

es o

ver

a gi

ven

inte

rval

. The

inte

rval

is u

sual

ly a

tim

e in

terv

al, s

uch

as

the

num

ber

of p

eopl

e en

teri

ng a

sto

re d

urin

g on

e ho

ur. T

he d

istr

ibut

ion

is a

di

scre

te d

istr

ibut

ion,

so

x m

ust

be a

who

le n

umbe

r.

The

prob

abili

ty o

f an

even

t oc

curr

ing

x ti

mes

ove

r a

give

n in

terv

al is

λx ·

e-

λ

−

x!

,

whe

re λ

is t

he a

vera

ge n

umbe

r of

tim

es t

he e

vent

occ

urs

duri

ng t

he in

terv

al.

A b

ank

man

ager

det

erm

ined

tha

t an

ave

rage

of 9

.2 c

usto

mer

s us

e a

cert

ain

ATM

eve

ry h

our.

You

can

cre

ate

a Po

isso

n di

stri

buti

on b

y us

ing

9.2

for

λ.

For

exam

ple,

the

pro

babi

lity

that

5 c

usto

mer

s us

e th

e A

TM in

a g

iven

hou

r

is g

iven

by

9.25 ·

e-

9.2

−

5!

. U

se y

our

calc

ulat

or t

o ve

rify

P(5

) ≈ 0

.055

or

abou

t 5.

5%.

On

a gr

aphi

ng c

alcu

lato

r, y

ou c

an a

lso

find

the

prob

abili

ty b

y us

ing

the

pois

sonp

df(

com

man

d fo

und

by p

ress

ing

[DIS

TR].

In t

he p

aren

thes

es, e

nter

λ

, x.

1. T

he b

ank

man

ager

det

erm

ined

tha

t an

ave

rage

of 1

.8 c

usto

mer

s in

quir

e ab

out

open

ing

a ne

w a

ccou

nt a

t a

cert

ain

bran

ch e

very

hou

r. C

ompl

ete

the

tabl

e.

2. N

otic

e th

at u

nlik

e a

bino

mia

l dis

trib

utio

n, x

has

no

uppe

r lim

it. T

he t

able

in

Exe

rcis

e 2

stop

ped

at x

= 5

, but

it is

pos

sibl

e fo

r 6

or m

ore

peop

le t

o in

quir

e ab

out

open

ing

a ne

w a

ccou

nt. H

ow c

an y

ou fi

nd P

(x >

2)?

S

ub

tra

ct

P(0

) +

P(1

) +

P(2

) fr

om

1.

3. O

n av

erag

e, t

he b

ank

rece

ives

4.2

onl

ine

loan

app

licat

ions

per

day

. Fin

d ea

ch p

roba

bilit

y.

a. T

he b

ank

rece

ives

3 o

nlin

e lo

an a

pplic

atio

ns o

ne d

ay.

18

.5%

b. T

he b

ank

rece

ives

4 o

nlin

e lo

an a

pplic

atio

ns o

ne d

ay.

19

.4%

c. T

he b

ank

rece

ives

5 o

nlin

e lo

an a

pplic

atio

ns o

ne d

ay.

16

.3%

d. T

he b

ank

rece

ives

6 o

nlin

e lo

an a

pplic

atio

ns o

ne d

ay.

11

.4%

e. T

he b

ank

rece

ives

7 o

nlin

e lo

an a

pplic

atio

ns o

ne d

ay.

6.9

%

4. M

ake

a co

njec

ture

abo

ut t

he g

ener

al s

hape

of a

ny P

oiss

on d

istr

ibut

ion.

It

is

po

sit

ive

ly s

ke

we

d.

Th

e p

ea

k i

s a

t λ

.

11-2

Enri

chm

ent

Th

e P

ois

son

Dis

trib

uti

on

X0

12

34

5

P(X

)0

.16

50

.29

80

.26

80

.16

10

.07

20

.02

6

005_

042_

PC

CR

MC

11_8

9381

2.in

dd15

3/25

/09

12:3

7:28

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


An

swer

s

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

16

Gle

ncoe

Pre

calc

ulus

The

Nor

mal

Dis

trib

utio

n A

nor

mal

dis

trib

utio

n is

a c

onti

nuou

s pr

obab

ility

dis

trib

utio

n. T

he g

raph

of a

nor

mal

dis

trib

utio

n is

sym

met

ric

and

bell-

shap

ed. I

t ap

proa

ches

but

nev

er t

ouch

es t

he x

-axi

s. I

t in

clud

es 1

00%

of

the

data

, so

the

area

und

er t

he c

urve

is 1

. The

z-v

alue

rep

rese

nts

the

num

ber

of s

tand

ard

devi

atio

ns t

hat

a gi

ven

data

val

ue is

from

the

mea

n.

z =

X -

μ

−

σ

, w

here

X is

the

dat

a va

lue,

μ is

the

mea

n, a

nd σ

is t

he s

tand

ard

devi

atio

n.

The

stan

dard

nor

mal

dis

trib

utio

n ha

s a

mea

n of

0 a

nd a

sta

ndar

d de

viat

ion

of 1

.

On

his

last

20

airl

ine

trip

s, a

n em

ploy

ee h

ad a

n av

erag

e la

yove

r of

82

min

utes

wit

h a

stan

dard

dev

iati

on o

f7.

5 m

inut

es. F

ind

the

num

ber

of la

yove

rs t

hat

wer

e le

ss t

han

75 m

inut

es.

Firs

t, fin

d th

e z-

valu

e.

z =

X -

μ

−

σ

F

orm

ula

fo

r z

-va

lue

s

=

75 -

82

−

7.5

or

abou

t -

0.93

X

= 7

5,

μ =

82

, a

nd

σ =

7.5

Use

a g

raph

ing

calc

ulat

or t

o fin

d th

e ar

ea u

nder

the

cur

ve t

hat

is t

o th

e le

ft o

f 75.

Pre

ss

2n

d [D

ISTR

] and

cho

ose

norm

alcd

f(.

Ent

er t

he lo

wer

val

ue (y

ou c

an u

se -

4 in

stea

d of

neg

ativ

e in

finit

y) a

nd t

he u

pper

val

ue a

s -

0.93

. The

res

ulti

ng a

rea

is 0

.176

. Th

is m

eans

tha

t ab

out

17.6

% o

f the

dat

a va

lues

are

less

tha

n-

0.93

sta

ndar

d de

viat

ions

from

the

mea

n.

Bec

ause

the

re a

re 2

0 fli

ghts

, abo

ut 2

0 � 0

.176

or

abou

t 4

fligh

ts h

ad la

yove

r ti

mes

tha

t w

ere

less

tha

n 75

min

utes

.

Exer

cise

s

1. A

t a

rest

aura

nt, t

he a

vera

ge t

ime

betw

een

whe

n an

ord

er is

pla

ced

and

whe

n th

e en

tree

is s

erve

d is

12.

5 m

inut

es w

ith

a st

anda

rd d

evia

tion

of

1.2

min

utes

. Out

of 1

00 r

ando

mly

sel

ecte

d cu

stom

ers,

how

man

y w

ill b

e se

rved

the

ir e

ntre

es w

ithi

n 14

min

utes

of o

rder

ing?

ap

pro

xim

ate

ly 8

9 c

us

tom

ers

2. M

rs. Q

uan,

a fu

ll pr

ofes

sor

at a

com

mun

ity

colle

ge, e

arns

a s

alar

y of

$4

8,60

0. T

he a

vera

ge s

alar

y fo

r a

full

prof

esso

r at

the

col

lege

is $

52,0

00

wit

h a

stan

dard

dev

iati

on o

f $36

00. H

ow m

any

of t

he 4

5 fu

ll pr

ofes

sors

ea

rn le

ss t

han

Mrs

. Qua

n?

ap

pro

xim

ate

ly 8

pro

fes

so

rs

3. D

urin

g on

e O

ctob

er, t

he a

vera

ge w

ater

tem

pera

ture

of a

pon

d w

as 5

3.2°

w

ith

a st

anda

rd d

evia

tion

of 2

.3°. H

ow m

any

days

was

the

tem

pera

ture

gr

eate

r th

an 5

0°?

ap

pro

xim

ate

ly 2

8 d

ay

s

Stud

y Gu

ide

and

Inte

rven

tion

Th

e N

orm

al

Dis

trib

uti

on

11-3

Exam

ple

005_

042_

PC

CR

MC

11_8

9381

2.in

dd16

3/25

/09

12:3

7:32

PM


Lesson X-3

NA

ME

DA

TE

PE

RIO

D

Lesson 11-3

Ch

ap

ter

11

17

Gle

ncoe

Pre

calc

ulus

Prob

abili

ty a

nd t

he N

orm

al D

istr

ibut

ion

The

area

und

er t

he

norm

al c

urve

cor

resp

onds

to

the

prob

abili

ty o

f dat

a va

lues

falli

ng w

ithi

n a

give

n in

terv

al.

T

he a

vera

ge m

onth

ly t

empe

ratu

res

for

a ci

ty f

or o

ne

year

wer

e no

rmal

ly d

istr

ibut

ed w

ith

μ =

65°

and

σ =

5. F

ind

P(5

0 <

X <

70)

. Use

a g

raph

ing

calc

ulat

or t

o sk

etch

the

co

rres

pond

ing

area

und

er t

he c

urve

.

Find

the

z-v

alue

s fo

r X

= 5

0 an

d X

= 7

0.

z =

X

- μ

−

σ

z

= X

- μ

−

σ

=

50

- 6

5 −

5 =

-3

= 70

- 6

5 −

5 =

1

Find

the

are

a be

twee

n z

= -

3 an

d z

= 1

. On

your

cal

cula

tor,

pre

ss

2n

d

[DIS

TR] a

nd c

hoos

e Sh

adeN

orm

und

er t

he D

RA

W m

enu.

Ent

er t

he lo

wer

an

d up

per

z-va

lues

and

pre

ss e

nter

.

P(50

< X

< 7

0) ≈

84%

Ther

efor

e, a

ppro

xim

atel

y 84

% o

f the

tem

pera

ture

s w

ere

betw

een

50°

and

70°.

Exer

cise

s 1.

The

ave

rage

age

of t

he s

wim

mer

s on

a m

aste

r sw

im t

eam

is n

orm

ally

di

stri

bute

d w

ith

μ =

56

and

σ =

4. F

ind

each

pro

babi

lity.

Use

a g

raph

ing

calc

ulat

or t

o sk

etch

the

cor

resp

ondi

ng a

rea

unde

r th

e cu

rve.

a. P

(53

< X

< 5

9)

54

.7%

b. P

(X <

53)

2

2.7

%

2. S

tude

nts

who

sco

re in

the

bot

tom

5%

of a

phy

sica

l edu

cati

on t

est

will

be

enro

lled

in a

sup

plem

enta

l phy

sica

l edu

cati

on p

rogr

am. T

he s

core

s of

all

of t

he s

tude

nts

who

too

k th

e te

st a

re n

orm

ally

dis

trib

uted

wit

h μ

= 12

2.6

and

σ =

18.

Wha

t is

the

gre

ates

t sc

ore

that

a s

tude

nt w

ho e

nrol

led

in t

he

supp

lem

enta

l pro

gram

cou

ld h

ave

rece

ived

? 9

2

11-3

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Th

e N

orm

al

Dis

trib

uti

on

Exam

ple

[ –4,

4] s

cl: 1

by

[ 0, 0

.5] s

cl: 0

.125

[ –4,

4] s

cl: 1

by

[ 0, 0

.5] s

cl: 0

.125

005_

042_

PC

CR

MC

11_8

9381

2.in

dd17

11/1

9/09

6:40

:36

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

18

Gle

ncoe

Pre

calc

ulus

1. T

REES

The

hei

ghts

of 2

00 t

rees

in a

nur

sery

are

nor

mal

ly d

istr

ibut

ed

wit

h a

mea

n of

120

inch

es a

nd a

sta

ndar

d de

viat

ion

of 1

6 in

ches

.

a. A

ppro

xim

atel

y ho

w m

any

tree

s ar

e m

ore

than

136

inch

es t

all?

3

2 t

ree

s

b. W

hat

perc

ent

of t

he t

rees

are

bet

wee

n 88

inch

es a

nd 1

04 in

ches

tal

l?

13

.6%

Fin

d ea

ch o

f th

e fo

llow

ing.

2. z

if X

= 6

5, μ

= 5

0, a

nd σ

= 1

0 3.

X if

z =

-0.

4, μ

= 4

0, a

nd σ

= 5

1

.5

3

8

Fin

d th

e in

terv

al o

f z-

valu

es a

ssoc

iate

d w

ith

each

are

a.

4. t

he m

iddl

e 60

% o

f the

dat

a -

0.8

4 <

z <

0.8

4

5. t

he o

utsi

de 3

0% o

f the

dat

a z

< -

1.0

4 a

nd

z >

1.0

4

6. D

OG

S Th

e w

eigh

ts o

f the

42

full-

grow

n G

erm

an s

heph

erds

at

a ke

nnel

are

nor

mal

ly

dist

ribu

ted.

The

mea

n w

eigh

t is

86

poun

ds a

nd t

he s

tand

ard

devi

atio

n is

3 p

ound

s.

a. D

eter

min

e th

e nu

mbe

r of

Ger

man

she

pher

ds t

hat

wei

gh m

ore

than

82

poun

ds.

ab

ou

t 3

8 d

og

s

b. H

ow m

any

Ger

man

she

pher

ds w

eigh

less

tha

n 88

pou

nds?

a

bo

ut

31

do

gs

7. H

OTE

LS T

he p

rice

s of

roo

ms

at h

otel

s ar

ound

an

airp

ort

are

norm

ally

di

stri

bute

d w

ith

μ =

$12

0 an

d σ

= $

20. F

ind

each

pro

babi

lity.

a. T

he c

ost

of a

roo

m is

gre

ater

tha

n $1

50.

6

.7%

b. T

he c

ost

of a

roo

m is

bet

wee

n $1

10 a

nd $

130.

3

8.3

%

c. T

he c

ost

of a

roo

m is

bet

wee

n $9

0 an

d $1

00.

9

.2%

d. I

f onl

y th

e m

ost

expe

nsiv

e 10

% o

f the

roo

ms

are

avai

labl

e, w

hat

is t

he

leas

t am

ount

you

will

pay

for

a ro

om?

$

14

6

8. E

XA

MS

A s

tude

nt s

core

d 65

on

a bi

olog

y ex

am w

ith

μ =

50

and

σ =

10.

Sh

e sc

ored

30

on a

lite

ratu

re e

xam

wit

h μ

= 2

5 an

d σ

= 5

. Com

pare

her

sc

ores

on

each

tes

t. A

ssum

e th

at b

oth

sets

of s

core

s w

ere

norm

ally

di

stri

bute

d. S

he

did

be

tte

r o

n h

er

bio

log

y e

xa

m.

Th

e z

-sc

ore

fo

r th

e b

iolo

gy

ex

am

wa

s 1

.5 a

nd

th

e z

-sc

ore

fo

r th

e

lite

ratu

re e

xa

m w

as

1.

11-3

Prac

tice

Th

e N

orm

al

Dis

trib

uti

on

005_

042_

PC

CR

MC

11_8

9381

2.in

dd18

3/25

/09

12:3

7:44

PM


Lesson X-3

NA

ME

DA

TE

PE

RIO

D

Lesson 11-3

Ch

ap

ter

11

19

Gle

ncoe

Pre

calc

ulus

1. R

EAL

ESTA

TE T

he a

vera

ge p

rice

of a

on

e-be

droo

m c

ondo

min

ium

list

ed b

y a

real

tor

is $

145,

500

wit

h a

stan

dard

de

viat

ion

of $

1500

. The

pri

ces

are

norm

ally

dis

trib

uted

. Det

erm

ine

the

prob

abili

ty t

hat

a ra

ndom

ly s

elec

ted

cond

omin

ium

cos

ts b

etw

een

$143

,580

an

d $1

47,4

20.

