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Stat Camp for theMBA Program

Daniel SolowLecture 1 

Exploratory Data Analysis

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What is Statistics?Statistics is the art and science of collecting, analyzing, presenting and interpreting data, which are information you have or can obtaininformation you have or can obtain.Business Statistics helps managers make more informed decisions.

Descriptive Statistics

Inferential Statistics

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Describes properties of large data sets with a few summary numbers or graphs.

Helps you make decisions when you can obtain only a portion of the desired data.

Where Is Statistics Needed?• Market survey/research

A k t k t h i 19% ith– A market survey says your market share is 19% with margin of error of 3%. What does this mean?

• Manpower planning– A bank wants to know how many tellers they should

have during the busiest time on a given day?• Quality control

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Quality control– A machine is set to produce parts with a length of 2

inches. A part just produced has a length of 2.1 inches. Should you stop the production and reset the machine?

Where Is Statistics Needed?

• ForecastingH h l I t t t ?– How much sales can I expect next quarter?

• Premiums and Warranties– What should the insurance premium be for a

particular class of customers?– You have just introduced a new automobile tire in

the market. How many miles of warranty should you offer on this product?

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you offer on this product?• Fun and Games

– I bet that “this class has at least two persons with the same birthday (day and month)”. Should you take this bet?

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Example 1: Suppose you want to know the average length of iron bars produced by your machine.

Inferential Statistics

Population: All iron bars produced on that machine

In such situations, there are a large number of items you are interested in, which is called the population.

Population:Length of the bar.

Average length of all iron bars

All iron bars produced on that machine.Number of interest for each item:Parameter: = .

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Every item in the population has a number of interest.You want to know the value of one number associated with the whole population, called the parameter.

Inferential Statistics• Example 2: You want to know your “market

share” (the fraction of customers that purchaseshare (the fraction of customers that purchase your product).– Population:– Number Associated with Each Item in the

Population:

All people that buy this product.

1 if that person buys your product

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– Parameter: = fraction of the population thatbuys your product.

1, if that person buys your product0, if that person does not buy your product

Inferential Statistics• In general, you can never know the value of the

parameter of a population (why?).– Because there are too many items in the population.

• In such cases, you should compute your best estimate (statistic) from a “manageable” subset of data (sample) collected randomly from the population.

parameter is unknown

Population

statisticbest

estimate

Random Sample

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unknown

statisticestimate

sample

Example 1 (Iron Bars):– Collect a sample of n iron bars (iron bar i has a length xi).– Compute the following statistic (sample mean):

Inferential Statistics

Compute the following statistic (sample mean):

Example 2 (Market Share):– Collect a sample of n people from the population of people

that buy the product (each person i has a value xi of 1 or 0).C h f ll i i i ( l )

.... mean sample 21n

xxxX n

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– Compute the following statistic (sample proportion):

. proportion sample ny

s

y = number in the sample who buy your product

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Data

• Data are information that are collected, summarized and analyzed for presentation and interpretation.y p p

• Cross-Sectional: Data collected at the same point in time.

• Time Series: Data collected over several time periods.

• Example: The Data Files web site on the first page

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Example: The Data Files web site on the first page of these notes has the following file shadow02.xls with data on certain stocks.

Company ExchangeTicker Symbol

Market Cap ($ millions)

Price/Earnings Ratio

1 DeWolfe Companies AMEX DWL 36.40 8.402 North Coast Energy OTC NCEB 52.50 6.203 Hansen Natural Corp. OTC HANS 41.10 14.604 MarineMax, Inc. NYSE HZO 111.50 7.205 Nanometrics Incorporated OTC NANO 228.60 38.00

Exchange Classes:

OTC

AMEX

Qualitative Quantitative

6 TeamStaff, Inc. OTC TSTF 92.10 33.507 Environmental Tectonics AMEX ETC 51.10 35.808 Measurement Specialties AMEX MSS 101.80 26.809 SEMCO Energy, Inc. NYSE SEN 193.40 18.70

10 Party City Corporation OTC PCTY 97.20 15.9011 Embrex, Inc. OTC EMBX 136.50 18.9012 Tech/Ops Sevcon, Inc. AMEX TO 23.20 20.7013 ARCADIS NV OTC ARCAF 173.40 8.8014 Qiao Xing Universal Tele. OTC XING 64.30 22.1015 Energy West Incorporated OTC EWST 29.10 9.7016 Barnwell Industries Inc AMEX BRN 27 30 7 40

NYSE

Mkt Cap Classes:

0-50

50 100

10

16 Barnwell Industries, Inc. AMEX BRN 27.30 7.4017 Innodata Corporation OTC INOD 66.10 11.0018 Medical Action Industries OTC MDCI 137.10 26.9019 Instrumentarium Corp. OTC INMRY 240.90 3.6020 Petroleum Development OTC PETD 95.90 6.1021 Drexler Technology Corp. OTC DRXR 233.60 45.6022 Gerber Childrenswear Inc. NYSE GCW 126.90 7.9023 Gaiam, Inc. OTC GAIA 295.5 68.224 Artesian Resources Corp. OTC ARTNA 62.80 20.50 25 York Water Company OTC YORW 92.20 22.90

50-100

100-150150-200

200-250

Data SetsAs shown on the previous slide, • Elements: Entities on which data are

collected (the 25 different companies in the shadow-stocks example).

• Variable: A characteristic of the elements you are interested in and whose value varies (Exchange, Ticker Symbol, and so on).

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( g , y , )• Class: A group consisting of one or more

values for a variable.

Types of Statistical Data• Qualitative (non-numeric)

– Nominal – values cannot be compared in terms pof order (color, stock exchange, and so on)

– Ordinal – values can be compared in terms of order (rank, quality level, satisfaction)

• Quantitative (numeric)– Interval – difference between values is

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meaningful (birth year, customer arrival time)– Ratio – ratio of two values is meaningful

(income, age, height, inventory level)

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Example: MBA SURVEY Identify the Data Type

• What is your height in inches? RATIO

• What is your gender?

• Attitude toward this Course on 1 to 6 scale:

1 = seriously worried (strongly dreading this),

6 th d & fid t ( t t t)

NOMINAL

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6 = enthused & confident (eager to start)

• Do you smoke?

• WWW purchases (in $) over past year.

ORDINAL

NOMINAL

RATIO

Descriptive Statistics• Descriptive statistics is the art of

summarizing a data set using either:– Graphical Methods (Charts)– Graphical Methods (Charts)

• Note: Software packages have extensive graphical capabilities.

– Numerical Methods• Note: Software packages will do this for you too.

– Used all the time in annual reports, news

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articles, research studies.– Different for qualitative and quantitative data.

Summarizing Qualitative Data• File SoftDrink.xls

Coke Classic Sprite Pepsi-Cola Frequency DistributionVariable: Soft Drink

Frequency Distribution: A table listing the number of elements in each class.

Coke Classic Sprite Pepsi ColaDiet Coke Coke Classic Coke ClassicPepsi-Cola Diet Coke Coke ClassicDiet Coke Coke Classic Coke ClassicCoke Classic Diet Coke Pepsi-ColaCoke Classic Coke Classic Dr. PepperDr. Pepper Sprite Coke ClassicDiet Coke Pepsi-Cola Diet CokePepsi-Cola Coke Classic Pepsi-ColaPepsi-Cola Coke Classic Pepsi-Cola

Frequency Distribution

Value Frequ-ency

Coke Classic 19Diet Coke 8D P 5

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Coke Classic Coke Classic Pepsi-ColaDr. Pepper Pepsi-Cola Pepsi-ColaSprite Coke Classic Coke ClassicCoke Classic Sprite Dr. PepperDiet Coke Dr. Pepper Pepsi-ColaCoke Classic Pepsi-Cola SpriteCoke Classic Diet Coke

Dr. Pepper 5Pepsi-Cola 13Sprite 5Total 50

Using SPSS for Frequency  Table(See the files UsingSPSS_Intro.ppt and UsingSPSS_DescriptiveStats.ppt)

To Open an EXCEL file:

Click on

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•Click on file/open/data.

