Elementary Statistics

5 Normal Probability Distributions

Larson Farber

Introduction to Introduction to Introduction to Introduction to

Section 5.1

Introduction to Introduction to Introduction to Introduction to Normal DistributionsNormal DistributionsNormal DistributionsNormal Distributions

Properties of a Normal Distribution

x• The mean, median, and mode are equal

• Bell shaped and is symmetric about the mean

• The total area that lies under the curve is one or 100%

x

x

Inflection pointInflection point

Properties of a Normal Distribution

• As the curve extends farther and farther away from the mean, it gets closer and closer to the x-axis but never

touches it.• The points at which the curvature changes are calle d

inflection points. The graph curves downward betwee n the inflection points and curves upward past the inflec tion

points to the left and to the right.

x

Means and Standard Deviations

12 15 1810 11 13 14 16 17 19 20

Curves with different means, same standard deviatio n

2012 15 1810 11 13 14 16 17 19 21 229

12 15 1810 11 13 14 16 17 19 20

Curves with different means, different standard dev iations

Empirical Rule

About 68% of the area lies within 1 standard deviation of the mean

68%

About 95% of the area lies within 2 standard

deviations

About 99.7% of the area lies within 3 standard deviations of the mean

4.2 4.5 4.8 5.13.93.63.3

Determining Intervals

x4.2 4.5 4.8 5.13.93.63.3

An instruction manual claims that the assembly time for a product is normally distributed with a mean of 4.2 hours

and standard deviation 0.3 hour. Determine the interval in which 95% of the assembly times fall.

4.2 – 2 (0.3) = 3.6 and 4.2 + 2 (0.3) = 4.8.95% of the assembly times will be between 3.6 and 4.8 hrs.

95% of the data will fall within 2 standard deviations of the mean.

The Standard The Standard The Standard The Standard

Section 5.2

The Standard The Standard The Standard The Standard Normal DistributionNormal DistributionNormal DistributionNormal Distribution

The Standard ScoreThe standard score , or z-score , represents the number of standard deviations a random variable x falls from the mean.

The test scores for a civil service exam are normally The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of:(a) 161 (b) 148 (c) 152

(a) (b) (c)

The Standard Normal Distribution

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

Using z-scores any normal distribution can be transformed into the standard normal distribution.

–4 –3 –2 –1 0 1 2 3 4 z

Cumulative Areas

Thetotalarea

underthe curve

.

• The cumulative area is close to 1 for z-scores close to 3.49.

0 1 2 3–1–2–3 zis one.

• The cumulative area is close to 0 for z-scores close to –3.49.• The cumulative area for z = 0 is 0.5000.

Find the cumulative area for a z-score of –1.25.

Cumulative Areas

0.1056

0 1 2 3–1–2–3 z

Read down the z column on the left to z = –1.25 and across to the column under .05. The value in the cell is 0.10 56, the

cumulative area.

The probability that z is at most –1.25 is 0.1056.

Finding ProbabilitiesTo find the probability that z is less than a given value, read the cumulative area in the table corresponding to that z-score.

Find P(z < –1.45).

P (z < –1.45) = 0.0735

0 1 2 3–1–2–3 z

Read down the z-column to –1.4 and across to .05. The cumulative area is 0.0735.

P (z < –1.45) = 0.0735

Finding ProbabilitiesTo find the probability that z is greater than a given value, subtract the cumulative area in the table from 1.

Find P(z > –1.24).

0 1 2 3–1–2–3 z

P(z > –1.24) = 0.8925

The cumulative area (area to the left) is 0.1075. So the area to the right is 1 – 0.1075 = 0.8925.

0.10750.8925

Finding ProbabilitiesTo find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from

the larger.

Find P(–1.25 < z < 1.17).

1. P(z < 1.17) = 0.8790 2. P(z < –1.25) = 0.1056

3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734

0 1 2 3–1–2–3 z

Summary

0 1 2 3-1-2-3 zTo find the probability is greater

To find the probability that z is lessthan a given value, read thecorresponding cumulative area.

0 1 2 3-1-2-3 z

To find the probability is greater than a given value, subtract the cumulative area in the table from 1.

0 1 2 3-1-2-3 z

To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger.

Normal DistributionsNormal DistributionsNormal DistributionsNormal Distributions

Section 5.3

Normal DistributionsNormal DistributionsNormal DistributionsNormal DistributionsFinding ProbabilitiesFinding ProbabilitiesFinding ProbabilitiesFinding Probabilities

Probabilities and Normal Distributions

If a random variable, x is normally distributed, the probability that x will fall within an interval is equal to the area under the curve in the interval.IQ scores are normally distributed with a mean of 1 00 and a standard deviation of 15. Find the probabilit y that a person selected at random will have an IQ score les s than 115.