8

0%

2. C

OM

MU

TIN

G T

he a

vera

ge t

imes

spe

nt

com

mut

ing

to w

ork

in a

cer

tain

cit

y ar

e no

rmal

ly d

istr

ibut

ed w

ith

a m

ean

of25

.5 m

inut

es a

nd a

sta

ndar

d de

viat

ion

of 6

.1 m

inut

es. W

hat

is t

he p

roba

bilit

y th

at a

ran

dom

ly s

elec

ted

com

mut

e to

w

ork

take

s lo

nger

tha

n a

half

hour

?

2

3.0

%

3. S

IGH

TSEE

ING

The

tim

es p

eopl

e sp

end

view

ing

cert

ain

anci

ent

ruin

s ar

e no

rmal

ly d

istr

ibut

ed w

ith

a m

ean

of

96 m

inut

es w

ith

a st

anda

rd d

evia

tion

of

17 m

inut

es.

a. F

ind

the

prob

abili

ty t

hat

a si

ghts

eer

will

spe

nd a

t le

ast

two

hour

s at

the

ru

ins.

7

.9%

b. F

ind

the

prob

abili

ty t

hat

a si

ghts

eer

will

spe

nd a

t m

ost

80 m

inut

es a

t th

e ru

ins.

1

7.3

%

c.

If a

tou

r bu

s dr

ops

off a

gro

up o

f si

ghts

eers

at

9 A.M

., w

hat

tim

e sh

ould

th

e bu

s pi

ck u

p th

e si

ghts

eers

? E

xpla

in.

S

am

ple

an

sw

er:

Ab

ou

t 9

5%

of

the

sig

hts

ee

rs w

ill

sp

en

d l

es

s

tha

n 1

24

min

ute

s a

t th

e r

uin

s,

so

th

e b

us

sh

ou

ld r

etu

rn a

t 1

1:0

5 A

.M.

4. T

RAIN

ING

To

qual

ify fo

r a

secu

rity

po

siti

on, c

andi

date

s m

ust

take

a p

hysi

cal

fitne

ss t

est.

The

scor

es o

n th

e te

st a

re

norm

ally

dis

trib

uted

wit

h a

mea

n of

400

an

d a

stan

dard

dev

iati

on o

f 100

.

a. C

andi

date

s sc

orin

g in

the

top

3%

are

la

ter

recr

uite

d as

tra

iner

s in

the

pr

ogra

m. W

hat

is t

he m

inim

um s

core

a

cand

idat

e ne

eds

in o

rder

to

be

recr

uite

d la

ter

as a

tra

iner

?

5

89

b. C

andi

date

s sc

orin

g in

the

bot

tom

1.5

%

mus

t re

take

the

phy

sica

l tra

inin

g pr

ogra

m. W

hat

is t

he m

inim

um s

core

a

cand

idat

e w

ould

nee

d to

avo

id

reta

king

the

tra

inin

g pr

ogra

m?

1

82

5. J

UIC

E Th

e am

ount

of j

uice

pou

red

into

bot

tles

in a

fact

ory

is n

orm

ally

di

stri

bute

d w

ith

a m

ean

of 1

6 ou

nces

and

a st

anda

rd d

evia

tion

of 0

.3 o

unce

. A

shi

pmen

t co

ntai

ns 2

80 b

ottl

es.

a. H

ow m

any

bott

les

are

expe

cted

to

cont

ain

mor

e th

an 1

6.5

ounc

es o

f ju

ice?

a

bo

ut

13

bo

ttle

s

b. H

ow m

any

bott

les

are

expe

cted

to

cont

ain

less

tha

n 15

.75

ounc

es o

f ju

ice?

a

bo

ut

57

bo

ttle

s

11-3

Wor

d Pr

oble

m P

ract

ice

Th

e N

orm

al

Dis

trib

uti

on

005_

042_

PC

CR

MC

11_8

9381

2.in

dd19

11/1

4/09

4:59

:36

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

20

Gle

ncoe

Pre

calc

ulus

Bef

ore

grap

hing

cal

cula

tors

, stu

dent

s ha

d to

find

the

are

a un

der

a no

rmal

dis

trib

utio

n cu

rve

by u

sing

a t

able

of v

alue

s. Y

ou c

an s

till

find

thes

e ta

bles

in m

any

stat

isti

cs b

ooks

.It

is g

ood

to k

now

how

to

use

them

, in

case

you

find

you

rsel

f in

a si

tuat

ion

wit

hout

a

grap

hing

cal

cula

tor.

Her

e is

par

t of

a s

tand

ard

norm

al p

roba

bilit

y ta

ble.

It

give

s th

e ar

ea t

hat

is b

etw

een

the

mea

n, 0

, and

the

z-s

core

. To

find

the

area

bet

wee

n a

z-sc

ore

of 0

and

a z

-sco

re o

f 1.3

5, fi

nd

1.3

in t

he v

erti

cal c

olum

n an

d 0.

05 in

the

top

row

. It

is 0

.411

5.

To fi

nd t

he a

rea

that

is t

o th

e le

ft o

f a c

erta

in p

osit

ive

z-sc

ore,

rem

embe

r to

add

the

are

ath

at is

to

the

left

of t

he m

ean,

0.5

.

Bec

ause

a n

orm

al d

istr

ibut

ion

is s

ymm

etri

c ab

out

the

mea

n, y

ou c

an u

se t

he s

ame

char

t to

find

the

are

a be

twee

n a

nega

tive

z-s

core

and

the

mea

n. F

or e

xam

ple,

the

area

bet

wee

n a

z-sc

ore

of -

1.35

and

0 is

the

sam

e as

bet

wee

n 0

and

1.35

: 0.4

115.

Use

the

tab

le t

o an

swer

eac

h qu

esti

on.

1. W

hat

is t

he a

rea

betw

een

a z-

scor

e of

0 a

nd a

z-s

core

of 1

.02?

0

.34

61

2. W

hat

is t

he a

rea

betw

een

a z-

scor

e of

1.1

8 an

d a

z-sc

ore

of 1

.43?

Exp

lain

how

you

foun

d it

. 0

.04

26

: S

ub

tra

ct

the

are

a b

etw

ee

n 0

an

d 1

.18

fro

m t

he

are

a b

etw

ee

n 0

an

d 1

.43

: 0

.42

36

- 0

.38

10

= 0

.04

26

.

3. F

ind

the

area

to

the

righ

t of

a z

-sco

re o

f 1.1

7. E

xpla

in h

ow y

ou fo

und

it.

0.1

21

; S

ub

tra

ct

the

are

a i

n t

he

ch

art

fro

m 0

.5:

0.5

- 0

.37

90

= 0

.12

1.

4. W

hat

is t

he a

rea

to t

he le

ft o

f a z

-sco

re o

f 1.4

1?

0.9

20

7

5. W

hat

is t

he a

rea

betw

een

a z-

scor

e of

-1.

25 a

nd a

z-s

core

of 0

? 0

.39

44

6. W

hat

is t

he a

rea

to t

he le

ft o

f a z

-sco

re o

f -1.

33?

0.0

91

8

7. W

hat

is t

he a

rea

betw

een

a z-

scor

e of

-1.

05 a

nd a

z-s

core

of 1

.38?

0

.76

93

11-3

Enri

chm

ent

Fin

din

g A

reas

by

Usi

ng

a T

ab

le

z.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9

1.0

.34

13

.34

38

.34

61

.34

85

.35

08

.35

31

.35

54

.35

77

.35

99

.36

21

1.1

.36

43

.36

65

.36

86

.37

08

.37

29

.37

49

.37

70

.37

90

.38

10

.38

30

1.2

.38

49

.38

69

.38

88

.39

07

.39

25

.39

44

.39

62

.39

80

.39

97

.40

15

1.3

.40

32

.40

49

.40

66

.40

82

.40

99

.41

15

.41

31

.41

47

.41

62

.41

77

1.4

.41

92

.42

07

.42

22

.42

36

.42

51

.42

65

.42

79

.42

92

.43

06

.43

19

005_

042_

PC

CR

MC

11_8

9381

2.in

dd20

11/1

9/09

6:41

:03

PM


Lesson X-3

NA

ME

DA

TE

PE

RIO

D

Lesson 11-3

Ch

ap

ter

11

21

Gle

ncoe

Pre

calc

ulus

The

grap

h sh

own

at t

he r

ight

is k

now

n as

the

sta

ndar

d no

rmal

curv

e. T

he s

tand

ard

norm

al c

urve

is t

he g

raph

of f

(x) =

1

−

√

��

2π

e

-

x2 −

2 .

You

can

use

a g

raph

ing

calc

ulat

or t

o in

vest

igat

e pr

oper

ties

of

this

func

tion

and

its

grap

h. E

nter

the

func

tion

for

the

norm

al

curv

e in

the

Y =

list

of a

gra

phin

g ca

lcul

ator

.

1. T

he s

tand

ard

norm

al c

urve

mod

els

a pr

obab

ility

dis

trib

utio

n. A

s a

resu

lt, p

roba

bilit

ies

for

inte

rval

s of

x-v

alue

s ar

e eq

ual t

o ar

eas

of

regi

ons

boun

ded

by t

he c

urve

, the

x-a

xis,

and

the

ver

tica

l lin

es

thro

ugh

the

endp

oint

s of

the

inte

rval

s. T

he c

alcu

lato

r ca

n ap

prox

imat

e th

e ar

eas

of s

uch

regi

ons.

To

find

the

area

of t

he r

egio

n bo

unde

d by

the

cur

ve, t

he x

-axi

s, a

nd t

he v

erti

cal l

ines

x =

-1

and

x =

1, g

o to

the

CA

LC

men

u an

d se

lect

7: ∫

f(x)

dx. M

ove

the

curs

or t

o

the

poin

t w

here

x =

-1.

Pre

ss E

NT

ER

. The

n m

ove

the

curs

or t

o th

e

poin

t w

here

x =

1 a

nd p

ress

EN

TE

R. T

he c

alcu

lato

r w

ill s

hade

the

re

gion

and

dis

play

its

appr

oxim

ate

area

. Wha

t nu

mbe

r do

es t

he

calc

ulat

or d

ispl

ay fo

r th

e ar

ea o

f the

sha

ded

regi

on?

0.6

82

68

94

9

2. E

nter

2

nd

[DR

AW

] 1. T

his

caus

es t

he c

alcu

lato

r to

cle

ar t

he s

hadi

ng a

nd

redi

spla

y th

e gr

aph.

Fin

d th

e ar

ea o

f the

reg

ion

boun

ded

by t

he c

urve

, th

e x-

axis

, and

the

ver

tica

l lin

es x

= -

2 an

d x

= 2

. 0.9

5449974

3. F

ind

the

area

of t

he r

egio

n bo

unde

d by

the

cur

ve, t

he x

-axi

s, a

nd t

he v

erti

cal

lines

x =

-3

and

x =

3.

0.9

97

30

02

4. W

itho

ut u

sing

a c

alcu

lato

r, e

stim

ate

the

area

of t

he r

egio

n bo

unde

d by

the

cu

rve,

the

x-a

xis,

and

the

ver

tica

l lin

es x

= -

4 an

d x

= 4

to

four

dec

imal

pl

aces

. Ver

ify y

our

conj

ectu

re.

Sa

mp

le a

ns

we

r: 1

; s

ee

stu

de

nts

’ w

ork

.

11-3

Grap

hing

Cal

cula

tor

Activ

ityTh

e S

tan

dard

No

rmal

Cu

rve

[ -4.

7, 4

.7] s

cl: 1

by

[ -0.

2, 0

.5] s

cl: 0

.1

005_

042_

PC

CR

MC

11_8

9381

2.in

dd21

11/1

4/09

5:03

:38

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

22

Gle

ncoe

Pre

calc

ulus

11-4

Stud

y Gu

ide

and

Inte

rven

tion

Th

e C

en

tral

Lim

it T

heo

rem

The

Cent

ral L

imit

The

orem

The

Cen

tral

Lim

it T

heor

em s

tate

s th

at a

s th

e sa

mpl

ing

size

n in

crea

ses:

•

the

shap

e of

the

dis

trib

utio

n of

the

sam

ple

mea

ns o

f a p

opul

atio

n w

ith

mea

n μ

and

stan

dard

dev

iati

on σ

will

app

roac

h a

norm

al d

istr

ibut

ion

and

•

the

corr

espo

ndin

g di

stri

buti

on w

ill h

ave

a m

ean

μ a

nd s

tand

ard

devi

atio

n σ

−

x =

σ

−

√

�

n

.

The

z-va

lue

for

a sa

mpl

e m

ean

in a

pop

ulat

ion

is g

iven

by

z =

−

x -

μ

−

σ −

x , w

here

−

x is

the

sam

ple

mea

n, μ

is t

he m

ean

of t

he p

opul

atio

n, a

nd σ

−

x is

the

stan

dard

err

or.

A

stu

dy w

as d

one

in w

hich

par

ents

rep

orte

d th

e nu

mbe

r of

hou

rs p

er d

ay t

hat

thei

r ch

ildr

en, b

etw

een

the

ages

of

2.0

and

5.9,

wat

ched

tel

evis

ion

or v

ideo

s. T

he m

ean

age

of t

he c

hild

ren

was

3.3

yea

rs w

ith

a st

anda

rd d

evia

tion

of

0.9

year

. Ass

ume

that

the

va

riab

le is

nor

mal

ly d

istr

ibut

ed. I

f a

rand

om s

ampl

e of

30

chil

dren

in

thi

s st

udy

is s

elec

ted,

fin

d th

e pr

obab

ilit

y th

at t

he m

ean

age

is

less

tha

n 3

year

s.

The

dist

ribu

tion

of s

ampl

e m

eans

will

be

appr

oxim

atel

y no

rmal

wit

h μ

= 3

.3 a

nd

σ

−

x =

0.9

−

√

��

30

or

abou

t 0.

164.

Fin

d th

e z-

valu

e.

z =

−

x -

μ

−

σ −

x

z-va

lue

fo

r a

sa

mp

le m

ea

n

z =

3 -

3.3

−

0.16

4 or

abo

ut -

1.83

−

x =

3,

μ =

3.3

, σ

−

x =

0.1

64

The

area

to

the

left

of a

z-v

alue

of -

1.83

is 0

.033

59.

The

prob

abili

ty t

hat

the

mea

n ag

e of

the

sam

ple

is le

ss t

han

3 ye

ars

or P

( −

x <

3) i

s ab

out

3.36

%.

Exer

cise

s 1.

The

mea

n pi

tch

of t

he e

xper

t sl

opes

at

the

ski r

esor

ts in

a c

erta

in r

egio

n is

25˚

wit

h a

stan

dard

dev

iati

on o

f 3˚. A

ssum

e th

at t

he v

aria

ble

is

norm

ally

dis

trib

uted

. If 2

0 ex

pert

slo

pes

are

chos

en a

t ra

ndom

, wha

t is

th

e pr

obab

ility

tha

t th

e m

ean

of t

he s

lope

s w

ill b

e gr

eate

r th

an

26.3

˚?