•Under Files of Type use .xls files.

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SPSS OutputDRINK

19 38 0 38 0 38 0Coke ClassicValidFrequency Percent Valid Percent

CumulativePercent

19 38.0 38.0 38.08 16.0 16.0 54.05 10.0 10.0 64.0

13 26.0 26.0 90.05 10.0 10.0 100.0

50 100.0 100.0

Coke ClassicDiet CokeDr. PepperPepsi-ColaSpriteTotal

Valid

The relative frequency table shows the proportion

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q y p p(or fraction) of elements in each class. You can display both the frequency and relative frequency tables in a graphical form for easy visualization.

•Click on G h /L

Using SPSS for a Bar GraphBar Graph: A graph with the classes on the x-axis and the frequencies (or percentages) on the y-axis.

Graphs/Legacy Dialogs/Bar.

•Click on Simple then Define.

•Drag the var. to the Category

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the Category axis and click either N of Cases or % of Cases.

SPSS Output

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•Click on

Using SPSS for a Pie ChartPie Chart: A circle having one “slice” for each class, with the size of each slice proportional to the relative frequency of that value..

•Click on Graphs/Legacy Dialogs/Pie.

•Click Define

•Move the var. into the Slice By

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into the Slice By box and click % of Cases.

•Click OK

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SPSS Output

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Summarizing Quantitative Data• With quantitative data, the classes have to be

determined by the statistician. Given the minimum and maximum data values:and maximum data values:– Determine the number of non-overlapping classes

(usually 5 – 20).• Too few classes: variation does not show.• Too many classes: too much detail.

– The class widths and class limits are then determined from the number of classes

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from the number of classes.

min

[ ][ ][ ][ ][ ]

width max

lower limit upper limit

Graphical Methods for Summarizing Quantitative Data

• Tabular Summaries– Frequency Distributions

• Number of items in each class• Relative Frequency (Percentage)• Cumulative (everything up to a certain value)

• Graphical Summaries

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p– Histograms (like a bar chart)

Example: Audit Times

• File audit.xls• Here 5 classes are used so

min• Here, 5 classes are used, so

max

Class Width = (max – min) / classes= (33 – 12) / 5= 4.2 5 (round up)

Class Limits shows 10 14

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maxClass Limits shows the smallest and largest values in the class.

10-1415-1920-2425-2930-34

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Frequency Table• The frequency table is constructed by

counting how many data items fall withincounting how many data items fall within each class (relative frequency table for percentages).

Audit time (days) Frequency Rel. Frequency (%)

10-14 4 20%

25

15-19 8 40%

20-24 5 25%

25-29 2 10%

30-34 1 5%

Histogram

• A histogram is a plot of a frequency distribution.– Classes on the x-axis.C sses o e s.– Frequencies or relative frequencies on the y-axis.

• Similar to bar graph, only now the bars are not separated.

• In SPSS: Choose Graph/Legacy Dialogs/Histogram, move the variable to the Variable box, and then customize the plot.

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p• In EXCEL: First create a column of “bins” (upper

class limits), then choose Tools/Data Analysis/Histogram.

Histogram of Audit Times

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EXCEL Histogram of Audit Times

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Numerical Summaries of Data• Location, Average, Central Tendency

– Mean– Median, Percentiles, Quartiles– Mode

• Variation (how spread out the numbers are)– Range– Variance, Standard Deviation

• Shape

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• Shape– Skewness

MEAN

• MEAN = Arithmetic Average

valuesof# valuesof sum =Mean X

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Example: Invention Development Time (Develop.xls)

Invention     Development                  Time

Automatic Transmission 16B ll i t P 7

Invention Development  Time

Radar 35Ballpoint Pen 7Filter Cigarettes 2Frozen Foods 15Helicopter 37Instant Coffee 22Minute Rice 18Nylon 12Photography 56

Radio 24Roll-On Deodorant 7Telegraph 18Television 63Transistor 16Video Cassette Recorder 6Xerox Copying 15Zipper 30

399301562716

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167.2218399

1830156...2716 X

In Excel: AVERAGE(range)

An invention on average takes 22.167 years to develop.

MEDIAN  (splits data in half)

• MEDIAN = middle value when data values are sorted from low to high...– 50% of values are below the median, 50% are

above the median.– If sample size (n) is even, the median is the mean

of the two middle values.

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• What is the median development time?

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Example: Invention Development Time

Invention Concept Realize DevTimeFilter Cigarettes 1953 1955 2Video Cassette 1950 1956 6 n

Median = (16+18)/2 17

Video Cassette 1950 1956 6Ballpoint Pen 1938 1945 7Roll-on Deodorant 1948 1955 7Nylon 1927 1939 12Frozen Foods 1908 1923 15Xerox Copying 1935 1950 15Automatic Transmissi 1930 1946 16Transistor 1940 1956 16Minute Rice 1931 1949 18Telegraph 1820 1838 18

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18 nn

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= 17Telegraph 1820 1838 18Instant Coffee 1934 1956 22Radio 1890 1914 24Zipper 1883 1913 30Radar 1904 1939 35Helicopter 1904 1941 37Photography 1782 1838 56Television 1884 1947 63

In Excel:MEDIAN(range)

Mean vs. Median

• The mean is the most commonly used measure f l tiof location.

• However the mean is affected by extremely large or small values.

• In those cases the median may be a more reliable measure of location

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reliable measure of location.

Example: Salaries

M 65 400Employee Salary • Mean = 65,400

• Median = 32,000

Employee Salary

John 30,000

Doe 32,000

Smith 32,000

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Perry 33,000

Sweeney 200,000

Histogram

Example: Invention Development Time

Median = 17

12345

Freq

uenc

y

Median 17

0

5 10 15 20 25 30 35 40 45 50 55 60

Mor

e

Bin

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Mean = 22.167

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SYMMETRIC DATA

50% 50%

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Mean = Median

MedianMean

RIGHT SKEWED DATA

d

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Long Right Hand TailMean > Median

Median Mean

LEFT SKEWED DATA

MedianMean

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Long Left Hand TailMean < Median

MedianMean

PercentilesThink about your numerical data values lying on a line:

pth per-centile

The p-percentile is a number such that:

At least p % are ≤

•At least p% of your data values are ≤ that number and

At least 100 p % are ≥

•At least (100 p)% of your data values are ≥ that

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Example: The 90th percentile on the GMAT is a score so that at least 90% of people’s GMAT scores are ≤ that number and at least 10% are ≥ that number .

At least (100 p)% of your data values are ≥ that number.

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Quartiles• Q1 = First quartile = 25th percentile = a value so

that at least 25% of the elements are that value and at least 75% are ≥ that value.

• Q2 = Second quartile = 50th percentile = = a value so that at least 50% of the elements are that value and at least 50% are ≥ that value

• Q3 = Third quartile = 75th percentile = a value so that at least 75% of the elements are that value

= the median..

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that at least 75% of the elements are that value and at least 25% are ≥ that value.

Percentiles in EXCEL: (file salary.xls)

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Percentiles in SPSS(File salary.xls)

Analyze; Descriptive Statistics; 123 Frequencies; then move the desired variable to the Variable(s) box; then click on Statistics; then click Percentile(s) and type your desired percentiles and Add; then click Continue and OK.