115100

than 115.

To find the area in this interval, first find the s tandard score equivalent to x = 115.

Probabilities and Normal Distributions

115100Standard Normal

Find P(x < 115).

Normal Distribution

SA

ME

0 1

Find P(z < 1).

Standard Normal Distribution

P(z < 1) = 0.8413, so P(x <115) = 0.8413

SA

ME

SA

ME

Monthly utility bills in a certain city are normall y distributed with a mean of $100 and a standard devi ation of $12. A utility bill is randomly selected. Find t he probability it is between $80 and $115.

P(80 < x < 115)

Normal Distribution

Application

P(80 < x < 115)

P(–1.67 < z < 1.25)0.8944 – 0.0475 = 0.8469

The probability a utility bill is between $80 and $115 is 0.8469.

Normal DistributionsNormal DistributionsNormal DistributionsNormal Distributions

Section 5.4

Normal DistributionsNormal DistributionsNormal DistributionsNormal DistributionsFinding ValuesFinding ValuesFinding ValuesFinding Values

From Areas to z-ScoresFind the z-score corresponding to a cumulative area of 0.9803 .

z = 2.06 correspondsroughly to the

98th percentile.

zLocate 0.9803 in the area portion of the table. Read the

values at the beginning of the corresponding row an d at the top of the column. The z-score is 2.06.

–4 –3 –2 –1 0 1 2 3 4

0.9803

Finding z-Scores from Areas

Find the z-score corresponding to the 90th percentile.

z0

.90

z0

The closest table area is .8997. The row heading is 1.2 and column heading is .08. This corresponds to z = 1.28.

A z-score of 1.28 corresponds to the 90th percentile.

Find the z-score with an area of .60 falling to its right.

.60.40

z

Finding z-Scores from Areas

0 zz

With .60 to the right, cumulative area is .40. The closest area is .4013. The row heading is 0.2 and column heading is .05. The z-score is 0.25.

A z-score of 0.25 has an area of .60 to its right. It also corresponds to the 40th percentile

Find the z-score such that 45% of the area under the curve falls between – z and z.

.45.275.275

Finding z-Scores from Areas

0 z–z

The area remaining in the tails is .55. Half this a rea isin each tail, so since .55/2 = .275 is the cumulati ve area for the negative z value and .275 + .45 = .725 is the cumulative area for the positive z. The closest table area is .2743 and the z-score is 0.60. The positive zscore is 0.60.

From z-Scores to Raw Scores

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard score of:

To find the data value, x when given a standard score, z:

Find the test score for a person with a standard score of:(a) 2.33 (b) –1.75 (c) 0

(a) x = 152 + (2.33)(7) = 168.31

(b) x = 152 + (–1.75)(7) = 139.75

(c) x = 152 + (0)(7) = 152

Finding Percentiles or Cut-off ValuesMonthly utility bills in a certain city are normall y distributed with a mean of $100 and a standard deviation of $12 . What is the smallest utility bill that can be in the top 10 % of the bills?

10%90%

$115.36 is the smallestvalue for the top 10%.

10%

Find the cumulative area in the table that is close st to 0.9000 (the 90th percentile.) The area 0.8997 corr esponds to a z-score of 1.28.

x = 100 + 1.28(12) = 115.36.

z

To find the corresponding x-value, use

The Central Limit The Central Limit The Central Limit The Central Limit

Section 5.5

The Central Limit The Central Limit The Central Limit The Central Limit TheoremTheoremTheoremTheorem

Sampling DistributionsA sampling distribution is the probability distribu tion of a sample statistic that is formed when samples of siz e n are repeatedly taken from a population. If the sample s tatistic is the sample mean, then the distribution is the sampl ing distribution of sample means.

Sample

Sample

Sample

The sampling distribution consists of the values of the sample means,

SampleSample

Sample

The Central Limit TheoremIf a sample n � 30 is taken from a population withany type distribution that has a mean =and standard deviation =

x

the sample means will have a normal distributionand standard deviation

The Central Limit Theorem

x

If a sample of any size is taken from a population with a normal distribution with mean = and standard

deviation =

the distribution of means of sample size n, will be normalwith a mean

standard deviation

x

ApplicationThe mean height of American men (ages 20-29) is inches. Random samples of 60 such men are selected. Find the mean and standard deviation (standard error) of the sampling distribution.