2.6

2%

2. A

vet

erin

aria

n re

port

s th

at t

he a

vera

ge a

ge o

f the

cat

s th

at s

he t

reat

s is

96

mon

ths

wit

h a

stan

dard

dev

iati

on o

f 16

mon

ths.

If a

ran

dom

sam

ple

of

36 o

f her

cat

pat

ient

s is

sel

ecte

d, fi

nd t

he p

roba

bilit

y th

at t

he m

ean

age

is

betw

een

90 a

nd 1

00 m

onth

s. 9

2.1

%

Exam

ple

005_

042_

PC

CR

MC

11_8

9381

2.in

dd22

11/1

4/09

5:06

:47

PM


Lesson X-4

NA

ME

DA

TE

PE

RIO

D

Lesson 11-4

Ch

ap

ter

11

23

Gle

ncoe

Pre

calc

ulus

11-4

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Th

e C

en

tral

Lim

it T

heo

rem

The

Nor

mal

App

roxi

mat

ion

The

norm

al d

istr

ibut

ion

can

be u

sed

to a

ppro

xim

ate

a bi

nom

ial d

istr

ibut

ion

if:

•

the

orig

inal

var

iabl

e is

nor

mal

ly d

istr

ibut

ed o

r n

≥ 3

0 an

d

•

np ≥

5 a

nd n

q ≥

5,

whe

re n

is t

he n

umbe

r of

tri

als,

p is

the

pro

babi

lity

of s

ucce

ss, a

nd q

is t

hepr

obab

ility

of f

ailu

re.

Whe

n us

ing

the

norm

al d

istr

ibut

ion

to a

ppro

xim

ate

a bi

nom

ial d

istr

ibut

ion,

the

cont

inui

ty c

orre

ctio

n fa

ctor

mus

t be

use

d. T

his

invo

lves

add

ing

0.5

unit

to

or

subt

ract

ing

0.5

unit

from

a g

iven

dis

cret

e bo

unda

ry.

T

en p

erce

nt o

f th

e m

embe

rs o

f a

golf

leag

ue a

re y

oung

erth

an 3

0. I

f 20

0 m

embe

rs a

re r

ando

mly

sel

ecte

d, f

ind

the

prob

abil

ity

that

mor

e th

an 1

0 w

ill b

e yo

unge

r th

an 3

0.

Step

1

Find

the

mea

n μ

and

sta

ndar

d de

viat

ion

σ o

f the

bin

omia

l di

stri

buti

on.

μ =

np

σ =

√

��

np

q

=

200

· 0.

10 o

r 20

= √

��

��

��

�

200

· 0.

10 ·

0.90

or

abou

t 4.

24

Si

nce

np =

20

and

nq =

180

, the

nor

mal

dis

trib

utio

n ca

n be

use

d.

Step

2

Wri

te t

he p

robl

em in

pro

babi

lity

nota

tion

: P ( X

> 1

0) .

Step

3

Rew

rite

the

pro

blem

wit

h th

e co

ntin

uity

fact

or in

clud

ed. S

ince

the

qu

esti

on is

ask

ing

for

the

prob

abili

ty t

hat

mor

e th

an 1

0 m

embe

rs

will

be

youn

ger

than

30,

add

0.5

to

10.

P (

X >

10)

= P

( X >

10

+ 0

.5)

or

P (X

> 1

0.5)

Step

4

Find

the

cor

resp

ondi

ng z

-val

ue fo

r X

= 1

0.5.

z =

X -

μ

−

σ

= 10

.5 -

20

−

4.24

≈ -

2.24

Step

5

Find

the

cor

resp

ondi

ng a

rea

abov

e z

= -

2.24

. It

isab

out

0.98

7. T

he p

roba

bilit

y is

abo

ut 9

8.7%

.

Exer

cise

s 1.

A s

tudy

foun

d th

at 4

0% o

f all

of a

tow

n’s

citi

zens

app

rove

of a

new

tra

in s

tati

on.

If 5

0 ci

tize

ns a

re r

ando

mly

sel

ecte

d, fi

nd t

he p

roba

bilit

y th

at fe

wer

tha

n 20

cit

izen

s ap

prov

e of

a n

ew t

rain

sta

tion

. 4

4.2

%

2. A

bas

ebal

l pla

yer

gets

a h

it 3

2% o

f the

tim

e. F

ind

the

prob

abili

ty t

hat

the

play

er w

ill g

et a

t le

ast

26 h

its

in h

is n

ext

100

tim

es a

t ba

t. 9

1.8

%

Exam

ple

005_

042_

PC

CR

MC

11_8

9381

2.in

dd23

11/1

4/09

5:07

:35

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

24

Gle

ncoe

Pre

calc

ulus

11-4

Prac

tice

Th

e C

en

tral

Lim

it T

heo

rem

1. N

URS

ING

The

mea

n sa

lary

for

nurs

es in

a c

ity

is $

52,1

29 w

ith

a st

anda

rd d

evia

tion

of $

1800

. Wha

t is

the

pro

babi

lity

that

the

mea

n sa

lary

for

a ra

ndom

ly s

elec

ted

grou

p of

50

nurs

es in

the

cit

y is

gre

ater

tha

n $5

2,50

0?

7

.21

%

2. U

TILI

TIES

The

ave

rage

ele

ctri

c bi

ll in

a r

esid

enti

al a

rea

in J

une

is $

72. A

ssum

eth

is v

aria

ble

is n

orm

ally

dis

trib

uted

wit

h a

stan

dard

dev

iati

on o

f $6.

Fin

d th

epr

obab

ility

tha

t th

e m

ean

elec

tric

bill

for

a ra

ndom

ly s

elec

ted

grou

p of

15

resi

dent

sis

less

tha

n $7

5.

9

7.3

8%

3. S

HO

ES T

he p

rice

s of

sho

es in

a s

tore

are

nor

mal

ly d

istr

ibut

ed w

ith

a m

ean

of$9

3 an

d a

stan

dard

dev

iati

on o

f $18

. If n

ine

pair

s of

sho

es a

re r

ando

mly

sel

ecte

d,fin

d th

e pr

obab

ility

tha

t th

e m

ean

cost

is b

etw

een

$100

and

$11

0.

1

1.8

6%

4. C

OLL

EGE

Of t

he t

otal

pop

ulat

ion

at a

sm

all c

olle

ge, 2

0% a

re fr

om t

he M

id-A

tlan

tic

stat

es. I

f 200

stu

dent

s ar

e ra

ndom

ly s

elec

ted,

find

the

pro

babi

lity

that

at

leas

t50

are

from

the

Mid

-Atl

anti

c st

ates

.

4

.64

%

5. W

EATH

ER K

yle

has

rese

arch

ed t

he a

vera

ge a

nnua

l pre

cipi

tati

on in

his

cit

y fo

rse

vera

l yea

rs a

nd c

alcu

late

d th

e m

ean

as 1

9.32

inch

es. A

ssum

e th

e av

erag

epr

ecip

itat

ion

is n

orm

ally

dis

trib

uted

and

the

sta

ndar

d de

viat

ion

is 2

.44

inch

es.

a. I

f one

yea

r fr

om t

he t

ime

peri

od t

hat

Kyl

e re

sear

ched

is r

ando

mly

sel

ecte

d,w

hat

is t

he p

roba

bilit

y th

at t

he p

reci

pita

tion

is m

ore

than

18

inch

es?

7

0.5

4%

b. I

f fiv

e ye

ars

from

the

tim

e pe

riod

tha

t K

yle

rese

arch

ed a

re r

ando

mly

sel

ecte

d,w

hat

is t

he p

roba

bilit

y th

at t

he m

ean

prec

ipit

atio

n is

mor

e th

an 1

8 in

ches

?

8

8.6

9%

6. G

ROCE

RY E

ight

y-fiv

e pe

rcen

t of

the

sho

pper

s at

a g

roce

ry s

tore

hav

e a

freq

uent

-buy

er c

ard.

Thi

rty-

five

shop

pers

are

ran

dom

ly s

elec

ted

for

a ta

ste

test

. Wha

t is

the

pro

babi

lity

that

at

leas

t 25

and

at

mos

t 30

of t

he t

aste

test

ers

have

a fr

eque

nt-b

uyer

car

d?

6

3.4

%

Fin

d th

e m

inim

um s

ampl

e si

ze n

eede

d fo

r ea

ch p

roba

bili

ty s

o th

at t

he n

orm

al

dist

ribu

tion

can

be

used

to

appr

oxim

ate

the

bino

mia

l dis

trib

utio

n.

7. p

= 0

.6

8. p

= 0

.15

1

3

3

4

Ch

ap

ter

11

24

Gle

ncoe

Pre

calc

ulus

005_

042_

PC

CR

MC

11_8

9381

2.in

dd24

11/1

4/09

5:10

:57

PM


Lesson X-4

NA

ME

DA

TE

PE

RIO

D


Lesson 11-4

NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

25

Gle

ncoe

Pre

calc

ulus

Ch

ap

ter

11

25

Gle

ncoe

Pre

calc

ulus

Wor

d Pr

oble

m P

ract

ice

Th

e C

en

tral

Lim

it T

heo

rem

11-4

1. H

EATI

NG

Wor

kers

at

a pu

blic

uti

litie

s co

mpa

ny s

urve

yed

200

of t

heir

cus

tom

ers

and

foun

d th

at t

he m

ean

tem

pera

ture

at

whi

ch t

he c

usto

mer

s’ th

erm

osta

ts w

ere

set

in J

anua

ry w

as 7

0˚ w

ith

a st

anda

rd

devi

atio

n of

1.8

˚. I

f 30

cust

omer

s ar

e ra

ndom

ly s

elec

ted

for

a fo

llow

-up

surv

ey, f

ind

the

prob

abili

ty t

hat

the

mea

n te

mpe

ratu

re a

t w

hich

the

y se

t th

eir

ther

mos

tats

in J

anua

ry is

less

than

69.

5˚.

6

.30

%

2. M

ILK

A fo

od s

tudy

in a

cit

y fo

und

that

th

e pr

ice

of a

gal

lon

of w

hole

milk

in it

s st

ores

was

$3.

72. T

his

vari

able

is

norm

ally

dis

trib

uted

wit

h a

stan

dard

de

viat

ion

of $

0.08

.

a. I

f a g

allo

n of

who

le m

ilk is

ran

dom

ly

sele

cted

from

one

sto

re, f

ind

the

prob

abili

ty t

hat

it c

osts

less

tha

n $3

.70.

4

0.1

3%

b. I

f a g

allo

n of

milk

is r

ando

mly

se

lect

ed fr

om e

ach

of 4

0 di

ffere

nt

stor

es, f

ind

the

prob

abili

ty t

hat

the

mea

n co

st o

f the

sam

ple

is le

ss t

han

$3.7

0.

6

.17

%

3. B

ICY

CLES

Thi

rty-

seve

n pe

rcen

t of

the

re

side

nts

in a

nei

ghbo

rhoo

d ow

n bi

cycl

es.

If a

gro

up o

f 40

resi

dent

s ar

e ra

ndom

ly

sele

cted

, wha

t is

the

pro

babi

lity

that

at

leas

t ha

lf ow

n bi

cycl

es?

6

.17

%

4. T

ESTI

NG

The

ave

rage

tim

e it

tak

es a

gr

oup

of s

tude

nts

to c

ompl

ete

a re

adin

g te

st is

46.

2 m

inut

es w

ith

a st

anda

rd

devi

atio

n of

8 m

inut

es. T

he t

imes

are

no

rmal

ly d

istr

ibut

ed.

a. A

gro

up o

f 10

stud

ents

is r

ando

mly

se

lect

ed. F

ind

the

prob

abili

ty t

hat

the

mea

n ti

me

to c

ompl

ete

the

test

is

mor

e th

an 4

5 m

inut

es.

6

8.2

2%

b. H

ow d

oes

your

ans

wer

to

part

ach

ange

if t

he g

roup

con

sist

s of

15 s

tude

nts?

T

he

pro

ba

bil

ity

in

cre

as

es

by

3.7

%.

5. C

AFE

TERI

A C

olle

ge fr

eshm

en w

ere

aske

d if

they

ate

bre

akfa

st o

n Su

nday

m

orni

ng in

the

caf

eter

ia. T

he g

raph

sh

ows

the

perc

ent

of m

ales

and

fem

ales

w

ho s

aid

yes.

Fem

aleM

ale

4060 20 0

Percent

80

a. O

ut o

f 30

rand

omly

sel

ecte

d fe

mal

e fr

eshm

en, w

hat

is t

he a

ppro

xim

ate

prob

abili

ty t

hat

at le

ast

25 o

f the

m

ate

brea

kfas

t in

the

caf

eter

ia o

n Su

nday

mor

ning

?

a

bo

ut

46

%

b. O

ut o

f 30

rand

omly

sel

ecte

d m

ale

fres

hmen

, wha

t is

the

app

roxi

mat

e pr

obab

ility

tha

t at

leas

t 25

of t

hem

at

e br

eakf

ast

in t

he c

afet

eria

on

Sund

ay m

orni

ng?

a

bo

ut

3%

005_

042_

PC

CR

MC

11_8

9381

2.in

dd25

11/1

7/09

11:4

2:42

AM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

26

Gle

ncoe

Pre

calc

ulus

11-4

Exp

eri

en

ce t

he C

en

tral

Lim

it T

heo

rem

The

Cen

tral

Lim

it T

heor

em is

bes

t ap

prec

iate

d w

hen

expe

rien

ced.

Thi

s ac

tivi

tyw

ill a

llow

you

thi

s pr

ivile

ge. Y

ou m

ay w

ish

to w

ork

wit

h a

part

ner.

Si

mul

ate

spin

ning

a s

pinn

er w

ith

equa

l sec

tion

s la

bele

d 1–

10tw

o ti

mes

by

sele

ctin

g th

e ra

ndIn

t( c

omm

and

from

the

PR

B m

enu

afte

r pr

essi

ng M

AT

H. T

he s

cree

n sh

own

indi

cate

sth

e sp

inne

r la

nded

on

7 on

the

firs

t sp

in a

nd 1

on

the

seco

nd s

pin.

The

mea

n of

the

se s

pins

is 4

.

1.

Com

plet

e 25

sim

ulat

ions

and

list

the

mea

ns o

f the

spi

ns.

An

sw

ers

wil

l v

ary

.

2.

Find

the

mea

n an

d st

anda

rd d

evia

tion

of t

he m

eans

in E

xerc

ise

1.

An

sw

ers

wil

l v

ary

.

Now

sim

ulat

e sp

inni

ng t

he s

ame

spin

ner

four

tim

es.

3.

Com

plet

e 25

sim

ulat

ions

and

list

the

mea

ns o

f the

spi

ns.

An

sw

ers

wil

l v

ary

.

4.

Find

the

mea

n an

d st

anda

rd d

evia

tion

of t

he m

eans

in E

xerc

ise

3.

An

sw

ers

wil

l v

ary

.

Now

sim

ulat

e sp

inni

ng t

he s

ame

spin

ner

six

tim

es.

5.

Com

plet

e 25

sim

ulat

ions

and

list

the

mea

ns o

f the

spi

ns.

An

sw

ers

wil

l v

ary

.

6.

Find

the

mea

n an

d st

anda

rd d

evia

tion

of t

he m

eans

in E

xerc

ise

5.

An

sw

ers

wil

l v

ary

.

7. W

hat

do y

ou n

otic

e ab

out

the

mea

ns in

Exe

rcis

es 2

, 4, a

nd 6

?

Th

ey

ap

pro

ac

h t

he

me

an

of

the

da

ta s

et,

5.5

.

8.