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MODE

• The mode of a variable is the value or categoryThe mode of a variable is the value or category that occurs most often in the batch of data.

• A data set can have more than one mode (bimodal, trimodal).

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Example: Invention Development Time

Modes: 7, Invention Concept Realize DevTimeFilter Cigarettes 1953 1955 2Vid C tt 1950 1956 6 15,

16,18

Video Cassette 1950 1956 6Ballpoint Pen 1938 1945 7Roll-on Deodorant 1948 1955 7Nylon 1927 1939 12Frozen Foods 1908 1923 15Xerox Copying 1935 1950 15Automatic Transmissi 1930 1946 16Transistor 1940 1956 16Minute Rice 1931 1949 18

In Excel:MODE(range),

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Telegraph 1820 1838 18Instant Coffee 1934 1956 22Radio 1890 1914 24Zipper 1883 1913 30Radar 1904 1939 35Helicopter 1904 1941 37Photography 1782 1838 56Television 1884 1947 63

( g ),which returnsonly one of

these values.

Do It Yourself Example: Blood Problem

Suppose that the number of units per day of whole blood used in transfusions at a hospital over theblood used in transfusions at a hospital over the previous 11 days is:25, 18, 61, 12, 18, 15, 20, 25, 17, 19, 28.Use the file blood.xls and Excel to:•Find and interpret the mean, median and mode(s).

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Is the Mean Enough? In the Blood Problem, an average of 23.45 pints of

blood are used on a day. Question: Does this mean you should have exactly

23.45 pints of blood available? Answer: Because the amount of blood you need

varies, that is, there is variation in the blood data. Question: How much variation is there?

No. Why not?

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Question: How much variation is there? Answer: What is needed is a numerical value to

represent how much variation there is in the data. Example: Range = Largest Value – Smallest Value

Variance• Variance is a number ≥ 0 that measures how close

the data values are to the mean .

V i ll V i l• Variance is generally a relative measure.• More reliable measure of variation than the range.• Uses all the data.• There are two different formulas, depending on

µ Var. is small µ Var. is larger

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whether you are computing the population variance or sample variance (see the handout formulas.pdf).

• Consider the following example for managing the amount of blood at a hospital (file blood.xls).

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Example: Blood Problem (blood.xls)

Day Pints Used Mean Deviation (Dev.)^21 25 23.4545455 1.545454545 2.388429752 18 23.4545455 -5.454545455 29.75206613 61 23.4545455 37.54545455 1409.661164 12 23.4545455 -11.45454545 131.2066125 18 23.4545455 -5.454545455 29.75206616 15 23.4545455 -8.454545455 71.47933887 20 23.4545455 -3.454545455 11.93388438 25 23.4545455 1.545454545 2.388429759 17 23 4545455 6 454545455 41 661157

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2S

9 17 23.4545455 -6.454545455 41.66115710 19 23.4545455 -4.454545455 19.842975211 28 23.4545455 4.545454545 20.661157

Population Variance = 160.975207Sample Variance = 177.072727

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Population Variance• = population mean• xi = value of the ith item

( ) d i i f th i f• (xi –) = deviation of ith item from • (xi –)2 = square deviation of ith item• Variance = average of the square deviations:

22 21 22 ... Nx x x

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• In Excel: VAR.P(range)

1 22 N

N

Sample Variance

X• xi = value of the ith item• = sample mean

• (xi – )2 = square deviation • (xi – ) = deviation of ith item from X

X

xi value of the i item

• Sample Variance = ... 22

12 XxXxS n

X

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1

12

nS n

• In Excel: VAR.S(range)

Standard Deviation• Square root of the variance.• Expressed in the same units as the data.• More intuitive measure of variabilityMore intuitive measure of variability.• Blood Problem

– Sample Variance = S2 = 177.07– Sample Standard Deviation = S = = 13.31

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• In Excel: Sample Std. Dev. = STDEV.S(range)Pop. Std. Dev. = STDEV.P(range)

• Under circumstances you will learn soon, the std. dev. has a useful interpretation)

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Using EXCEL and SPSS to Compute Descriptive Statistics

•Both EXCEL and SPSS can automatically compute all of the descriptive statistics.p p•In EXCEL:

– Tools/Data Analysis/Descriptive Statistics

•In SPSS: – Analyze/Descriptive Statistics/Frequencies

Click on the “Statistics” box and select all of the

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– Click on the Statistics box and select all of the descriptive statistics you want (including the percentiles).

•EXCEL and SPSS are now illustrated on the data in the file salary.xls.

Descriptive Statistics in ExcelTo compute descriptive statistics in EXCEL, in the Data tab, use the Data-Analysis add-in and choose D i ti St ti tiDescriptive Statistics:

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EXCEL Salary Example

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Descriptive Statistics in SPSSTo compute descriptive statistics in SPSS, use the Analyze/Descriptive Statistics/Frequencies and then on the bottom of the screen, click on Statistics and ,choose the statistics you want reported:

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SPSS Salary ExampleStatistics

Starting SalaryStarting Salary12

02940.002905.00

2880165.653

27440.909615

2710

ValidMissing

N

MeanMedianModeStd. DeviationVarianceRangeMi i

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27103325

352802905.003139.75

MinimumMaximumSum

5085

Percentiles

Relationship BetweenTwo Variables

• So far you have seen ways to analyze information about a single variable.

• One is often interested in the relationshipbetween two or more variables.

• Examples of relationships– Advertisement expenditures and sales

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Advertisement expenditures and sales.– Company profits and stock price.– Home size and sales price.

Example: Stereo Store• File stereo.xls

• Is there any relationship between the number of commercials and the sales levels?Week Commercials Sales ($100)

1 2 501 2 502 5 573 1 414 3 545 4 546 1 387 5 638 3 48

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8 3 489 4 5910 2 46

Scatter Diagrams in Excel• In Excel, select the two columns of data;

click on the Insert tab; then on the Scatter icon; then on the top left diagramicon; then on the top left diagram.

• Number of commercials on the x-axis.• Sales levels on the y-axis.

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Scatter Diagrams in SPSS• Plot of two variables on the same graph.• In SPSS, choose Graphs/Legacy Dialogs/ Scatter

then choose Simple and click on Define• Number of commercials on the x-axis.• Sales levels on the y-axis.

70

60

61Commercials

6543210

SALE

S

50

40

30

Covariance and Correlation The sample and population covariance of two

variables X and Y are numbers whose sign have the following meaning:g g COV(X,Y) > 0 means that the two variables tend to

move in the same direction—if one increases (decreases), then the other increases (decreases).

COV(X,Y) < 0 means that the two variables tend to move in opposite directions—if one increases (decreases), then other decreases (increases).

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(decreases), then other decreases (increases).

The value of the covariance is hard to interpret, so the covariance is converted to a number between −1 and +1 called the correlation of X and Y that indicates how strongly X and Y are correlated.

Covariance and Correlation• For two variables X and Y for which you have n

pairs of data in the form (x1, y1), …, (xn, yn), the i d l i d bcovariance and correlation are computed by:

n

i

YiXiXY n

yx1

))((

Population Sample

COV(X, Y):

n

i

iiXY n

YyXxS1 1

))((

XY XYS

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COR(X, Y):YX

XYXY

YX

XYXY SS

Sr

Note: COVARIANCE.P and COVARIANCE.S in Excel compute the population and sample covariance XY. CORREL computes the sample correlation = population correlation.

Cov. and Correlation in EXCEL

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Cov and Correlation in SPSS• In SPSS, choose Analyze/Correlate/Bivariate.• On the next menu, click on Options.• Select Cross-Product Deviations and Covariances.• Click Continue and, on the previous menu, OK.