Distribution of means of sample size 60 , will be normal.

mean

Standard deviation

69.2

Interpreting the Central Limit Theorem

The mean height of American men (ages 20-29) is = 69.2”. If a random sample of 60 men in this age grou p is selected, what is the probability the mean height for the sample is greater than 70”? Assume the standard deviation is 2.9”. Since n > 30 the sampling distribution of will be norma l

Find the z-score for a sample mean of 70:

standard deviation

mean

Since n > 30 the sampling distribution of will be norma l

Interpreting the Central Limit Theorem

2.14z

There is a 0.0162 probability that a sample of 60 men will have a mean height greater than 70”.

Application Central Limit TheoremDuring a certain week the mean price of gasoline in California was $1.164 per gallon. What is the probability that the mean price for the sample of 38 gas stations in California is betw een $1.169 and $1.179? Assume the standard deviation = $0.049.

mean

Since n > 30 the sampling distribution of will be norma l

standard deviation

mean

Calculate the standard z-score for sample values of $1.169 and $1.179.

z

Application Central Limit Theorem

P( 0.63 < z < 1.90)

= 0.9713 – 0.7357

= 0.2356

.63 1.90

z

The probability is 0.2356 that the mean for the

sample is between $1.169 and $1.179.

Normal Approximation Normal Approximation Normal Approximation Normal Approximation

Section 5.6

Normal Approximation Normal Approximation Normal Approximation Normal Approximation to the Binomialto the Binomialto the Binomialto the Binomial

Binomial Distribution Characteristics

• There are a fixed number of independent trials. ( n)

• Each trial has 2 outcomes, Success or Failure.

• The probability of success on a single trial is p and

the probability of failure is q. p + q = 1the probability of failure is q. p + q = 1

• We can find the probability of exactly x successes out

of n trials. Where x = 0 or 1 or 2 … n.

• x is a discrete random variable representing a count

of the number of successes in n trials .

Application34% of Americans have type A+ blood. If 500 Americans are sampled at random, what is the probability at least 300 have type A+ blood?

Using techniques of Chapter 4 you could calculate t he probability that exactly 300, exactly 301… exactly 5 00 Americans have A + blood type and add the probabilities.Americans have A + blood type and add the probabilities.

Or…you could use the normal curve probabilities to approximate the binomial probabilities.

If np � 5 and nq � 5, the binomial random variable x is approximately normally distributed with mean

Why Do We Require np � 5 and nq � 5?

0 1 2 3 4 5

n = 5p = 0.25, q = .75

np =1.25 nq = 3.75

n = 20p = 0.25

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

p = 0.25np = 5 nq = 15

n = 50p = 0.25

np = 12.5 nq = 37.5

0 10 20 30 40 50

Binomial Probabilities

The binomial distribution is discrete with a probab ility histogram graph. The probability that a specific va lue ofx will occur is equal to the area of the rectangle wi th midpoint at x.If n = 50 and p = 0.25 find

Add the areas of the rectangles with midpoints atAdd the areas of the rectangles with midpoints atx = 14, x = 15, x = 16.

14 15 16

0.111 0.0890.065

0.111 + 0.089 + 0.065 = 0.265

Correction for Continuity Use the normal approximation to the binomial to find .

14 15 16

Values for the binomial random variable xare 14, 15 and 16.

Correction for Continuity Use the normal approximation to the binomial to find .

14 15 16

The interval of values under the normal curve is

To ensure the boundaries of each rectangle are included in the interval, subtract 0.5 from a left-hand boundary and add 0.5 to a right-hand boundary.

Normal Approximation to the Binomial

Use the normal approximation to the binomial to fin d

Find the mean and standard deviation using binomial distribution formulas.

.

Adjust the endpoints to correct for continuity P .

Convert each endpoint to a standard score.

ApplicationA survey of Internet users found that 75% favored government regulations of “junk” e-mail. If 200 Int ernet users are randomly selected, find the probability t hat fewer than 140 are in favor of government regulation.

Since np = 150 � 5 and nq = 50 � 5 use the normal approximation to the binomial.

The binomial phrase of “fewer than 140” means0, 1, 2, 3…139.

Use the correction for continuity to translate to t he continuous variable in the interval . Find P( x < 139.5).

ApplicationA survey of Internet users found that 75% favored government regulations of “junk” e-mail. If 200 Int ernet users are randomly selected, find the probability t hat fewer than 140 are in favor of government regulation.

Use the correction for continuity P( x < 139.5).

P( z < -1.71) = 0.0436

The probability that fewer than 140 are in favor of government regulation is approximately 0.0436.