Wha

t do

you

not

ice

abou

t th

e st

anda

rd d

evia

tion

s in

Exe

rcis

es 2

, 4, a

nd 6

?

Th

ey

ge

t s

ma

lle

r a

s t

he

nu

mb

er

of

sp

ins

in

cre

as

es

.

9.

Find

σ

−

√

�

n

for

n =

2, 4

, and

6 g

iven

σ =

2.8

7. W

hat

do y

ou n

otic

e?

Th

ey

are

clo

se

to

th

e s

am

ple

sta

nd

ard

de

via

tio

ns

fo

r 2

, 4

, a

nd

6 s

pin

s.

10. I

f n =

1, t

hat

is t

he s

pinn

er is

spu

n on

ce, w

hat

is t

he d

istr

ibut

ion

of t

he s

ampl

es?

How

doe

s it

cha

nge

as n

incr

ease

s?

Th

e d

istr

ibu

tio

n i

s u

nif

orm

; it

be

co

me

s n

orm

al

as

n i

nc

rea

se

s.

Enri

chm

ent

005_

042_

PC

CR

MC

11_8

9381

2.in

dd26

3/25

/09

12:3

8:18

PM


Lesson X-5

NA

ME

DA

TE

PE

RIO

D

Lesson 11-5

Ch

ap

ter

11

27

Gle

ncoe

Pre

calc

ulus

Nor

mal

Dis

trib

utio

n Th

e co

nfid

ence

leve

l c is

the

pro

babi

lity

that

an

inte

rval

es

tim

ate

cont

ains

a g

iven

par

amet

er, s

uch

as a

pop

ulat

ion

mea

n. T

he m

axim

um e

rror

of

esti

mat

e E

is t

he m

axim

um d

iffer

ence

bet

wee

n th

e po

int

esti

mat

e an

d th

e ac

tual

val

ue o

f th

e pa

ram

eter

. A c

onfi

denc

e in

terv

al C

I is

a s

peci

fic in

terv

al e

stim

ate

of a

par

amet

er.

For

a po

pula

tion

mea

n:

CI

= −

x ±

E o

r −

x ±

z ·

σ

−

√

�

n

.

z is

the

cri

tica

l val

ue t

hat

corr

espo

nds

to a

cer

tain

con

fiden

ce le

vel.

The

mos

t co

mm

on a

re

show

n be

low

.

Co

nfi

de

nc

e L

ev

el

90

%9

5%

99

%

z-V

alu

e1

.64

51

.96

02

.57

6

Whe

n σ

is u

nkno

wn,

the

sam

ple

stan

dard

dev

iati

on s

can

be

subs

titu

ted

for

σ, p

rovi

ded

the

sam

ple

size

n is

gre

ater

tha

n or

equ

al t

o 30

.

In

a s

ampl

e of

50

peop

le w

ho b

uy m

agaz

ines

, a r

esea

rche

r fi

nds

the

mea

n am

ount

spe

nt p

er m

onth

to

be $

12. A

ssum

e a

stan

dard

dev

iati

on o

f $4

.50.

Fin

d th

e 95

% c

onfi

denc

e in

terv

al f

or t

he m

ean

amou

nt s

pent

for

m

agaz

ines

eac

h m

onth

.C

I =

−

x ±

z ·

σ

−

√

�

n

C

on

fide

nce

In

terv

al f

or

the

Me

an

= 1

2 ±

1.9

6 · 4.

50

−

√

��

50

−

x =

12

, z

= 1

.96

, σ

= 4

.50

, a

nd

n =

50

≈ 1

2 ±

1.2

5 S

imp

lify.

Add

and

sub

trac

t th

e m

argi

n of

err

or.

Left

Bou

ndar

y: 1

2 -

1.2

5 =

10.

75

Rig

ht B

ound

ary:

12

+ 1

.25

= 1

3.25

The

95%

con

fiden

ce in

terv

al is

10.

75 <

μ <

13.

25. T

here

fore

, wit

h 95

% c

onfid

ence

, th

e m

ean

amou

nt s

pent

on

mag

azin

es p

er m

onth

is b

etw

een

$10.

75 a

nd $

13.2

5.

Exer

cise

s

The

num

ber

of d

ays

wit

h te

mpe

ratu

res

abov

e fr

eezi

ng f

or a

sam

ple

of 3

5 ci

ties

ha

d a

mea

n of

190

.7 d

ays

and

a sa

mpl

e st

anda

rd d

evia

tion

of

54.2

day

s.

1. F

ind

the

95%

con

fiden

ce in

terv

al fo

r th

e m

ean

num

ber

of d

ays

wit

h te

mpe

ratu

res

abov

e fr

eezi

ng.

1

72

.74

< μ

< 2

08

.66

2. F

ind

the

90%

con

fiden

ce in

terv

al fo

r th

e m

ean

num

ber

of d

ays

wit

h te

mpe

ratu

res

abov

e fr

eezi

ng.

1

75

.63

< μ

< 2

05

.77

Stud

y Gu

ide

and

Inte

rven

tion

Co

nfi

den

ce I

nte

rvals

Exam

ple

11-5

005_

042_

PC

CR

MC

11_8

9381

2.in

dd27

3/25

/09

12:3

8:25

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

28

Gle

ncoe

Pre

calc

ulus

t-D

istr

ibut

ion

To fi

nd a

con

fiden

ce in

terv

al w

hen

σ is

unk

now

n an

d th

e sa

mpl

e si

ze

n is

less

tha

n 30

, use

the

t-d

istr

ibut

ion,

pro

vide

d th

e va

riab

le is

app

roxi

mat

ely

norm

ally

di

stri

bute

d. T

he fo

rmul

a fo

r us

ing

the

t-di

stri

buti

on t

o co

nstr

uct

a co

nfid

ence

inte

rval

is

CI

= −

x ±

t ·

s −

√

�

n

,

whe

re −

x is

the

sam

ple

mea

n, t

is a

cri

tica

l val

ue w

ith

n -

1 d

egre

es o

f fre

edom

, s is

the

sa

mpl

e st

anda

rd d

evia

tion

, and

n is

the

sam

ple

size

.

T

he m

ean

pric

e of

9 r

ando

mly

sel

ecte

d te

levi

sion

s at

an

elec

tron

ics

stor

e is

$78

3 w

ith

a st

anda

rd d

evia

tion

of

$116

. Fin

d th

e 90

% c

onfi

denc

e in

terv

al

of t

he m

ean

pric

e of

all

of

the

tele

visi

ons

at t

he s

tore

. Ass

ume

that

the

var

iabl

e is

no

rmal

ly d

istr

ibut

ed.

Bec

ause

the

pop

ulat

ion

stan

dard

dev

iati

on is

unk

now

n an

d n

< 3

0, u

se t

he t-

dist

ribu

tion

. B

ecau

se n

= 9

, the

re a

re 9

- 1

or

8 de

gree

s of

free

dom

.

To fi

nd t

he t-

valu

e, d

eter

min

e th

e ar

ea in

the

low

er t

ail o

f the

di

stri

buti

on. S

ince

100

% -

90%

or

10%

is in

the

tai

ls, t

hen

5%

is in

eac

h ta

il. P

ress

2

nd

DIS

TR

and

cho

ose

invT

(. Ty

pe t

hear

ea t

o th

e le

ft o

f eac

h cr

itic

al v

alue

, as

a de

cim

al, f

ollo

wed

by

the

degr

ees

of fr

eedo

m.

CI

= −

x

± t

·

s −

√

�

n

C

on

fide

nce

inte

rva

l usi

ng

t-d

istr

ibu

tion

≈

783

± 1

.860

· 11

6 −

√

�

9 −

x =

78

3,

t ≈ 1

.86

0,

s =

11

6,

an

d n

= 9

≈

783

± 7

1.92

S

imp

lify.

Ther

efor

e, t

he 9

0% c

onfid

ence

inte

rval

is $

711.

08 <

μ <

$85

4.92

.

Exer

cise

s

1. A

t a

man

ufac

turi

ng p

lant

, the

thr

eads

on

ten

rand

omly

sel

ecte

d sc

rew

s ha

ve a

m

ean

dept

h of

0.3

2 in

ch a

nd a

sta

ndar

d de

viat

ion

of 0

.03

inch

. Fin

d th

e 95

%

conf

iden

ce in

terv

al o

f the

mea

n de

pth

of a

ll th

e sc

rew

s, a

ssum

ing

that

the

var

iabl

e is

no

rmal

ly d

istr

ibut

ed.

0

.30

< μ

< 0

.34

2. A

n en

viro

nmen

tal s

tudy

invo

lves

cou

ntin

g th

e nu

mbe

r of

ligh

t bu

lbs

in r

ando

mly

sel

ecte

d ro

oms

of a

bui

ldin

g. A

bui

ldin

g m

anag

er c

ount

s th

e bu

lbs

in 1

6 ro

oms

and

finds

the

mea

n nu

mbe

r of

bul

bs t

o be

21

wit

h a

stan

dard

dev

iati

on o

f 4.8

. Fin

d th

e 99

% c

onfid

ence

in

terv

al o

f the

mea

n nu

mbe

r of

bul

bs in

all

the

room

s of

the

bui

ldin

g, a

ssum

ing

the

vari

able

is n

orm

ally

dis

trib

uted

.

1

7.4

6 <

μ <

24

.54

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Co

nfi

den

ce I

nte

rvals

Exam

ple

11-5

005_

042_

PC

CR

MC

11_8

9381

2.in

dd28

3/25

/09

12:3

8:29

PM


Lesson X-5

NA

ME

DA

TE

PE

RIO

D

Lesson 11-5

Ch

ap

ter

11

29

Gle

ncoe

Pre

calc

ulus

1. C

HIL

D C

ARE

A r

ando

m s

ampl

e of

50

pare

nts

of y

oung

chi

ldre

n fo

und

that

the

y sp

end

a m

ean

of $

7648

eac

h ye

ar fo

r ch

ild c

are.

The

sta

ndar

d de

viat

ion

of t

he s

ampl

e w

as $

630.

Fi

nd t

he 9

0% c

onfid

ence

inte

rval

of t

he m

ean

annu

al c

ost

of c

hild

car

e.

$7

50

1 <

μ <

$7

79

5

2. F

ITN

ESS

Twen

ty-e

ight

peo

ple

who

enr

olle

d in

a fi

tnes

s pr

ogra

m lo

st a

mea

n of

14

.3 p

ound

s w

ith

a st

anda

rd d

evia

tion

of 2

pou

nds.

Fin

d th

e 95

% c

onfid

ence

inte

rval

of

the

mea

n w

eigh

t lo

ss in

pou

nds

of a

ll of

the

mem

bers

enr

olle

d in

the

pro

gram

. 1

3.5

< μ

< 1

5.1

3. C

OM

MU

TE T

he a

vera

ge n

umbe

r of

min

utes

it t

akes

8 p

eopl

e to

com

mut

e to

and

from

w

ork

duri

ng r

ush

hour

is s

how

n. A

ssum

e th

at t

he t

imes

are

nor

mal

ly d

istr

ibut

ed.

Co

mm

uti

ng

Tim

e (

min

ute

s)

55

70

60

55

60

56

55

60

a. D

ecid

e th

e ty

pe o

f dis

trib

utio

n th

at c

an b

e us

ed t

o es

tim

ate

the

com

mut

ing

tim

e m

ean.

Exp

lain

you

r re

ason

ing.

Us

e a

t-d

istr

ibu

tio

n b

ec

au

se

σ i

s u

nk

no

wn

an

d n

< 3

0.

b. C

alcu

late

the

mea

n an

d st

anda

rd d

evia

tion

to

the

near

est

hund

redt

h.

58

.88

min

; 5

.08

min

c. C

onst

ruct

a 9

0% c

onfid

ence

inte

rval

for

the

aver

age

com

mut

ing

tim

e in

min

utes

for

a co

mm

uter

from

thi

s ci

ty.

55

.48

< μ

< 6

2.2

8

4. T

RAIN

ING

In

a su

rvey

of 4

42 e

mpl

oyee

s at

a c

all c

ente

r, t

he m

ean

tim

e th

at e

mpl

oyee

s fe

lt w

as n

eede

d fo

r ad

equa

te t

rain

ing

for

thei

r jo

bs w

as 7

day

s. T

he s

ampl

e st

anda

rd

devi

atio

n w

as 1

.5 d

ays.

Con

stru

ct a

98%

con

fiden

ce in

terv

al fo

r th

e am

ount

of t

rain

ing

tim

e th

at e

mpl

oyee

s fe

lt w

as a

dequ

ate

to b

egin

the

ir jo

bs.

6.8

< μ

< 7

.2

Det

erm

ine

whe

ther

the

nor

mal

dis

trib

utio

n or

t-d

istr

ibut

ion

shou

ld b

e us

ed fo

r ea

ch q

uest

ion.

The

n fi

nd e

ach

conf

iden

ce in

terv

al g

iven

the

follo

win

g in

form

atio

n.

5. 9

5%;

−

x =

115

, s =

6, n

= 6

t-

dis

trib

uti

on

; 1

08

.7 <

μ <

12

1.3

6. 9

6%;

−

x =

18.

5, s

= 1

.2, n

= 4

0 n

orm

al

dis

trib

uti

on

; 1

8.1

1 <

μ <

18

.89

7. 9

9%;

−

x =

236

, σ =

8, n

= 4

5 n

orm

al

dis

trib

uti

on

; 2

32

.9 <

μ <

23

9.1

8. F

OO

D T

he o

wne

rs o

f a s

andw

ich

shop

wan

t to

find

the

95%

con

fiden

ce in

terv

al o

f the

tr

ue m

ean

cost

of a

ham

burg

er in

the

ir c

ity.

How

larg

e sh

ould

the

ir s

ampl

e be

if t

hey

wan

t to

be

accu

rate

to

wit

hin

15 c

ents

? In

an

earl

ier

surv

ey, t

he s

tand

ard

devi

atio

n of

th

e pr

ice

was

26

cent

s.

at

lea

st

12

ob

se

rva

tio

ns

9. S

PORT

S A

tea

cher

wan

ts t

o es

tim

ate

the

aver

age

num

ber

of h

ours

per

wee

k th

at h

er

stud

ents

are

at

prac

tice

or

at g

ames

for

a sp

orts

tea

m. T

he s

tand

ard

devi

atio

n fr

om a

pr

evio

us y

ear

was

6.2

hou

rs. H

ow la

rge

of a

sam

ple

mus

t sh

e se

lect

if s

he w

ants

to

be

99%

con

fiden

t of

find

ing

the

aver

age

amou

nt o

f tim

e st

uden

ts s

pent

par

tici

pati

ng in

sp

orts

wit

hin

1.5

hour

s?

at

lea

st

11

4 o

bs

erv

ati

on

s

Prac

tice

Co

nfi

den

ce I

nte

rvals

11-5

005_

042_

PC

CR

MC

11_8

9381

2.in

dd29

11/1

4/09

5:24

:25

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

30

Gle

ncoe

Pre

calc

ulus

1. R

EAD

ING

A s

ampl

e of

nor

mal

ly

dist

ribu

ted

read

ing

scor

es o

f for

ty

eigh

th-g

rade

stu

dent

s ha

s a

mea

n of

82

and

a s

tand

ard

devi

atio

n of

15.

Fin

d th

e 95

% c

onfid

ence

inte

rval

for

the

mea

n of

all

of t

he r

eadi

ng s

core

s.