XYr

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XYS

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Stat Camp for theMBA Program

Daniel Solow

Lecture 2

66

Probability

Motivation• You often need to make decisions under

uncertainty, that is, facing an unknown future.y, , g• Examples:

– How many computers should I produce this month?– What premium should I charge a class of customers

for a particular type of insurance policy?Th h i i

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• The answers to such questions requires knowledge of probability that is, the study of likelihood of certain events occurring.

Probability• Probability is a number that measures the

likelihood that an event will happen.• Useful as an indicator of the uncertainty

associated with an event.• Scale from 0 to 1.

– Probability = 0: the event will certainly not happen

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happen.– Probability = 1: the event will certainly happen.– Probability = 0.5: the event is equally likely to

happen or not happen.

Experiments and Outcomes• Experiment: A situation in which an action

could be repeated many times, each resulting i f iblin one of many possible outcomes or sample points. Exactly one of these outcomes will occur, but it is not known which. For example:

Experiment OutcomesToss a coin Head, Tail

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Toss a coin Head, Tail

Roll a die 1, 2, 3, 4, 5, 6

Sales Call Sale, No sale

Dow Jones tomorrow All positive numbers

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Assigning Probabilities• Assume an experiment has n possible

outcomes E1, E2,…, En .outcomes E1, E2,…, En . • The probability assigned to each outcome

must be a number between 0 and 1.

• The sum of the probabilities of all outcomes 1Pr0 iE

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pmust equal 1.

1)Pr(...)Pr()Pr( 21 nEEE

Assigning Probabilities

• Depending on the situation, you can obtain the probabilities of the outcomes of an experimentprobabilities of the outcomes of an experiment from:– Assumption that all outcomes are equally likely

(classical method)– Experience from past data (relative frequency

method)

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method)– Experience, intuition or personal judgment

(subjective method)

Classical Method

• In many situations it is reasonable to assume that all n outcomes of an experiment arethat all n outcomes of an experiment are equally likely to occur.

• Then each outcome has probability equal to 1/n (why)?

• Examples:

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– Toss a coin: P(H) = P(T) = 1/2.– Roll a die: P(1) = P(2) = … = P(6) = 1/6.

Relative Frequency Method

• In some experiments, there are past data il bl f hi h ti t thavailable, from which you can estimate the

proportion of time each outcome has occurred if the experiment is repeated a large number of times.

• This proportion is used as an estimate of the

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This proportion is used as an estimate of the probability of the outcome.

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Example• When asking about a person’s attitude on a new law, the

outcome could be: disagree (D), neutral (N), agree (A), or uninformed (U).

• One way to assign probabilities to these four outcomes is to use the results of a survey, such as the following:

AttitudeNumber of Responses

RelativeFrequency

Disagree (D) 52 52/127=0.41Neutral (N) 12 12/127=0.09A (A) 37 37/127 0 29

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• P(D) = 0.41, P(N) = 0.09, P(A) = 0.29, P(U) = 0.21

Agree (A) 37 37/127=0.29Uninformed (U) 26 26/127=0.21Total 127 1

Subjective Method

• Appropriate when– Equally-likely assumption is not appropriate.– No data are available.

• One can use other available information, experience, intuition, judgment.

• In this case the probability of each outcome

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• In this case, the probability of each outcome expresses the person’s subjective degree of belief about the likelihood of the outcome.

Example

• What is the probability that I will get that jobWhat is the probability that I will get that job offer?– My personal belief of the chance is 50%.

P(offer) = 0.5, P(no offer) = 0.5.– The career office believes that the chance is 70%.

P(offer) = 0.7, P(no offer) = 0.3

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( ) , ( )– I am more pessimistic than the career office.

Events• The outcomes are the simplest elements

associated with an experiment.associated with an experiment.• You are often interested in finding the

probabilities of more complicated events related with this experiment.

• In probability language, events are

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collections of outcomes.

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Example 1: Rolling a Die

• Experiment: Roll a die• Outcomes: 1, 2, …, 6• Events:

– A = the outcome is less than 3 = {1, 2}– B = the outcome is even = {2, 4, 6}

C th t i t l th 4 {4 5 6}

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– C = the outcome is not less than 4 = {4, 5, 6}

Example 2: Sporting Event

• Experiment: One of the following eight cities will be• Experiment: One of the following eight cities will be chosen to host a sports contest:

New York (N), London (L), Paris (P), Tokyo (T),Beijing (B), Sydney (S), Madrid (M), Chicago (C)

• Events:– A = contest is in Asia = {T B}

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A contest is in Asia– B = contest is in an English-speaking city– C = contest is not in Europe

{T, B}= {N, L, S, C}

= {N, T, B, S, C}

Example 3: Playing Cards

• A deck of cards consists of 52 cards, arranged in four suits: spades (S) clubs(C) diamonds (D) andfour suits: spades (S), clubs(C), diamonds (D), and hearts (H).

• Spades and clubs are black, diamonds and hearts are red.

• Each suit has 13 cards arranged in order:1(ace), 2, 3,…, 10, J(ack) ,Q(ueen), K(ing)

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• A card is denoted by the card number and the suit:1C = ace of clubs, JH = jack of hearts, and so on.

face cards

Example 3 (cont)

• Experiment: Draw a card at random• 52 Outcomes:

1S,…,KS,1C,…,KC,1D,…,KD,1H,…,KH• Events:

– A = draw a king = {KS, KC, KD, KH}B draw a red two {2H 2D}

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– B = draw a red two = {2H, 2D}– C = draw a club face card = {JC, QC, KC}

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Probabilities of Events

• The probability of an event can be computed b ddi th b biliti f ll tby adding up the probabilities of all outcomes included in that event.

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Example 1: Rolling a Die• Pr(1) = Pr(2) = … = Pr(6) = 1/6.

– A = the outcome is less than 3 = {1, 2}{ , }

– B = the outcome is even = {2, 4, 6}

C = the outcome is not less than 4 = {4 5 6}

P(A) = P(1) + P(2) = 1/6 + 1/6 = 1/3.

P(B) = P(2) + P(4) + P(6) = ½.

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– C = the outcome is not less than 4 = {4, 5, 6}

P(C) = P(4) + P(5) + P(6) = ½.

Example 2: Sporting Event

• Assume: – P(N) = 0.2, P(L) = 0.1, P(P) = 0.1, P(T) = 0.1,

P(B) 0 05 P(S) 0 15 P(M) 0 05 P(C) 0 25P(B) = 0.05, P(S) = 0.15, P(M) = 0.05, P(C) = 0.25.• Then

– A = contest is in Asia = {T, B}

– B = contest in an English-speaking city = {N, L, S, C}

P(A) = P(T)+P(B) = 0.1 + 0.05 = 0.15.

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– C = contest not in Europe = {N, T, B, S, C}P(B) = 0.2 + 0.1 + 0.15 +0.25 = 0.70.

P(C) = 0.2 + 0.1 + 0.05 + 0.15 + 0.25 = 0.75.

Example: Playing Cards

• P(any outcome) = 1/52.A d ki {KS KC KD KH}– A = draw a king = {KS, KC, KD, KH}

– B = draw a red two = {2H, 2D}

P(A) = 4/52 = 1/13 = 0.077.

P(B) = 2/52 = 1/26 = 0.038.

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– C = draw a club face = {JC, QC, KC}

P(C) = 3/52 = 0.058.

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Complement of an Event• The complement of an event A is the event that

A does not happen and thus contains all t th t t t i d i Aoutcomes that are not contained in A.

• The complement of A is written as Ac.• If A happens, Ac does not happen and vice versa.• Complement Law: P(A) = 1 – P(Ac).

N t If h A i d i h

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• Note: If the event A you are interested in has many outcomes but Ac does not, then, to compute P(A), it is easier to find P(Ac) and then

P(A) = 1 – P(Ac).

Examples• Example 1: Suppose A is the event that the

weekly sales exceed $2,000 and P(A) = 0.75.• Then Ac is the event that the weekly sales do

not exceed $2,000 and P(Ac) = 0.25.• Example 2: When you pick a card at

random, what is the probability that you do not pick an ace?

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not pick an ace?• Answer: Let A = not pick and ace so

Ac = pick an ace =P(A) = 1 – P(Ac) = 1 – (4/52) = 48/52.

{AC, AD, AH, AS} and so

Intersection of Two Events• If A and B are two events, you are often

interested in the probability that both A and B occur simultaneouslyoccur simultaneously.

• The event A and B is called the intersection of A and B, written A B, and consists of outcomes that are in both A and B simultaneously.

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When you see the word AND think of .

ABA B

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Intersection of Two Events• Example: What is the probability of drawing a

red king?A d ki {KS KC KH KD}– A = draw a king = {KS, KC, KH, KD}

– B = draw a red card = {1H,2H,3H,4H,5H,6H,7H,8H,9H,10H,JH,QH,KH,1D,2D,3D,4D,5D,6D,7D,8D,9D,10D,JD,QD, KD}.

– A B = draw a king and red card = {KD, KH}.

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P(drawing a red king) = P(drawing a red card and a king) = P(A B) = {KD, KH} = 1 / 26.

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Mutually Exclusive Events

• If A B contains no outcomes, then the t A d B ll d t ll l ievents A and B are called mutually exclusive.

• If A and B are mutually exclusive it is not possible that both A and B will happen.

• If A and B are mutually exclusive, then

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P(A B) = 0.

Example

• A = draw a king = {KS, KC, KH, KD}• B = draw a queen = {QS, QC, QH, QD}• A B contains no outcomes.• If you draw a single card, it is not possible that

the card will be both a king and a queen.

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Union of Two Events• If A and B are two events, you are often

interested in probability that either A or B (or both) occur simultaneously.( ) y

• In math terms, this is called the union of A and B and is denoted by A B.

Example: Roll a die. A = the outcome is less than 3 = {1, 2}.

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B = the outcome is even = {2, 4, 6}. AB = the outcome is less than 3 or even (or

both) = {1, 2, 4, 6}.

The Addition Law• The event A B consists of all the

outcomes that belong to both A and B.• When you see the word OR think of . The probability of A B can be computed

by adding the probabilities of the individual outcomes in A B, OR with the following addition law (whichever is easier):

93AA B: B

P(A B) = P(A) + – P(A B)

A B

addition law (whichever is easier):

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Union of Mutually Exclusive Events

• If A and B are mutually exclusive, you know th t P(A B) 0that P(A B) = 0.

• Therefore, for mutually exclusive events,P(A B) = P(A) + P(B) – P(A B)

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= P(A) + P(B)

Example• Roll a die

– A = the outcome is less than 3 = {1, 2}– B = the outcome is even = {2, 4, 6}– AB={2}AB {2}– AB = {1, 2, 4, 6}

• Probabilities– P(A) = 1/3– P(B) = 1/2– P(AB) = 1/6

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– P(AB) = 4/6 = 2/3.• From the addition law:

– P(AB) = P(A) + P(B) – P(AB)= 1/3 + 1/2 – 1/6 = 2/3which agrees with the direct calculation.

Example

• You draw a single card from a deck.• You win if you draw a king or a red card.• What is the probability you win?

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Answer• A = draw a king

P(A) = 4/52.• B = draw a red card

• AB = red king

• AB = king or red card

( )

P(B) = 26/52.

P(AB) = 2 /52

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• AB = king or red card P(AB) = P(A) + P(B) – P (AB) =4/52 + 26/52 – 2/52 = 28/52 = 0.538

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Conditional Probability• Sometimes the probability of an event changes

when you get information about another related event.

• Example: You roll a die and don’t see the outcome.– What is the probability of the outcome being a 2?– If I tell you that the outcome was odd, what is the

probability of a 2 now? 0

1/6

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– If I tell you that the outcome was even, what is the probability of a 2 now?

• This is the conditional probability of an event given that another event happened.

1/3

Example: Police Force• A police force consists of 960 men an 240 women

officers. Last year 288 men and 36 women were promoted. Women officers complained of discri-mination The administration said that this was duemination. The administration said that this was due to the low number of women officers in the force.

• Do you think the discrim. complaint is justified?• Approach: Compare P(promotion given a man)

and P(promotion given a woman).Probability Model

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Probability ModelExperiment: Select an officer at randomOutcomes: Any one of 1200 officers,

all equally likely.

Events and Their Probabilities• Events:

– M = man officer, – W = woman officer

P(M) = 960/1200 = 0.8P(W) = 240/1200 = 0.2W woman officer,

– A = promoted officer, – Ac = non-promoted officer, – MA = male prom. off.,

P(W) 240/1200 0.2P(A) = 324/1200 = 0.27

P(Ac) = 0.73P(MA) = 288/1200 = 0.24

Men Women Totals

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Men Women TotalsPromoted

NotTotals 960 240 1200

288672

36204 876

324

Conditional ProbabilityMen Women Totals

PromotedN t

288672

36204 876

324

• Select an officer at random.• P(selected officer is promoted) = P(A) =

NotTotals 960 240 1200

672 204 876

324/1200 = 0.27

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Suppose the selected off. is a man.

P(A | M) = 288/960 = 0.3

Suppose the selected off. is a woman.

P(A | W) = 36/240 = 0.15

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Conclusion

• We have found that – P(A|M) = 0.3: 30% of men officers were

promoted.– P(A|W) = 0.15: 15% of women officers were

promoted.• Is the complaint justified in your opinion? YES!

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p j y p

Computing Conditional Prob.• The general formula (Baye’s Theorem) for the

conditional probability of an event A given h h d ithat event B has occurred is:

)()()|(

BPBAPBAP

• Police Example:240)( MAP

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3.08.0

24.0)(

)()|(

MP

MAPMAP

This is consistent with our original calculation.

Independent Events

• Two events A and B are independent if the b bilit f A d t h ithprobability of A does not change with

information about B, that is, ifP(A|B) = P(A)

and vice versaP(B|A) = P(B).

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• If this relation is not true, then the two events are dependent.

( | ) ( )

Example: Police Force

• You have found that – P(A|M) = 0 3P(A|M) 0.3– P(A) = 0.27

• Therefore, P(A|M) P(A), and the events A and M are dependent.

• This means that the probability of an officer

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p ybeing promoted is influenced by whether this officer is a man.

• This justifies the discrimination claim.

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Multiplication Law

• Assume A and B are independent. ThenP(A|B) P(A)P(A|B) = P(A).

• However, you also know thatP(A|B) = P(AB) / P(B).

• Then you find thatP(AB) / P(B) = P(A),

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orP(AB) = P(A) P(B).

Independence• It is often reasonable to assume that two events are

independent because of their natureindependent, because of their nature.• For example, if you roll two dice.

– A = 1st roll is a 2.– B = 2nd roll is a 5.

• Because there is no relationship between the first and the second roll, you may assume that A and B

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are independent.• Then the multiplication law implies

P(AB) = P(A) P(B) = (1/6) (1/6) = 1/36.

Summary Example• Bob and Jon live together and each has a car that works,

respectively, 60% and 90% of the time.• A potential employer has said she will hire them if they have

one working car at least 95% of the time. Will they get the job?g• State what you are looking for as a prob. question.