7

7.4

< μ

< 8

6.6

2. C

HO

LEST

ERO

L Th

e se

rum

cho

lest

erol

le

vel w

as c

olle

cted

for

a gr

oup

of52

5 co

llege

wom

en. T

he m

ean

of t

he

sam

ple

was

191

.7

mg

−

100

mL w

ith

a

stan

dard

dev

iati

on o

f 41.

0.

a. C

onst

ruct

a 9

0% c

onfid

ence

leve

l for

th

e m

ean

seru

m c

hole

ster

ol le

vel.

1

88

.8 <

μ <

19

4.6

b. C

onst

ruct

a 9

5% c

onfid

ence

leve

l for

th

e m

ean

seru

m c

hole

ster

ol le

vel.

1

88

.2 <

μ <

19

5.2

c. S

uppo

se y

ou h

ear

a cl

aim

tha

t th

e m

ean

seru

m c

hole

ster

ol le

vel f

or

wom

en in

col

lege

is 2

00. W

hat

wou

ld

be y

our

reac

tion

bas

ed o

n yo

ur

answ

ers

to p

arts

a a

nd b

? W

hy?

S

am

ple

an

sw

er:

Th

e c

laim

is

in

co

rre

ct;

ne

ith

er

of

the

in

terv

als

co

nta

ins

20

0,

so

a

lev

el

of

20

0 i

s n

ot

lik

ely

.

3. C

AR

POLL

UTI

ON

Sev

en c

ars

wer

e te

sted

for

nitr

ogen

-oxi

de e

mis

sion

s.

The

resu

lts

are

show

n in

the

tab

le.

Em

iss

ion

s (

gra

ms

pe

r m

ile

)

0.0

50

.12

0.1

60

.15

0.1

40

.19

0.1

4

a. F

ind

the

mea

n of

the

dat

a.

0.1

36

b. F

ind

the

stan

dard

dev

iati

on o

f th

e da

ta.

0.0

44

c. F

ind

the

99%

con

fiden

ce in

terv

al o

f th

e m

ean

nitr

ogen

-oxi

de e

mis

sion

s.

0.0

74

7 <

μ <

0.1

96

7

4. F

UEL

CO

NSU

MPT

ION

The

mea

n an

d st

anda

rd d

evia

tion

for

city

and

hig

hway

fu

el c

onsu

mpt

ion

in m

iles

per

gallo

n fo

r 33

ran

dom

ly s

elec

ted

pre-

owne

d ca

rs o

n a

deal

er’s

lot

is s

how

n. A

ssum

e th

e va

riab

les

are

norm

ally

dis

trib

uted

.

−

x s

City

21

.35

4.1

3

Hig

hw

ay

29

.65

3.6

5

a. F

ind

the

98%

con

fiden

ce in

terv

al

for

the

mea

n fu

el c

onsu

mpt

ion

in

the

city

. 1

9.6

77

< μ

< 2

3.0

23

b. F

ind

the

98%

con

fiden

ce in

terv

al fo

r th

e m

ean

fuel

con

sum

ptio

n on

the

hi

ghw

ay.

28

.17

2 <

μ <

31

.12

8

c.

Com

pare

the

con

fiden

ce in

terv

als

in

part

s a

and

b.

S

am

ple

an

sw

er:

Th

e h

igh

wa

y

mil

ea

ge

is

co

ns

ide

rab

ly h

igh

er

tha

n t

he

cit

y m

ile

ag

e,

alt

ho

ug

h

the

ra

ng

e o

f v

alu

es

is

sm

all

er

for

hig

hw

ay

th

an

cit

y,

as

wo

uld

b

e e

xp

ec

ted

giv

en

th

at

the

s

tan

da

rd d

ev

iati

on

sh

ow

s l

es

s

va

ria

bil

ity

fo

r h

igh

wa

y m

ile

ag

e.

5. I

NTE

LLIG

ENCE

QU

OTI

ENT

Supp

ose

man

ager

s of

a c

orpo

rati

on w

ant

to

esti

mat

e th

e IQ

sco

re fo

r th

eir

empl

oyee

s. H

ow m

any

empl

oyee

s m

ust

be r

ando

mly

sel

ecte

d fo

r IQ

tes

ts if

the

m

anag

ers

wan

t to

be

95%

con

fiden

t th

at

the

mea

n is

wit

hin

2 IQ

poi

nts

of t

he

popu

lati

on m

ean?

The

y kn

ow fr

om

prev

ious

stu

dies

tha

t th

e st

anda

rd

devi

atio

n is

15

poin

ts.

a

t le

as

t 2

17

em

plo

ye

es

Wor

d Pr

oble

m P

ract

ice

Co

nfi

den

ce I

nte

rvals

11-5

005_

042_

PC

CR

MC

11_8

9381

2.in

dd30

11/1

4/09

5:22

:49

PM


Lesson X-5

NA

ME

DA

TE

PE

RIO

D

Lesson 11-5

Ch

ap

ter

11

31

Gle

ncoe

Pre

calc

ulus

Th

e P

op

ula

tio

n C

orr

ecti

on

Facto

rTh

e fo

rmul

a fo

r th

e m

axim

um e

rror

of e

stim

ate,

E =

z ·

σ

−

√

�

n

, as

sum

es a

ver

y la

rge

popu

lati

on o

r sa

mpl

ing

wit

h re

plac

emen

t. If

the

pop

ulat

ion

size

is N

and

the

sam

ple

num

ber

is n

, the

form

ula

shou

ld b

e m

odifi

ed w

ith

a co

rrec

tion

fact

or if

n >

0.0

5N.

The

popu

lati

on c

orre

ctio

n fa

ctor

is √

��

�

N

- n

−

N -

1 .

Ther

efor

e, if

n >

0.0

5N, t

he m

axim

um

erro

r of

est

imat

e is

E =

z ·

σ

−

√

�

n

· √

��

�

N

- n

−

N -

1 .

Det

erm

ine

if t

he p

opul

atio

n co

rrec

tion

fac

tor

wou

ld b

e us

ed w

hen

dete

rmin

ing

a co

nfid

ence

inte

rval

. Jus

tify

you

r an

swer

.

1. 2

9 of

the

140

em

ploy

ees

in a

com

pany

are

sur

veye

d

y

es

; n

= 2

9,

0.0

5N

= 7

, a

nd

29

> 7

2. 7

5 of

the

tow

n’s

resi

dent

s ar

e su

rvey

ed a

nd t

he t

own

popu

lati

on is

299

8

n

o;

n =

75

, 0

.05N

= 1

49

.9,

an

d 7

5 <

14

9.9

3. 1

0 of

the

80

page

s pr

inte

d by

a n

ew p

rint

er a

re e

xam

ined

y

es

; n

= 1

0,

0.0

5N

= 4

, a

nd

10

> 4

4. T

here

are

250

stu

dent

s in

a s

choo

l. A

sam

ple

of 3

5 ra

ndom

ly s

elec

ted

stud

ents

fin

ds t

hat

thei

r m

ean

daily

stu

dy t

ime

is 5

2 m

inut

es w

ith

a st

anda

rd d

evia

tion

of 3

.3 m

inut

es. F

ind

the

95%

con

fiden

ce in

terv

al fo

r th

e m

ean

daily

stu

dy t

ime

of a

ll of

the

stu

dent

s. R

ound

E t

o th

e ne

ares

t hu

ndre

dth.

5

0.9

8 <

μ <

53

.02

5. W

hat

wou

ld b

e th

e co

nfid

ence

inte

rval

in E

xerc

ise

4 if

ther

e w

ere

750

stud

ents

in

the

sch

ool?

5

0.9

1 <

μ <

53

.09

6. F

orty

of t

he 1

60 p

otat

oes

in a

bin

are

ran

dom

ly s

elec

ted

for

wei

ght.

The

mea

nw

eigh

t of

the

sam

ple

is 1

50 g

ram

s w

ith

a st

anda

rd d

evia

tion

of 6

.5 g

ram

s. F

ind

the

90%

con

fiden

ce in

terv

al fo

r th

e m

ean

wei

ght

of a

ll of

the

pot

atoe

s. R

ound

Eto

the

nea

rest

hun

dred

th. A

ssum

e th

e va

riab

le is

nor

mal

ly d

istr

ibut

ed.

1

48

.53

< μ

< 1

51

.47

Enri

chm

ent

11-5

005_

042_

PC

CR

MC

11_8

9381

2.in

dd31

3/25

/09

12:3

8:43

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

32

Gle

ncoe

Pre

calc

ulus

Hyp

othe

ses

A h

ypot

hesi

s te

st a

llow

s yo

u to

eva

luat

e a

clai

m a

bout

a p

opul

atio

n.

The

null

hyp

othe

sis

or H

0 sta

tes

that

the

re is

not

a s

igni

fican

t di

ffere

nce

betw

een

the

sam

ple

valu

e an

d th

e po

pula

tion

par

amet

er a

nd it

alw

ays

cont

ains

a s

tate

men

t of

equ

alit

y.Th

e al

tern

ativ

e hy

poth

esis

or

Ha s

tate

s th

at t

here

is a

sig

nific

ant

diffe

renc

e be

twee

nth

e sa

mpl

e va

lue

and

the

popu

lati

on p

aram

eter

and

it a

lway

s co

ntai

ns a

sta

tem

ent

of

ineq

ualit

y.

For

eac

h st

atem

ent,

wri

te t

he n

ull a

nd a

lter

nati

ve h

ypot

hese

s.

Stat

e w

hich

hyp

othe

sis

repr

esen

ts t

he c

laim

.

a. A

doc

tor

clai

ms

that

the

pul

se r

ate

of a

pat

ient

cha

nges

fro

m 8

2 be

ats

per

min

ute

afte

r ta

king

a n

ew m

edic

atio

n.

Th

is c

laim

bec

omes

μ ≠

82

and

is t

he a

lter

nati

ve h

ypot

hesi

s si

nce

it in

clud

es a

n in

equa

lity

sym

bol.

The

com

plem

ent

is μ

= 8

2.

H0:

μ =

82

Ha:

μ ≠

82

(cla

im)

b. A

tra

ck m

embe

r cl

aim

s th

at h

e ca

n ru

n at

leas

t 10

mil

es t

hat

day.

Th

is c

laim

bec

omes

μ ≥

10

and

is t

he n

ull h

ypot

hesi

s si

nce

it in

clud

es a

n eq

ualit

y sy

mbo

l. Th

e co

mpl

emen

t is

μ <

10.

H0:

μ ≥

10

(cla

im)

Ha:

μ <

10

c. A

con

trac

tor

clai

ms

that

inst

alli

ng a

par

ticu

lar

kind

of

insu

lati

on w

ill

low

er t

he a

vera

ge J

uly

cool

ing

bill

of

$68

in h

er a

rea.

Th

is c

laim

bec

omes

μ <

$68

and

is t

he a

lter

nati

ve h

ypot

hesi

s si

nce

it in

clud

esan

ineq

ualit

y sy

mbo

l. Th

e co

mpl

emen

t is

μ ≥

$68

.H

0: μ

≥ $

68

Ha:

μ <

$68

(cla

im)

Exer

cise

s

Wri

te t

he n

ull a

nd a

lter

nati

ve h

ypot

hese

s fo

r ea

ch s

tate

men

t. S

tate

whi

ch

hypo

thes

is r

epre

sent

s th

e cl

aim

.

1. O

n a

scal

e of

1 t

o 10

, pat

ient

s de

scri

be t

heir

anx

iety

leve

ls a

s 8

duri

ng d

enta

lpr

oced

ures

. A d

enta

l ass

ista

nt t

hink

s th

is le

vel i

s lo

wer

whe

n so

ft m

usic

ispl

ayed

dur

ing

the

proc

edur

es.

H0:

μ ≥

8;

Ha:

μ <

8 (

cla

im)

2. A

rea

l est

ate

agen

t cl

aim

s th

at t

he a

vera

ge h

ome

pric

e in

an

area

is $

250,

000.

H

0:

μ =

25

0,0

00

(c

laim

); H

a:

μ ≠

25

0,0

00

3. A

hik

er c

laim

s th

at t

he a

vera

ge t

rail

leng

th in

a p

ark

is a

t m

ost

10 m

iles.

H

0:

μ ≤

10

(c

laim

); H

a:

μ >

10

4. A

res

taur

ant

owne

r cl

aim

s th

at t

he a

vera

ge a

ge o

f din

ers

in a

cer

tain

are

a is

gr

eate

r th

an 4

0.

H0:

μ ≤

40

; H

a:

μ >

40

(c

laim

)

Stud

y Gu

ide

and

Inte

rven

tion

Hyp

oth

esi

s Test

ing

Exam

ple

11-6

005_

042_

PC

CR

MC

11_8

9381

2.in

dd32

3/25

/09

12:3

8:46

PM


Lesson X-6

NA

ME

DA

TE

PE

RIO

D

Lesson 11-6

Ch

ap

ter

11

33

Gle

ncoe

Pre

calc

ulus

Sign

ific

ance

and

Tes

ts T

o va

lidat

e a

clai

m, t

he n

ull h

ypot

hesi

s is

al

way

s te

sted

at

a ch

osen

leve

l of

sign

ific

ance

, α, w

hich

def

ines

the

are

a of

the

cri

tica

l reg

ion.

If t

he t

est

stat

isti

c z-

or

t-va

lue

is in

the

cri

tica

l reg

ion,

th

en H

0 sho

uld

be r

ejec

ted.

Oth

erw

ise,

H0 s

houl

d no

t be

rej

ecte

d.

A

n em

ploy

ee a

t a

spor

ting

goo

ds s

tore

cla

ims

that

the

ave

rage

pr

ice

of a

pai

r of

bas

ebal

l cle

ats

is le

ss t

han

$80.

Ano

ther

em

ploy

ee r

ando

mly

se

lect

s a

sam

ple

of 3

6 pa

irs

and

find

s −

x =

75

and

s =

19.

2. D

eter

min

e if

the

resu

lts

are

stat

isti

call

y si

gnif

ican

t at

α =

0.1

0.St

ep 1

Sta

te t

he n

ull a

nd a

lter

nati

ve h

ypot

hese

s an

d id

enti

fy t

he c

laim

.

H0:

μ ≥

$80

Ha:

μ <

80

(cla

im)

Step

2 D

eter

min

e th

e cr

itic

al v

alue

and

reg

ion.

Bec

ause

n ≥

30,

use

the

z-v

alue

. The

tes

t is

left

-tai

led

sinc

e μ

< 8

0.

All

of t

he c

riti

cal r

egio

n, w

ith

an a

rea

of 0

.10,

is t

o th

e le

ft o

f the

cr

itic

al v

alue

. Use

2

nd

DIS

TR

invN

orm

to

see

that

the

cri

tica

lva

lue

is -

1.28

. The

cri

tica

l reg

ion

is t

he a

rea

to t

he le

ft o

f z =

-1.

28.

Step

3 C

alcu

late

the

tes

t st

atis

tic.

Use

σ −

x = 19

.2

−

√

��

36

= 3

.2.

z =

−

x -

μ

−

σ −

x

Fo

rmu

la f

or

z-st

atis

tic

=

75 -

80

−

3.2

or

-1.

5625

−

x =

75

, μ

= 8

0,

an

d σ

−

x =

3.2

Step

4 A

ccep

t or

rej

ect

the

null

hypo

thes

is. H

0 is

reje

cted

sin

ce t

he t

est

stat

isti

c of

-1.

56 fa

lls w

ithi

n th

e cr

itic

al r

egio

n. T

here

is e

noug

h ev

iden

ce t

o su

ppor

t th

e cl

aim

tha

t th

e av

erag

e co

st o

f bas

ebal

l cle

ats

is le

ss t

han

$80.

Exer

cise

s

1. M

anag

ers

of a

larg

e de

part

men

t st

ore

clai

m t

hat

the

aver

age

sala

ry fo

r th

eir

part

-tim

e em

ploy

ees

is $

24,0

00. A

sam

ple

of 1

0 pa

rt-t

ime

empl

oyee

s ha

s a

mea

nsa

lary

of $

23,4

50 a

nd a

sta

ndar

d de

viat

ion

of $

1400

. Det

erm

ine

whe

ther

the

reis

eno

ugh

evid

ence

to

supp

ort

the

clai

m a

t α

= 0

.05.

t

= -

1.2

42

3,

nu

ll h

yp

oth

es

is i

s n

ot

reje

cte

d.

Th

e e

vid

en

ce

do

es

no

t re

jec

t th

e c

laim

th

at

the

me

an

is

$2

4,0

00

.

2. I

n E

xerc

ise

1, s

uppo

se t

hat

the

mea

n of

$23

,450

was

tak

en fr

om a

sam

ple

of50

par

t-ti

me

empl

oyee

s. W

ould

the

re b

e en

ough

evi

denc

e to

sup

port

the

cla

imat

α =

0.0

5?

z

= -

2.7

8,

nu

ll h

yp

oth

es

is i

s r

eje

cte

d.

Th

ere

wo

uld

be

en

ou

gh

ev

ide

nc

e

to r

eje

ct

the

cla

im.

11-6

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Hyp

oth

esi

s Test

ing

Exam

ple

-1.

28

Criti

cal r

egio

n

005_

042_

PC

CR

MC

11_8

9381

2.in

dd33

3/25

/09

12:3

8:51

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

34

Gle

ncoe

Pre

calc

ulus

Wri

te t

he n

ull a

nd a

lter

nati

ve h

ypot

hese

s fo

r ea

ch s

tate

men

t. S

tate

w

hich

hyp

othe

sis

repr

esen

ts t

he c

laim

.

1. A

flor

ist

clai

ms

that

a c

erta

in t

ype

of fl

ower

has

a v

ase

life

of a

t le

ast

7 da

ys.

H0:

μ ≥

7 (

cla

im);

Ha:

μ <

7

2. A

bra

nd o

f cer

eal c

laim

s th

at a

ser

ving

con

tain

s le

ss t

han

2 gr

ams

of

suga

r.H

0:

μ ≥

2;

Ha:

μ <

2 (

cla

im)

3. R

ober

t cl

aim

s th

at h

e sw

ims

100

laps

eac

h w

eek.

H0:

μ =

10

0 (

cla

im);

Ha:

μ ≠

10

0

For

eac

h cl

aim

k, u

se t

he s

peci

fied

info

rmat

ion

to c

alcu

late

the

tes

t st

atis

tic

and

dete

rmin

e w

heth

er t

here

is e

noug

h ev

iden

ce t

o re

ject

th

e nu

ll h

ypot

hesi

s. T

hen

mak

e a

stat

emen

t re

gard

ing

the

orig

inal

cl

aim

.

4. k

: μ ≥

60,

α =

0.1

0, −

x =

58.

88, s

= 5

.08,

n =

8

t =

-0

.62

4;

Th

e n

ull

h

yp

oth

es

is i

s n

ot

reje

cte

d;

the

cla

im i

s n

ot

reje

cte

d.

5. k

: μ =

8, α

= 0

.05,

−

x =

8.2

, s =

0.6

, n =

32

z =

1.8

9;

Th

e n

ull

h

yp

oth

es

is i

s n

ot

reje

cte

d;

the

cla

im i

s n

ot

reje

cte

d.

6. M

AIL

A m

ail c

arri

er c

laim

s th

at t

he a

vera

ge n

umbe

r of

pie

ces

of m

ail

rece

ived

dai

ly b

y ho

useh

olds

in a

cer

tain

nei

ghbo

rhoo

d is

7. S

ampl

e da

ta

for

15 h

ouse

hold

s is

col

lect

ed. T

he a

vera

ge n

umbe

r of

mai

l pie

ces

was

6.5

w

ith

a st

anda

rd d

evia

tion

of 0

.8.

a. I

s th

ere

enou

gh e

vide

nce

to r

ejec

t the

mai

l car

rier

’s cl

aim

at α

= 0

.05?

t

= -

2.4

2, n

ull

hyp

oth

esis

is r

eje

cte

d. T

here

is e

no

ug

h

evid

en

ce t

o r

eje

ct

the c

laim

.

b. I

s th

ere

enou

gh e

vide

nce

to r

ejec

t the

mai

l car

rier

’s cl

aim

at α

= 0

.01?

n

o;

P =

0.0

30

7. W

EDD

ING

S A

wed

ding

pla

nner

wan

ts t

o te

st t

he c

laim

tha

t th

e av

erag

e w

eddi

ng h

as 1

25 g

uest

s. I

n a

rand

om s

ampl

e of

35

wed

ding

s, h

e fo

und

the

mea

n to

be

110

gues

ts w

ith

a st

anda

rd d

evia

tion

of 3

0 gu

ests

. Fin

d th

e P-

valu

e an

d de

term

ine

whe

ther

the

re is

eno

ugh

evid

ence

to

supp

ort

the

clai

m a

t α

= 0

.01.

P

-va

lue

= 0

.00

3;

Th

ere

is

no

t e

no

ug

h e

vid

en

ce

to

su

pp

ort

th

e

cla

im.

Prac

tice

Hyp

oth

esi

s Test

ing

11-6

005_

042_

PC

CR

MC

11_8

9381

2.in

dd34

11/1

4/09

5:31

:54

PM


Lesson X-6

NA

ME

DA

TE

PE

RIO

D

Lesson 11-6

Ch

ap

ter

11

35

Gle

ncoe

Pre

calc

ulus

1. B

OTT

LED

WA

TER

The

aver

age

volu

me

in o

unce

s of

a r

ando

m s

ampl

e of

36

bot

tles

of w

ater

at

a pa

ckag

ing

plan

t w

as fo

und

to b

e 12

.19

ounc

es w

ith

a st

anda

rd d

evia

tion

of 0

.11

ounc

e. T

he

floor

sup

ervi

sor

mad

e th

e cl

aim

tha

t th

e m

ean

volu

me

was

gre

ater

tha

n 12

oun

ces.

Tes

t he

r cl

aim

at

α =

0.0

1.

a. W

rite

the

nul

l and

alt

erna

tive

hy

poth

eses

and

sta

te w

hich

hy

poth

esis

rep

rese

nts

the

clai

m.

H0:

μ ≤

12

; H

a:

μ >

12

(c

laim

)

b. I

s th

ere

enou

gh e

vide

nce

to r

ejec

t th

e nu

ll hy

poth

esis

? W

hy?

ye

s;

z =

10

.36

, w

hic

h i

s g

rea

ter

tha

n t

he

cri

tic

al

va

lue

of

2.3

3.

c. M

ake

a st

atem

ent

rega

rdin

g th

e or

igin

al c

laim

.

Th

ere

is e

no

ug

h e

vid

en

ce t

o

su

pp

ort

th

e c

laim

th

at

the m

ean

vo

lum

e o

f w

ate

r in

th

e b

ott

les is

g

reate

r th

an

12 o

un

ces.

2. T

EMPE

RATU

RE I

t is

a lo

ng-e

stab

lishe

d cl

aim

tha

t th

e m

ean

body

tem

pera

ture

fo

r hu

man

s is

98.

6°F.

In

a ra

ndom

sa

mpl

e of

10

peop

le, t

he m

ean

body

te

mpe

ratu

re w

as 9

8.3°

F w

ith

a st

anda

rd

devi

atio

n of

0.2

3°F.

Tes

t th

e cl

aim

at

α =

0.0

2.

a. W

hat

are

the

crit

ical

val

ues?

-

2.3

3 a

nd

2.3

3

b. W

hat

is t

he t

est

stat

isti

c?

-

4.1

2

c. I

s th

is e

vide

nce

enou

gh t

o su

ppor

tth

e cl

aim

tha

t th

e m

ean

body

te

mpe

ratu

re fo

r hu

man

s is

98.

6°?

n

o

3. P

ULS

E RA

TES

An

aero

bics

inst

ruct

or

had

a pu

lse

rate

of 1

10 b

eats

per

min

ute

afte

r a

war

m-u

p ro

utin

e w

ith

her

clas

s.

Stud

ents

rec

orde

d th

eir

puls

e ra

tes

at

the

sam

e ti

me.

The

ir r

ates

are

rec

orde

d in

the

tab

le b

elow

. The

inst

ruct

or c

laim

s th

at t

heir

ave

rage

pul

se r

ate

is lo

wer

th

an h

ers.

Tes

t he

r cl

aim

at

α =

0.0

5.

80

70

90

75

11

0

10

51

20

11

08

51

15

95

95

10

59

07

0

10

59

51

00

10

59

0

a. W

rite

the

nul

l and

alt

erna

tive

hy

poth

eses

and

sta

te w

hich

hy

poth

esis

rep

rese

nts

the

clai

m.

H0:

μ ≥

11

0;

Ha:

μ <

11

0 (

cla

im)

b. W

hat

are

the

crit

ical

val

ues

and

test

sta

tist

ic?

cri

tic

al

va

lue

: -

1.7

3;

tes

t s

tati

sti

c:

-4

.5

c.

Doe

s th

e da

ta s

uppo

rt t

he in

stru

ctor

’s cl

aim

s?

y

es

4. M

OU

NTA

IN C

LIM

BIN

G A

mou

ntai

n cl

imbi

ng in

stru

ctor

cla

ims

that

his

st

uden

ts t

ake

long

er t

han

5 m

inut

es

to p

ack

thei

r ba

ckpa

cks.

In

a ra

ndom

sa

mpl

e of

80

stud

ents

, the

ave

rage

ti

me

it t

ook

to p

ack

a ba

ckpa

ck w

as

5.3

min

utes

wit

h a

stan

dard

dev

iati

onof

1.7

min

utes

. Fin

d th

e P-

valu

e an

d de

term

ine

whe

ther

the

re is

eno

ugh

evid

ence

to

supp

ort

the

clai

m a

tα

= 0

.05.

P

-va

lue

= 0

.05

7;

Th

ere

is

no

t e

no

ug

h e

vid

en

ce

to

su

pp

ort

th

e

cla

im.

11-6

Wor

d Pr

oble

m P

ract

ice

Hyp

oth

esi

s Test

ing

005_

042_

PC

CR

MC

11_8

9381

2.in

dd35

11/1

4/09

5:32

:52

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

36

Gle

ncoe

Pre

calc

ulus

Hyp

oth

esi

s Test

ing

You

hav

e te

sted

hyp

othe

ses

by fi

ndin

g a

test

sta

tist

ic a

nd s

eein

g if

it fa

lls w

ithi

n th

e cr

itic

al

regi

on. A

lter

nati

vely

, you

can

tes

t a

hypo

thes

is b

y fin

ding

a c

onfid

ence

inte

rval

and

see

ing

if th

e cl

aim

falls

wit

hin

that

inte

rval

.

A c

ondo

min

ium

sup

erin

tend

ent

buys

5-p

ound

bag

s of

sal

t fo

r m

elti

ng ic

e.H

e w

eigh

s th

e sa

lt in

50

rand

omly

sel

ecte

d ba

gs a

nd f

inds

the

mea

n to

be

4.6

poun

ds w

ith

a st

anda

rd d

evia

tion

of

0.7

poun

d. U

se a

con

fide

nce

leve

lto

tes

t th

e su

peri

nten

dent

’s c

laim

tha

t th

e m

ean

wei

ght

is n

ot 5

pou

nds

at α

= 0

.05.

Wri

te t

he h

ypot

hese

s: H

0: μ

= 5

H

a: μ

≠ 5

Use

the

nor

mal

dis

trib

utio

n si

nce

n >

30.

Sin

ce α

= 0

.05,

the

con

fiden

ce le

vel i

s 1

- α

= 0

.05,

or

0.95

. The

z-v

alue

ass

ocia

ted

wit

h a

95%

con

fiden

ce le

vel i

s 1.

96.

Find

the

95%

con

fiden

ce in

terv

al fo

r th

e m

ean

wei

ght

of t

he s

alt

in t

he b

ags.

CI

= −

x ±

z ·

σ

−

√

�

n

C

on

fide

nce

inte

rva

l fo

r th

e m

ea

n

= 4

.6 ±

0.1

94

−

x =

4.6

, a

nd

z ·

σ

−

√

�

n = 1

.96

·

0.7

−

√

��

50

or

ab

ou

t 0

.19

4

So, t

he 9

5% c

onfid

ence

inte

rval

is 4

.406

< μ

< 4

.794

.

The

conf

iden

ce in

terv

al fo

r th

e po

pula

tion

doe

s no

t co

ntai

n th

e cl

aim

of 5

, so

we

reje

ct t

he

null

hypo

thes

is. E

vide

nce

supp

orts

the

cla

im t

hat

the

mea

n is

not

5 p

ound

s.

Exer

cise

s

1. S

OCC

ER A

soc

cer

coac

h cl

aim

s th

at t

he m

ean

wei

ght

of t

he p

laye

rs o

n th

e op

posi

ng

team

s is

200

pou

nds.

A r

ando

m s

ampl

e of

10

play

ers

from

opp

osin

g te

ams

has

a m

ean

of

198.

2 po

unds

and

a s

tand

ard

devi

atio

n of

3.3

pou

nds.

At

α =

0.0

5, c

an t

he c

laim

be

supp

orte

d? S

uppo

rt y

our

answ

er w

ith

a co

nfid

ence

inte

rval

.

T

he

da

ta s

up

po

rts

th

e c

laim

; 1

95

.84

< μ

< 2

00

.56

.

2. C

ALO

RIES

A t

each

er w

alke

d by

a v

endi

ng m

achi

ne a

nd c

laim

ed t

hat

the

aver

age

num

ber

of C

alor

ies

in t

he s

nack

s w

as 2

50. S

tude

nts

who

ove

rhea

rd h

er w

ante

d to

pr

ove

her

inco

rrec

t. Th

ey b

ough

t a

rand

om s

ampl

e of

8 s

nack

s an

d fo

und

the

mea

n nu

mbe

r of

Cal

orie

s of

tho

se s

nack

s to

be

210

wit

h a

stan

dard

dev

iati

on o

f 24.

At

α =

0.1

0, d

oes

the

data

sup

port

the

tea

cher

or

the

stud

ents

? Su

ppor

t yo

ur a

nsw

er

wit

h a

conf

iden

ce in

terv

al.

T

he

da

ta s

up

po

rts

th

e s

tud

en

ts;

19

3.9

2 <

μ <

22

6.0

8.

3. H

OM

EWO

RK A

tea

cher

cla

ims

her

stud

ents

spe

nd a

n av

erag

e of

45

min

utes

on

her

hom

ewor

k ea

ch n

ight

. To

test

the

cla

im, s

he a

sks

35 s

tude

nts

how

muc

h ti

me,

on

aver

age,

the

y sp

end

on h

er h

omew

ork

each

nig

ht a

nd t

he r

esul

ts s

how

a m

ean

of

40 m

inut

es w

ith

a st

anda

rd d

evia

tion

of 9

min

utes

. At

α =

0.0

5, d

oes

the

data

sup

port

he

r cl

aim

? Su

ppor

t yo

ur a

nsw

er w

ith

a co

nfid

ence

inte

rval

.

T

he

da

ta d

oe

s n

ot

su

pp

ort

he

r c

laim

; 3

7 <

μ <

43

.