Is P(that at least one car is working) 0.95?• Use probability theory to answer the question.

Events:B = Jon’s car works

y g j

A = Bob’s car worksP( ) = 0.9P( ) = 0.6

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= P(A) + P(B) P(AB) (Addition Law)(A & B are ind.)= 0.6

= 0.96.

P(at least one car is working)

They get the job.

0.6 (0.9) + 0.9

= P(A OR B) = P(AB)

Random Variables• A random variable (rv) is a quantity of interest:

– Whose value is uncertain (by “uncertain” is meant that there are many (at least two) possible values

– You cannot control the value that occurs.• Example 1: Y = the outcome of flipping a coin. • Random variables are used to help make decisions

in a problem involving uncertainty.

y ( ) pand you do not know which value will occur).

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p g y• Example 2: Roll a die once. If the outcome is 1 or 2

you lose $5, if 3 you lose $1, if 4 you win $2, if 5 or 6 you win $4.

• To decide, you need to identify appropriate rvs.Do you want to play this game?

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Identifying Random Variables and Their Possible Values

• The first two steps involved in working with aThe first two steps involved in working with a random variable are:

• Step 1: Identify the random variable.– Use a symbol and write the meaning of the variable,

including units.

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Example : Let X = $ earned in the dice game.• Step 2: List all possible values the r.v. can have.

Example: X = 5, 1, 2, or 4.

Types of Random Variables• Discrete Random Variable: A r.v. whose possible

values you can “count” (either finite or an infinite set of countable numbers, for example, 0,1,2,…).set of countable numbers, for example, 0,1,2,…).

Example Roll-and-Earn: X = −5, −1, 2, 4Can you count these possible values?

• Continuous Random Variable: A r.v. whose value is any number (including decimals and fractions) in an interval or a collection of intervals (infinite

Yes!

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an interval or a collection of intervals (infinite uncountable number of values).

Example: X = liters of water I drink today.Can you count these possible values?

[0, 5]No!

Poss. Val.

Examples of Random Variables• Number of heads in 50 tosses of a coin.

All possible values: (discrete)0 1 50– All possible values:• Number of customers who enter a store in a day.

– All possible values:• Number of cm of rain next month.

– All possible values:

(discrete)

(discrete)

(continuous)

0, 1, …, 50

0, 1, 2, …

[0, 30].

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• Time in minutes between two customers arriving at a bank.– All possible values:

( )[ ]

[0, ). (continuous)

• Number of defective products in a shipment of 100.

All possible values:

Examples of Random Variables

(discrete)0 1 2 100– All possible values: (discrete)

Quantity x of liquid inside a 12 oz can– All possible values: (continuous)

0, 1, 2, …, 100.

[0, 12]. Percentage x of a project completed by the

deadline

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deadline.– All possible values:

$ sales in a retail store tomorrow.– All possible values:

(continuous)

(discrete)

[0, 100].

$0.00 - $10000.00

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• The temperature at noon yesterday.• The temperature at noon tomorrow.

Which of the Following are RVs?(No)(Yes)

Time is critical!

• The age of a person chosen at random in this class.(It depends on timing.)

(Yes)

(No: there is only one value for that person’s age.)

If you have already selected the person:If you have not yet

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( )

All possible values:(discrete)

y yselected the person:

The (finite) list of agesof everyone in this class.

• The average of a population.

Which of the Following are RVs?(No, µ only has one val.)

(Depends on time.) The average of a sample of size 2.X

(Yes)

(No, there is only one value for that average.)

If you have already selected the sample:If you have not yet selected the sample:

The (finite) list of averagesof every group of size 2 in the population.

All possible values of :X

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G1Groups of size 2:

for the group:X A1

G2

A2

G3

A3(discrete, but with so many values that it is not practical to list them all)

RVs, Populations, and Sampling For any population, you can create the

following two discrete random variables:Y = the value of an item that will be chosen

randomly from the population.X

Note: The average of the pop is not a RV

= the average of a sample of size n before taking the sample.

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For any RV, you can create a sample of size n by observing and recording the value of the RV n separate times.

Note: The average of the pop. is not a RV.

When Something is Unknown

Alternative 1: Determine the value, however…If d i i h i ff h

Question: What can you do when a quantity of interest—such as the average of a population—is unknown?

If doing requires too much time, effort, money, then… Alternative 2: Estimate the value, for example, by…

Building a model! For example:Model 1: Take a sample of size n and use the average

from the sample as your best estimate of .

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p y Model 2: Think of as a discrete random variable with possible values: 20, 21, …, 30Model 3: Think of as a continuous random variable with possible values: [20, 30]

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Identifying Random Variables• INSURANCE PREMIUMS

Wh h ld h i i b fWhat should the insurance premium be for a particular class of customers?

• Question: Is the annual premium a r.v.?Answer: No, because you can control its value.L t C th $/ t b l i d b thi t f

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• Let C = the $/year to be claimed by this type of customer with possible values: (discrete)

0 – 100,000

Examples of Random Variables• WARRANTIES

G dTi h j i d d i i hGoodTire has just introduced a new tire in the car market. How many miles of warranty should the company offer?

• Qn: Is the warranty mileage a r.v.?• Ans: No because you can control its value

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• Ans: No, because you can control its value.• M = the number of miles such a tire is

expected to last, with possible values: (continuous)

[0, ).

Examples of Random Variables• PERSONNEL PLANNING

How many bank tellers should be working during theHow many bank tellers should be working during the busiest time of the day?

• A = the total number of customers that arrive during that period, with possible values:

• W = the number of minutes it takes a teller to serve a customer with possible values: [0 30]

0, 1, 2, … (discrete)

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customer, with possible values:(continuous)

[0, 30].

Probability Distribution• To “work” with a random variable, you must

know that variable’s probability distribution, which describes the probabilities of all the possible values of the random variable occurring.

• Note: Probability distributions are different for discrete RVs and continuous RVs

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discrete RVs and continuous RVs.

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Discrete Distributions

• For a discrete random variable X, the probability distribution is described by aprobability distribution is described by a probability density function that consists of:– The list of the possible values and, for each one,

the probability of that value occurring.• Notationally, if t is a possible value for the rv X

th th d it f ti i itt f ll

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then the density function is written as follows:

f(t) = P(X = t) = the probability that the random variable takes the value t.

Example 1• Toss a coin once.• X = number of heads.

P ibl l f X• Possible values of X: t 0 1

P(X = t) = f(t) 0.5 0.5• A valid probability density function for a

di t d i bl t ti f th

123

discrete random variable must satisfy the following two properties:– 0 f(t) 1, for each value of t.– f(t) = 1.

Example 2Toss a coin twice, and let X = number of heads.

Values of X Probability Density Function

f(0) =0.25

P(X = 0) = = P(T and T) P(no heads) = P(T) * P(T) = 0.5 (0.5) =

f(1) = P(X = 1) = = P(HT or TH) P(1 head)

0

1

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f( )0.5

( ) ( )( )0.25 + 0.25 == P(HT) + P(TH) =

f(2) =2 0.251 f(0) f(1) =1 0.25 0.5 =

Example 2 (cont’d)

• The probability density function can be shown i t bl hin a table or graph:

0 25

0.5f(t)

t 0 1 2f(t) 0.25 0.5 0.25

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0.25

0 1 t2

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Example 3: Roll and Earn

• Find the probability density function of X, the t d i th f ll iamount earned in the following game:

Roll a die once. If the outcome is 1 or 2 you lose $5, if 3 you lose $1, if 4 you win $2, if 5 or 6 you win $4.