Enri

chm

ent

11-6

005_

042_

PC

CR

MC

11_8

9381

2.in

dd36

11/1

9/09

7:40

:52

PM


Lesson X-7

NA

ME

DA

TE

PE

RIO

D

Lesson 11-7

Ch

ap

ter

11

37

Gle

ncoe

Pre

calc

ulus

11-7

Corr

elat

ion

To d

eter

min

e if

the

corr

elat

ion

coef

fici

ent

r is

sig

nific

ant,

perf

orm

a h

ypot

hesi

s te

st w

ith

H0 :

ρ =

0, a

nd t

as t

he t

est

stat

isti

c, t

= r

√

��

�

n

- 2

−

1 -

r2

, w

here

the

deg

rees

of f

reed

om is

n -

2.

The

tab

le s

how

s th

e nu

mbe

r of

abs

ence

s of

7 s

tude

nts

from

a p

sych

olog

y cl

ass

and

thei

r fi

nal g

rade

s.

Ab

se

nc

es

91

01

52

82

6

Gra

de

s7

05

84

39

07

88

77

9

a. M

ake

a sc

atte

r pl

ot o

f th

e da

ta a

nd id

enti

fy t

he r

elat

ions

hip.

T

hen

calc

ulat

e an

d in

terp

ret

the

corr

elat

ion

coef

fici

ent.

E

nter

the

dat

a in

to L

1 an

d L

2. T

urn

on P

lot1

and

cho

ose

a sc

atte

r pl

ot. T

he d

ata

appe

ars

to h

ave

a ne

gati

ve li

near

cor

rela

tion

.

Pr

ess

STA

T a

nd s

elec

t L

inR

eg(a

x +

b)

unde

r th

e C

alc

men

u.

The

corr

elat

ion

coef

ficie

nt r

is a

bout

-0.

9609

. Bec

ause

r is

clo

se

to -

1, t

his

sugg

ests

a s

tron

g ne

gati

ve c

orre

lati

on.

b. T

est

the

sign

ific

ance

of

the

corr

elat

ion

coef

fici

ent

from

pa

rt a

at

the

5% le

vel.

St

ep 1

Sta

te t

he h

ypot

hese

s.

H0:

ρ =

0

H

a: ρ ≠

0

St

ep 2

Det

erm

ine

the

crit

ical

val

ues

usin

g n

- 2

or

5 de

gree

s of

fr

eedo

m. T

he c

riti

cal v

alue

s ar

e ±

2.6.

St

ep 3

Cal

cula

te t

he t

est

stat

isti

c.

t = -

0.96

09 √

��

��

��

5

−

1 -

(-0.

9609

)2 ≈

-7.

7597

St

ep 4

Sin

ce t

< -

2.6,

rej

ect

the

null

hypo

thes

is. T

he e

vide

nce

supp

orts

a s

igni

fican

t co

rrel

atio

n be

twee

n th

e nu

mbe

r of

abs

ence

s an

d a

stud

ent’s

gra

de.

Exer

cise

s

The

tab

le s

how

s da

ta a

bout

gey

sers

at

Yel

low

ston

e N

atio

nal P

ark.

Du

rati

on

(m

in)

3.2

53

57

.51

7.5

3.5

63

51

06

.51

01

20

4

He

igh

t (f

t)1

84

25

15

02

07

57

51

01

00

16

06

05

25

75

5

Sour

ce: N

atio

nal P

ark

Serv

ice

1. M

ake

a sc

atte

r pl

ot o

f the

dat

a an

d id

enti

fy t

he r

elat

ions

hip.

Th

en c

alcu

late

and

inte

rpre

t th

e co

rrel

atio

n co

effic

ient

. It

ap

pe

ars

to

ha

ve

no

re

lati

on

sh

ip o

r a

we

ak

po

sit

ive

li

ne

ar

co

rre

lati

on

; r

= 0

.15

9;

litt

le t

o n

o c

orr

ela

tio

n

2. T

est

the

sign

ifica

nce

of t

he c

orre

lati

on c

oeffi

cien

t fr

om E

xerc

ise

1 at

the

5%

leve

l.

t =

0.5

58

; S

inc

e t

< 2

.2,

do

no

t re

jec

t n

ull

hy

po

the

sis

; n

o c

orr

ela

tio

n.

Stud

y Gu

ide

and

Inte

rven

tion

Co

rrela

tio

n a

nd

Lin

ear

Reg

ress

ion

Exam

ple

[ 1, 1

7] s

cl: 1

by

[ 35,

100

] scl:

5

[ –1,

40]

scl:

1 b

y [ –

1, 2

15] s

cl: 5

005_

042_

PC

CR

MC

11_8

9381

2.in

dd37

11/1

4/09

5:42

:10

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

38

Gle

ncoe

Pre

calc

ulus

Line

ar R

egre

ssio

n If

the

cor

rela

tion

bet

wee

n tw

o va

riab

les

is s

igni

fican

t, yo

u ca

n de

term

ine

the

leas

t-sq

uare

s re

gres

sion

line

, whi

ch is

a li

ne o

f be

st f

it. T

hen

you

can

use

the

equa

tion

of t

he li

ne t

o m

ake

pred

icti

ons

over

the

ran

ge o

f dat

a.

The

tab

le s

how

s th

e da

ta f

rom

the

pre

viou

s pa

ge, w

hich

sho

wed

a

sign

ific

ant

corr

elat

ion.

Fin

d th

e eq

uati

on o

f th

e re

gres

sion

line

for

the

dat

a.

Inte

rpre

t th

e sl

ope

and

inte

rcep

t in

con

text

. Use

the

equ

atio

n to

pre

dict

the

ex

pect

ed g

rade

for

a s

tude

nt w

ith

4 ab

senc

es a

nd s

tate

whe

ther

thi

s pr

edic

tion

is

rea

sona

ble.

Exp

lain

. Ab

se

nc

es

91

01

52

82

6

Gra

de

s7

05

84

39

07

88

77

9

Pres

s S

TA

T a

nd s

elec

t L

inR

eg(a

x +

b)

unde

r th

e C

alc

men

u.

The

equa

tion

is a

ppro

xim

atel

y y

= -

3.48

x +

97.

993.

The

slop

e in

dica

tes

that

for

ever

y ad

diti

onal

abs

ence

a s

tude

nt

has,

his

or

her

grad

e w

ill d

rop

by a

bout

3.5

poi

nts.

The

y-in

terc

ept

indi

cate

s th

at a

stu

dent

wit

h no

abs

ence

s w

ill h

ave

a gr

ade

of a

bout

98.

Eva

luat

e th

e re

gres

sion

equ

atio

n fo

r x

= 4

and

cal

cula

te y

.y

= -

3.48

x +

97.

993

Re

gre

ssio

n e

qu

atio

n

=

-3.

48(4

) + 9

7.99

3 x

= 4

=

84.

073

Sim

plif

y.

We

expe

ct t

hat

a st

uden

t w

ith

4 da

ys o

f abs

ence

s w

ould

hav

e a

final

gra

de o

f abo

ut 8

4.

This

is r

easo

nabl

e be

caus

e 4

is in

the

ran

ge o

f the

giv

en x

-val

ues

and

84 is

in t

he r

ange

of

the

giv

en y

-val

ues.

Exer

cise

s

A s

port

s w

rite

r ha

s al

read

y de

term

ined

tha

t th

ere

is a

neg

ativ

e co

rrel

atio

n be

twee

n th

e ba

ttin

g av

erag

e an

d th

e nu

mbe

r of

hom

e ru

ns f

or t

hird

bas

emen

w

hose

sta

tist

ics

are

show

n in

the

tab

le.

Av

era

ge

0.3

07

0.3

28

0.3

05

0.2

94

0.3

06

0.3

11

0.2

71

0.2

67

0.2

67

0.3

20

Ho

me

Ru

ns

96

11

83

17

65

78

10

35

12

26

85

48

58

1. F

ind

the

equa

tion

of t

he r

egre

ssio

n lin

e fo

r th

e da

ta.

y

= -

64

93

.7x

+ 2

14

8.8

2

2. U

se t

he e

quat

ion

to p

redi

ct t

he n

umbe

r of

hom

e ru

ns fo

r a

thir

d ba

sem

an w

ith

a ba

ttin

g av

erag

e of

0.3

50. S

tate

whe

ther

thi

s pr

edic

tion

is r

easo

nabl

e. E

xpla

in.

a

bo

ut

-1

24

ho

me

ru

ns

; n

ot

rea

so

na

ble

; T

he

nu

mb

er

of

ho

me

ru

ns

m

us

t b

e p

os

itiv

e.

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Co

rrela

tio

n a

nd

Lin

ear

Reg

ress

ion

11-7

Exam

ple

005_

042_

PC

CR

MC

11_8

9381

2.in

dd38

3/25

/09

12:3

9:18

PM


Lesson X-7

NA

ME

DA

TE

PE

RIO

D

Lesson 11-7

Ch

ap

ter

11

39

Gle

ncoe

Pre

calc

ulus

A s

uper

viso

r of

a c

lean

ing

busi

ness

has

the

dat

a in

the

tab

le t

hat

show

s th

e ag

e of

her

wor

kers

and

the

num

ber

of s

ick

days

the

y ta

ke e

ach

year

. She

won

ders

if

ther

e is

a s

igni

fica

nt li

near

rel

atio

nshi

p be

twee

n th

e ag

e of

an

empl

oyee

and

the

nu

mbe

r of

sic

k da

ys h

e or

she

tak

es e

ach

year

.

Ag

e3

82

56

01

84

55

41

92

24

33

4

Da

ys

91

22

16

45

15

17

31

1. M

ake

a sc

atte

r pl

ot o

f the

dat

a an

d id

enti

fy t

he r

elat

ions

hip.

Th

en c

alcu

late

and

inte

rpre

t th

e co

rrel

atio

n co

effic

ient

. T

he

da

ta a

pp

ea

r to

ha

ve

a n

eg

ati

ve

lin

ea

r c

orr

ela

tio

n,

r ≈

-0

.83

24

; T

he

co

rre

lati

on

co

eff

icie

nt

ind

ica

tes

a

fair

ly s

tro

ng

ne

ga

tiv

e l

ine

ar

co

rre

lati

on

.

2. D

eter

min

e if

the

corr

elat

ion

coef

ficie

nt is

sig

nific

ant

at t

he 1

%, 5

%, a

nd 1

0% le

vels

. E

xpla

in y

our

reas

onin

g.

t ≈

-4

.25

an

d t

< -

3.3

6,

t <

-2

.31

, t

< -

1.8

6,

so

th

e

nu

ll h

yp

oth

es

is i

s r

eje

cte

d a

nd

th

ere

is

a s

ign

ific

an

t c

orr

ela

tio

n.

3. I

f the

cor

rela

tion

is s

igni

fican

t at

the

10%

leve

l, fin

d th

e le

ast-

squa

res

regr

essi

on

equa

tion

and

inte

rpre

t th

e sl

ope

and

y-in

terc

ept

in c

onte

xt.

y =

-0

.34

8x

+ 2

0.8

67

; T

he

slo

pe

in

dic

ate

s t

ha

t fo

r e

ve

ry a

dd

itio

na

l y

ea

r o

lde

r, t

he

nu

mb

er

of

sic

k d

ay

s d

ec

rea

se

s b

y a

bo

ut

on

e-t

hir

d d

ay

; th

e y

-in

terc

ep

t is

no

t m

ea

nin

gfu

l. A

pe

rso

n 0

ye

ars

old

wo

uld

us

e 2

1 s

ick

da

ys

pe

r y

ea

r.

4. G

raph

and

ana

lyze

the

res

idua

l plo

t. T

he

re

sid

ua

ls a

pp

ea

r to

be

fa

irly

ra

nd

om

ly s

ca

tte

red

an

d m

os

tly

ce

nte

red

a

bo

ut

the

re

gre

ss

ion

lin

e y

= 0

. T

his

su

pp

ort

s t

he

c

laim

th

at

the

us

e o

f a

lin

ea

r m

od

el

is a

pp

rop

ria

te.

5. I

dent

ify a

ny in

fluen

tial

out

liers

. Des

crib

e th

e ef

fect

the

out

lier

has

on t

he s

tren

gth

of t

he c

orre

lati

on a

nd o

n th

e sl

ope

and

inte

rcep

t of

the

ori

gina

l re

gres

sion

line

. T

he

po

int

(34

, 1

) is

an

ou

tlie

r in

th

e y

-dir

ec

tio

n.

Th

e

co

rre

lati

on

is

str

on

ge

r w

ith

ou

t it

(-

0.9

38

). T

he

slo

pe

ch

an

ge

s v

ery

lit

tle

w

ith

ou

t it

, b

ut

the

y-i

nte

rce

pt

ch

an

ge

s s

om

e.

6. I

f any

dat

a w

ere

rem

oved

, rea

sses

s th

e si

gnifi

canc

e of

the

cor

rela

tion

at

the

10%

leve

l an

d, if

sti

ll ap

prop

riat

e, r

ecal

cula

te t

he r

egre

ssio

n eq

uati

on.

Wit

h (

34

, 1

) re

mo

ve

d,

t ≈

-7

.2 a

nd

t <

-1

.89

, s

o t

he

co

rre

lati

on

is

sti

ll

sig

nif

ica

nt

at

the

10

% l

ev

el.

^ y =

-0

.35

6x

+ 2

2.0

53

7. I

f app

ropr

iate

, pre

dict

the

num

ber

of s

ick

days

for

empl

oyee

s w

ho a

re 3

0, 5

0, a

nd 7

0 ye

ars

old.

Int

erpr

et y

our

resu

lts

and

stat

e w

heth

er t

he p

redi

ctio

ns a

re r

easo

nabl

e.

Exp

lain

you

r re

ason

ing.

1

1,

4,

an

d -

3 d

ay

s;

Th

e f

irs

t tw

o p

red

icti

on

s a

re

rea

so

na

ble

be

ca

us

e b

oth

va

ria

ble

s a

re i

n t

he

ra

ng

e o

f g

ive

n d

ata

; th

e l

as

t p

red

icti

on

is

no

t re

as

on

ab

le b

ec

au

se

70

is

ou

t o

f th

e g

ive

n x

-va

lue

s a

nd

th

e n

um

be

r o

f s

ick

da

ys

ca

nn

ot

be

ne

ga

tiv

e.

Prac

tice

Co

rrela

tio

n a

nd

Lin

ear

Reg

ress

ion

11-7

[ 10,

65]

scl:

1 b

y [ 0

, 20]

scl:

1

[ 10,

65]

scl:

1 b

y [ –

10, 1

0] s

cl: 1

005_

042_

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CR

MC

11_8

9381

2.in

dd39

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12:3

9:24

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

11

40

Gle

ncoe

Pre

calc

ulus

1. C

OLL

EGE

The

data

in t

he t

able

re

pres

ent

the

Am

eric

an C

olle

ge T

est

(AC

T) c

ompo

site

sco

res

and

grad

e po

int

aver

ages

(GPA

) of 2

0 ra

ndom

ly s

elec

ted

stud

ents

aft

er t

heir

firs

t se

mes

ter

in

colle

ge. A

col

lege

cou

nsel

or w

ants

to

dete

rmin

e if

ther

e is

a c

orre

lati

on

betw

een

AC

T sc

ores

and

firs

t se

mes

ter

GPA

s.