1 2 4

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Possible values for X: 5 1 2 4

Probabilities: 2/6 1/61 or 2 3

1/64

2/65 or 6

• In addition to the density function, you also want to find the expected value (mean) of the discrete

d i bl hi h i f th t l

Expected Value

random variable, which is a measure of the central location of the value of the random variable.

• Computed as the sum of the products of the possible values x and the corresponding probabilities:

E[X] t f(t)

127

X = E[X] = t f(t)

• Note: The mean of a r.v. is different from the mean of a population and the mean of a sample.

Why is the ExpectedValue Useful?

L t th X b d i bl

• Law of Large Numbers: If you observe the value of the random variable X a large number of times, the average of the

• Let the rv X = be your random variable.

128

, gobserved values will be very close to the expected value of the random variable X.

Example• Toss a coin twice and let X = number of heads.• The probability density function is:The probability density function is:

• The expected value is: E(X) = (0) (0.25) + (1)(0.5) +(2)(0.25) = 1

t 0 1 2f(t) 0.25 0.5 0.25

129

• If the two tosses are repeated many times and the number of heads recorded each time, the average number of heads per two-tosses will be close to 1.

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Computing Exp.Value in Excel• Roll and Earn

130

This means that, if you play the game many times and record the results, on average, you will lose $0.16 each time.

Example: Car Dealership

• Now that the new car models are available aNow that the new car models are available, a dealership has lowered the prices on last year’s models in order to clear its inventory. With prices slashed, a salesperson estimates the following probability distribution of X, the number of cars that person will sell next week.

t 0 1 2 3 4

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• Find the expected value of X. What does it mean?

t 0 1 2 3 4f(t) 0.05 0.15 0.35 0.25 0.2

Answer

t 0 1 2 3 4 f(t) 0 05 0 15 0 35 0 25 0 2

• E(X) = (0)(0.05) + (1)(0.15) + (2)(0.35) + (3)(0.25) + (4)(0.2) = 2.40

f(t) 0.05 0.15 0.35 0.25 0.2

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• If prices stay the same over several weeks, the average number of cars sold per week will be close to 2.40.

Variance and Standard Deviation of Random Variables• You also want to find the variance of the r.v., which

is a measure of how close the possible values are tois a measure of how close the possible values are to the expected value, .

The variance of a discrete random variable X is:µ Var. is small µ Var. is larger

2 = VAR(X) = (t – ) 2 f(t)

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• The standard deviation is the square root of the variance.

• Note: Variance and standard deviation for a r.v. are different from those for a population and sample.

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Example• Toss a coin twice and let X = number of heads.

Recall that the e pected al e is 1• Recall that the expected value is = 1.x x- (x- f(x) (x-f(x)0 -1 1.00 0.25 0.251 0 0.00 0.5 0.002 1 1.00 0.25 0.25

0.50 Var(X)

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• Variance: Var(X) = 2 = 0.50

( )

• Standard Deviation: = 0.50 = 0.707

Example• For the roll-and-earn game, find the

variance and standard deviation.R ll th t E(X) 0 16• Recall that E(X) = = 0.16.

x x- (x- f(x) (x-f(x)-5 -4.84 23.43 0.33 7.73-1 -0.84 0.71 0.17 0.122 2.16 4.67 0.17 0.794 4 16 17 31 0 33 5 71

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4 4.16 17.31 0.33 5.7114.35 Var(X)

• Variance: Var(X) = 2 = 14.35• Standard deviation: = 14.35 = 3.79

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Stat Camp for theMBA Program

Daniel Solow

Lecture 3

136

Lecture 3Random Variables and

Distributions

Probability Distribution• Recall that a random variable is a quantity of

interest whose value is uncertain and you cannot control it.control it. To use a r. v. to help solve a decision problem:

• Step 1: Identify the random variable, say X.• Step 2: List all possible values for X.• Step 3: Determine if X is discrete or continuous.

137

• Step 4: Identify the density function of X.• Step 5: Find E[ X ] = the expected value of X.• Step 6: Find VAR[ X ] = the variance of X (or the

STDEV[ X ]).

Example: Debbon Air Seat-Release Problem

• Debbon Air needs to make a decision about Flight 206 gto Miami, which is fully booked except that…

• 3 seats are reserved for last-minute customers (who pay $475 per seat), but the airline does not know if anyone will buy those seats.If th l th th k th ill b bl t

138

• If they release them now, they know they will be able to sell them all for $250 each.

• Debbon Air counts a $150 loss of goodwill for every last-minute customer turned away.

Debbon Air “Seat Release”

• Question: How many seats, if any, should Debbon Air l ?release?

• Question: On what basis that is, on what criterion are you going to make the final decision?

• Answer: Based on profits.

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• Approach: Find the expected profit when you release 0 seats, 1 seat, 2 seats, and 3 seats, and then…Choose the alternative that has the max. expected profit.

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Identifying Random Variables• Question: Can you identify any r.v.s that can

help you make this decision?• Let the r.v. X =• Possible values for X:

x 0 1 2 3

# of arriving last-minute customers

140

• Probability distribution for X:

f (x) 0.45 0.30 0.15 0.10

Identifying Random Variables

• Another random variable of interest is:R = net revenue (revenue minus loss

of goodwill) • However, this revenue depends on the number

of seats released, so, defineR t h i t

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Ri = net revenue when i seats arereleased (i = 0, 1, 2, 3).

“Debbon Air” Seat Release• What are the possible values for R3, that is,

what are the possible revenues when all 3what are the possible revenues when all 3 seats are released?

• The answer depends on how many last-minute customers (X) arrive, so: If X = 0: R3 = 3(250) = $750

142

3

If X = 1: R3 = If X = 2: R3 = If X = 3: R3 =

( )3(250) – 150 = $6003(250) – 2(150) = $4503(250) – 3(150) = $300

Debbon Air “Seat Release”

• Expected Value of R3 (revenue when 3 seats l d)are released):

x 0 1 2 3f(x) 0.45 0.3 0.15 0.1R (3 rel) 750 600 450 300

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• E(R3) = 750 (0.45) + 600 (0.3) + 450 (0.15) + 300 (0.1)

= 615.

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“Debbon Air” Seat Release

• How many seats should be released to maximize expected net revenue?

x 0 1 2 3f(x) 0.45 0.3 0.15 0.1 E(R)R (3 rel) 750 600 450 300 615.00R (2 rel) 500 975 825 675 708.75

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Two seats should be released.

R (1 rel) 250 725 1200 1050 615.00R (0 rel) 0 475 950 1425 427.50

A Binomial Experiment• In many applications, you will perform a binomial

experiment a number of times for which, each time, there are two possible outcomes:– A “success” or – A “failure”.

• Example: At a tire factory, you will examine a number of

Note: A “success” does not have to be a “good” outcome.

What you are then interested in is the probability of having k successes out of, say, n trials.

Qn: What is the prob. that there are 0 defects (“successes”) in, say, a sample of n = 20 tires? 145

tires and, for each one, determine if• There is a defect (“success”) or• There is no defect (“failure”).

Binomial ExperimentIn a binomial experiment, you must identify:• What constitutes a trial, a “success” and a “failure.”,• p = P(success) = the probability of a success occurring

in each trial (so, P(failure) = ).• n = the number of independent trials (repetitions).Then define the following r.v.:

X = number of successes out of n trials

1 – p

Possible Values0 1 2 n

discrete

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X number of successes out of n trials. 0, 1, 2, …, n.Density Function = Probabilities: get from EXCEL

[ ] (1 ).XSD X np p [ ]E X np

Example 1• Toss a fair coin 100 times.