AC

T2

71

81

71

5

GP

A3

.92

.93

.33

.0

AC

T2

22

01

72

1

GP

A3

.62

.72

.93

.4

AC

T2

51

72

51

8

GP

A3

.53

.14

.03

.0

AC

T2

31

92

02

9

GP

A3

.62

.63

.03

.4

AC

T2

32

82

22

0

GP

A1

.84

.03

.04

.0

a. C

alcu

late

and

inte

rpre

t th

e co

rrel

atio

n co

effic

ient

.

r ≈

0.4

52;

it s

ug

gests

th

at

the d

ata

h

as a

weak p

osit

ive c

orr

ela

tio

n

b. D

eter

min

e if

the

corr

elat

ion

coef

ficie

nt

is s

igni

fican

t at

the

1%

, 5%

, and

10%

le

vels

.

t ≈

2.1

5;

Becau

se t

< 2

.88,

t >

2.1

0,

an

d t

> 1

.73,

the c

orr

ela

tio

n

co

eff

icie

nt

is s

ign

ific

an

t at

the 5

%

an

d 1

0%

levels

.

c. G

raph

and

ana

lyze

the

res

idua

l plo

t.

T

he r

esid

uals

ap

pear

to b

e c

en

tere

d

ab

ou

t th

e r

eg

ressio

n lin

e y

= 0

, w

hic

h

mean

s a

lin

ear

reg

ressio

n m

od

el is

ap

pro

pri

ate

fo

r th

is d

ata

. T

here

ap

pears

to

be o

ne o

utl

ier

at

x =

23.

2.

SALE

S A

sal

es a

ssoc

iate

wan

ts t

o kn

ow

if th

ere

is a

rel

atio

nshi

p be

twee

n th

e av

erag

e nu

mbe

r of

tim

es h

is c

owor

kers

co

ntac

t cl

ient

s ea

ch m

onth

and

the

av

erag

e m

onth

ly s

ales

vol

ume

in

thou

sand

s of

dol

lars

. He

colle

cted

the

da

ta s

how

n in

the

tab

le.

Cli

en

ts2

12

34

85

04

6

Sa

les

30

30

95

11

08

0

Cli

en

ts1

25

51

45

01

6

Sa

les

15

13

02

59

03

0

a. M

ake

a sc

atte

r pl

ot o

f the

dat

a an

d id

enti

fy t

he r

elat

ions

hip.

The

n ca

lcul

ate

and

inte

rpre

t th

e co

rrel

atio

n co

effic

ient

.

T

here

ap

pears

to

be a

po

sit

ive l

inear

co

rrela

tio

n;

r ≈

0.9

73,

su

gg

esti

ng

a

str

on

g p

osit

ive l

inear

co

rrela

tio

n

b. D

eter

min

e if

the

corr

elat

ion

coef

ficie

nt is

sig

nific

ant

at t

he 1

%,

5%, a

nd 1

0% le

vels

. Exp

lain

you

r re

ason

ing.

t

≈ 1

1.9

; B

ecau

se t

> 3

.36,

t >

2.3

1,

an

d t

> 1

.86,

the c

orr

ela

tio

n

co

eff

icie

nt

is s

ign

ific

an

t at

all t

hre

e

levels

.

c. I

f the

cor

rela

tion

is s

igni

fican

t at

the

10

% le

vel,

find

the

leas

t-sq

uare

s re

gres

sion

equ

atio

n an

d in

terp

ret

the

slop

e an

d y-

inte

rcep

t in

con

text

.

y

= 2

.312x

- 1

3.9

58;

Th

e s

lop

e

su

gg

ests

th

at

every

clie

nt

co

nta

ct

per

mo

nth

co

rresp

on

ds t

o a

sale

s

incre

ase o

f 2.3

12 t

ho

usan

d d

olla

rs.

Th

e y

-in

terc

ep

t su

gg

ests

th

at

no

clie

nt

co

nta

ct

co

rresp

on

ds t

o a

sale

s

decre

ase o

f 13.9

58 t

ho

usan

d d

olla

rs.

11-7

Wor

d Pr

oble

m P

ract

ice

Co

rrela

tio

n a

nd

Lin

ear

Reg

ress

ion

[ 10,

34]

scl:

2 b

y [ –

2, 2

] scl:

1

[ 8, 6

0] s

cl: 2

by

[ –5,

150

] scl:

1

005_

042_

PC

CR

MC

11_8

9381

2.in

dd40

3/25

/09

12:3

9:30

PM


Lesson X-7

NA

ME

DA

TE

PE

RIO

D

Lesson 11-7

Ch

ap

ter

11

41

Gle

ncoe

Pre

calc

ulus

Mu

ltip

le R

eg

ress

ion

In m

ulti

ple

regr

essi

on, t

here

are

tw

o or

mor

e in

depe

nden

t va

riab

les

and

one

depe

nden

t va

riab

le. T

he m

ulti

ple

regr

essi

on e

quat

ion

is y

= a

+ b

1x1 +

b2x

2 + …

+ b

kxk,

whe

re

x 1, x 2,

…, x

k are

the

inde

pend

ent

vari

able

s.

The

stre

ngth

of t

he r

elat

ions

hip

betw

een

the

inde

pend

ent

vari

able

s an

d th

e de

pend

ent

vari

able

is m

easu

red

by t

he m

ulti

ple

corr

elat

ion

coef

fici

ent

R. T

his

valu

e ca

n ra

nge

from

0 t

o 1,

whe

re v

alue

s cl

oser

to

1 in

dica

te a

str

onge

r re

lati

onsh

ip t

han

thos

e cl

oser

to

0.

The

form

ula

for

a m

ulti

ple

corr

elat

ion

coef

ficie

nt w

ith

two

inde

pend

ent

vari

able

s is

R =

√

��

��

��

��

��

��

�

(r

y x 1 )

2 + (

r yx

2 ) 2 - 2

r yx1 ( r

yx2 ) (

r x 1x2 )

−−

√

��

��

1

- (

r x 1x2 ) 2

,

whe

re r

y x 1 is

the

cor

rela

tion

coe

ffici

ent

for

y an

d x 1;

r y x 2 is

the

cor

rela

tion

coe

ffici

ent

for

y an

d x 2

; and

r x 1x

2 is t

he c

orre

lati

on c

oeffi

cien

t fo

r x 1 a

nd x

2.

Hos

pita

l adm

inis

trat

ors

wis

h to

see

if a

nur

sing

ap

plic

ant’s

GP

A a

nd a

ge a

re r

elat

ed t

o th

e nu

rse’

s sc

ore

on t

he s

tate

nur

sing

boa

rd e

xam

. The

tab

le

show

s th

ese

stat

isti

cs f

or f

ive

nurs

es.

1. F

ind

the

valu

es o

f ry x

1 , r y x

2 , and

r x 1x

2 .

0.8

63

; 0

.69

3;

0.3

58

2. F

ind

and

inte

rpre

t R

. a

bo

ut

0.9

56

; th

ere

is

a s

tro

ng

lin

ea

r re

lati

on

sh

ip

am

on

g t

he

ap

pli

ca

nt’

s g

rad

e p

oin

t a

ve

rag

e,

ag

e,

an

d s

co

re o

n t

he

sta

te

bo

ard

ex

am

.

3. A

res

earc

her

is c

ondu

ctin

g a

stud

y to

see

if a

per

son’

s ag

e x 1 a

nd c

hole

ster

ol le

vel x

2 are

re

late

d to

his

or

her

syst

olic

blo

od p

ress

ure

y. F

ind

and

inte

rpre

t R

if r

y x 1 =

0.6

81,

r y x 2 =

0.8

72, a

nd r

x 1x2 =

0.7

46.

0.8

73

; S

inc

e R

is

fa

irly

clo

se

to

1,

we

ca

n s

ay

tha

t th

ere

is

a l

ine

ar

rela

tio

ns

hip

am

on

g a

ge

, c

ho

les

tero

l le

ve

l, a

nd

blo

od

p

res

su

re.

A t

rack

coa

ch w

ants

to

see

if t

he 5

k ti

mes

for

the

fi

rst

two

mee

ts o

f th

e se

ason

are

rel

ated

to

the

5k t

imes

at

the

stat

e m

eet.

The

tab

le s

how

s th

ese

stat

isti

cs, i

n m

inut

es, f

or f

ive

runn

ers.

4. F

ind

the

valu

es o

f r yx

1 , r yx

2 , and

r x 1x

2 .

0.8

16

; 0

.56

5;

0.8

00

5. F

ind

and

inte

rpre

t R

. a

bo

ut

0.8

29

; S

inc

e R

is

fa

irly

clo

se

to

1,

we

ca

n s

ay

th

at

the

re i

s a

lin

ea

r re

lati

on

sh

ip a

mo

ng

th

e t

hre

e r

ac

es

.

Enri

chm

ent

11-7

GP

A

x 1

Ag

e

x 2

Sta

te B

oa

rd

Ex

am

y

3.5

28

67

5

2.7

28

57

0

3.2

22

56

0

2.2

23

49

0

2.4

24

55

0

Me

et

1

x 1

Me

et

2

x 2

Sta

te M

ee

t

y

17

.41

6.5

16

.5

15

.41

6.1

15

.8

18

18

.21

6.9

16

.51

6.3

16

.5

17

.21

7.5

16

.1

005_

042_

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CR

MC

11_8

9381

2.in

dd41

3/25

/09

12:3

9:36

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

Chapter 11 Assessment Answer Key

Pdf Pass


Quiz 1 (Lessons 11-1 and 11-2) Quiz 3 (Lessons 11-5 and 11-6) Mid-Chapter TestPage 43 Page 44 Page 45

Quiz 2 (Lesson 11-3 and 11-4)

Page 43

Quiz 4 (Lesson 11-7)

Page 44

1.

2.

3.

4.

5.

D

-0.84 < z < 0.84

6.3%

35.8%

P(X < 15.5)

1.

2.

3.

4.

5.

0.607

9.61 < μ < 10.83

at least 35 boys

B

yes, t = 4.02, and 4.02 > 1.83.

1.

2.

3.

B

F

A

1.

2.

3.

4.

5.

[50, 100] scl: 10 by [0, 10] scl: 1

1.

2.

3.

4.

5.

4.

5.

6.

7.

[0, 15] scl: 1 by [0, 1] scl: 0.5

reasonably symmetric

2.57

C

use mean and std. dev. because data is symmetric;

mean: 76.9; std.dev.: 10.5

r ≈ 0.869; fairly strong positive linear correlationyes at all levels; t = 4.65 and t > 1.89, t > 2.36, and t > 3.5.

15

D

y = 1.07x - 1.05^

The median and spread is much greater for those who received the placebo.

z < -0.93 or z > 0.93

2.51 lb


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s

12.

13.

14.

15.

16.

17.

18.

G

D

G

D

J

B

G

B:

19.

20.

99.8%

F

D

Pdf Pass


Vocabulary Test Form 1Page 46 Page 47 Page 48

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

normal distribution

random variable

hypothesis test

confidence interval

t-distribution


z-value

As n increases, the distribution approaches a normal distribution, −

x approaches μ, and

σ −

x approaches σ −

√ � n .

A distribution that connects all possible values of X with their corresponding probabilities.

correlation coefficent

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

A

H

B

H

A

H

C

F

A

J

A


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.


12.

13.

14.

15.

16.

17.

18.

J

A

H

B

H

B

J

1.36

J

D

B:

19.

20.

1.22 B:

18.

19.

20. F

D

F

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

D

H

C

G

A

G

D

F

A

G

A

Pdf Pass


Form 2A Form 2BPage 49 Page 50 Page 51 Page 52

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

C

J

A

H

B

G

D

G

A

G

B

12.

13.

14.

15.

16.

17.

H

B

F

D

H

D


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s

Pdf Pass


Form 2CPage 53 Page 54

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

positively skewed

mean: 31.78; median: 21

6.1

2.40

-$4

201.125

about 29.7%

about 89

14.7%

[10, 90] scl: 5 by [0, 25] scl: 1

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

33.7%

0.14

5.06–5.34

28 clocks

H0: μ ≤ 0.910;

Ha: μ > 0.910

(claim)

crit. value: 1.70 test stat: 2.52

yes; 2.52 > 1.65

-0.824

0.8

yes; t = -4.6 and -4.6 < -2.23

y = -0.09x + 106.39^


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.


Pdf Pass


Form 2DPage 55 Page 56

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

negatively skewed

mean: 41.9; median: 44

3.95

1.85

-$1.75

134.075

24.8%

about 6 days

69.2%

[0, 100] scl: 10 by [0, 15] scl: 1

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

72.5%

0.06

2.54–2.66

44 people

H0: μ = 0.045 (claim);

Ha: μ ≠ 0.045

crit. value: -2.2 test stat: -0.94

no; -0.94 > -2.2

-0.896

1.41

yes; t = -6.4 and -6.4 < -2.23

y = -2.18x + 9.17^


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Form 3Page 57 Page 58

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Top 50% of NH speeds are greater than top 75% of MA speeds. NH has greater range and maximumsomewhat negatively skewed

47

621.4; 25

$105

12.9225

12%

about 4 sofas

42.1%

[55, 95] scl: 5 by [0, 1] scl: 0.5

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

40.8%

5.1

444.9–455.1

26 batteries

H0: μ ≤ 10 (claim);

Ha: μ > 10

crit. value: -1.73; test stat: -0.895; P-value: 0.81

no; -0.895 > -1.73 and 0.81 > 0.05.

0.764; slight positive linear

correlation

7 people

0.6395

Signif at 1% only; t = 3.348 and

t > 1.86 and 2.306, but t < 3.355


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Page 59, Extended-Response Test Sample Answers

1a. negatively skewed

1b. Use the five-number summary since the data are skewed: 64, 80, 87, 93, 101. While weights range from 64 to 101 pounds, the median weight is 87 pounds. The middle half of the data varies by 13 pounds.

2a. about 13 students; 94 is 1.5 standard deviations above the mean. The area to the right of 1.5 is about 6.7% and 6.7% of 200 is 13.4.

2b. 88; First, find the z-value, which has 70% of the distribution below it. This is about 0.524. Then solve the formula for z-values to find X.

2c. 13.8%; Using the Central Limit Theorem to adjust the standard deviation, 83 is 1.972 standard deviations below the mean and 84 is 0.986 standard deviations below the mean. The area between -1.972 and -0.986 is about 13.8% of the distribution.

3. H0: μ = 14; Ha: μ ≠ 14; The critical values are -2.14 and 2.14. The test statistic is t = -2.3. Because -2.3 < -2.14, it falls in the critical region. The evidence supports rejecting the company’s claim of a mean of 14 ounces.

4a. 0.965; There is a strong positive linear correlation.

4b. It is significant at all three levels because it is significant at 1%: t = 11.12 and the positive critical t-value for 1%, 9 degrees of freedom is 3.25.

4c. y^ = 0.6x + 2.53; The slope means that for every increase of one ounce in volume, the cost increases by $0.60. The intercept means 0 ounces of food cost $2.53, which does not make sense for this situation.

[60, 105] scl: 5 by [0, 12] scl: 1


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Standardized Test PracticePage 60 Page 61

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J


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Standardized Test Practice (continued)Page 62

19.

20.

21.

22.

23.

24.

25a.

25b.

25c.

25d.

25, 34, 43

1; 2π; 1 −

2π

; - π

−

2 ; 2

0; yes

(3, -1), (6, -1), y = 0, x = 0

0, π, 3π

−

4 ,

7π

−

4

H0: μ ≤ 20,000;

Ha: u > 20,000

(claim)

2.326

4.941

yes; 4.941 > 2.326

⎡

⎢

⎣

-15

26

-17

29 ⎤

⎦


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