A “ i l” i fli d “ ” h d• A “trial” is a flip and a “success” = a heads.• p = probability of success = P(head) = 0.5.• n = 100 independent coin tosses.• X = number of heads follows binomial

di t ib ti ith 100 d 0 5

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distribution with n = 100 and p = 0.5.Then, E(X) =

(100)(0.5)(0.5) = 25 = 5.

100 (0.5)= 50SD(X) =

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• A student takes a multiple choice test with 25 questions. Each question has 4 choices. Assume the student does not know the answer to any question and

Example 2

y qjust guesses.

• A “trial” is a question and a “success” is a correct answer.

• p = prob. of success = P(correct answer) = 0.25.• n = 25 independent questions.• X = number of correct answers follows the Binomial

148

• X = number of correct answers follows the Binomial distribution with n = 25 and p = 0.25.

• Then, E(X) =

SD(X) =

(25) (0.25) = 6.25.

(25) (0.25) (0.75) = 4.6875 = 2.17.

Binomial Random Variables • If n = 1, then the possible values for X are:

1 if the outcome is a successX =

1 if the outcome is a success0 if the outcome is a failure

Here, X is called a binomial random variable.• The density function is: 1 p, • The expected value of X is E(X) = np = p.

f(0) = f(1) = p.

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The expected value of X is E(X) np p.• The standard deviation of X is SD(X) =

.)1()1( pppnp

Binomial Probabilities in Excel • The Excel BINOMDIST function provides

two kinds of binomial probabilities.two kinds of binomial probabilities.• Suppose that the random variable X is

Binomial with parameters (n, p).• For k successes, where k is between 0 and n:

BINOMDIST(k FALSE)knk ppn

kXP

)1()(

150

= BINOMDIST(k, n, p, FALSE)

= BINOMDIST(k, n, p, TRUE)

ppk

kXP

)1()(

k

jjXPkXP

0)()(

Example 1: Bad Seafood• Consumer Reports (Feb. 1992) found widespread

contamination of seafood in supermarkets in NYC and Chicago.g

• 40% of the swordfish pieces for sale had a level of mercury above the maximum allowed by the Food & Drug Administration (FDA).

• Suppose a random sample of 12 swordfish pieces is selected.

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• What is the probability that exactly five of the pieces have mercury levels above the FDA maximum?

• What is the probability that at least 10 pieces are contaminated?

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Answer• Find P(exactly five of the pieces have mercury levels

above the FDA maximum)?• “Trial” is choosing a piece of fish and a “success” =• Trial is choosing a piece of fish and a success =

the piece is contaminated.• p = prob. of success = P(contamination) = 0.4.• n = 12 independent (why?) pieces.• X = number of contaminated pieces follows the

Binomial distribution with n = 12 and p = 0.4.

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Binomial distribution with n 12 and p 0.4.• From Excel:

P(X = 5) = BINOMDIST(5, 12, 0.4,FALSE) = 0.227

Answer (continued)• Find P(at least ten pieces are contaminated)• P(X 10) = P(X = 10 or 11 or 12) =• P(X 10) = P(X = 10 or 11 or 12) =

P(X = 10) + P(X = 11) + P(X = 12)• From Excel, you have

– P(X = 10) = BINOMDIST(10, 12, 0.4, FALSE) = 0.0025– P(X = 11) = BINOMDIST(11, 12, 0.4, FALSE) = 0.0003

P(X = 12) = BINOMDIST(12 12 0 4 FALSE) = 0

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– P(X = 12) = BINOMDIST(12, 12, 0.4, FALSE) = 0

• P(X 10) = 0.0025 + 0.003 + 0 = 0.0028

Another Way to Do This

• From the Complement Law you know that• From the Complement Law, you know that P(X 10) = 1 – P(X < 10)

= 1 – P(X 9)• From Excel, you have that

P(X 9) = BINOMDIST(9 12 0 4 TRUE) =

154

P(X 9) BINOMDIST(9, 12, 0.4,TRUE) 0.9972.

• Then P(X 10) = 1 – 0.9972 = 0.0028.

Example 2: Murder Trial• As the lawyer for a client accused of murder, you are

looking for ways to establish “reasonable doubt”. The prosecutor's case is based on the forensic evidence thatprosecutor's case is based on the forensic evidence that a blood sample from the crime scene matches the DNA of your client. It is known that 2% of the time DNA tests are in error.

• Suppose your client is guilty. If six laboratories in the

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Suppose your client is guilty. If six laboratories in the country are asked to perform a DNA test, what is the probability that at least one of them will make a mistake and conclude that your client is innocent?

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Answer• A “trial” is sending the DNA to a lab and a success =

lab makes error (finds no match)• p = prob. of success = P(error) = 0.02.

6 i d d t ( h ?) l b t t• n = 6 independent (why?) lab tests.• X = number of labs that make an error follows the

Binomial with n = 6 and p = 0.02.• P(X 1) = 1 – P(X < 1) = 1 – P(X = 0)• From Excel, P(X = 0) = 0.8858.• P(X 1) = 1 – 0.8858 = 0.1142

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( )• So there is an 11.42% probability that at least one lab

will find no DNA match.• Question: How many labs would you need to raise

this probability to 25%?

Example 3:Multiple-Choice Quiz

• A multiple-choice quiz has 15 questions. Each question has five possible answers, of which only one is correct.

• What is the expected number of correct answers by sheer guesswork?

• What is the standard deviation of the correct

157

answers by sheer guesswork?• What is the probability that sheer guesswork will

yield at least seven correct answers?

Answer

• A “trial” is answering a question and a success t= a correct answer.

• p = Prob. of success = P(correct answer) = 1/5 = 0.2.

• n = 15 independent (why?) answers.X b f t f ll

158

• X = number of correct answers follows a Binomial with n = 15 and p = 0.2.

• E(X) = np = 15 (0 2) = 3

Answer (continued)

• E(X) = np = 15 (0.2) = 3

• P(X 7) = 1 – P(X < 7) = 1 – P(X 6)P(X 6) BINOMDIST(6 15 0 2 TRUE)

SD(X) = n p (1 – p) = 15 (0.2) (0.8) = 2.4 = 1.55

159

• P(X 6) = BINOMDIST(6, 15, 0.2, TRUE)= 0.9819

• P(X 7) = 1 – 0.9819 = 0.0181 (1.81%)

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Covariance• When working with two random variables, say X

and Y, you are sometimes interested in the degree to which the values of X and Y are correlated—that is, as X increases, to what degree is it likely that Y increases (or decreases)?

• When X and Y are discrete RVs with n possible pairs of values, say (x1, y1), …, (xn, yn), and corresponding probabilities p1, …, pn, then the

160

covariance of X and Y, written COV(X, Y) or XY, is given by the following formula:

in

iiiXY pYEyXExYXCOV

1])[])([(),(

Covariance and Correlation•COV(X,Y) > 0 means that the two variables tend to move in the same direction—if one increases (decreases), then the other increases (decreases).COV(X,Y) < 0 means that the two variables tend to move in opposite directions—if one increases (decreases), then other decreases (increases). The value of the covariance is hard to interpret, so the

covariance is converted to the following number between −1 and +1 called the correlation of X and Y, written

161

COR(X, Y) or XY ( that indicates how strongly X and Y are correlated):

YX

XYXY

Note: Cov and correlation of RVs are different from cov. and correlation of samples and populations.

Example: Stocks and Bonds• Example: Suppose you are considering investing in both a

stock and a bond fund, and define the following RVs:S = annual rate of return on the stock fundB = annual rate of return on the bond fund

Possible Values (depend on the state of the economy):

Economy Stock Fund Bond Fund Prob.Recession −7% 17% 1/3Normal 12% 7% 1/3

162

Normal 12% 7% 1/3Boom 28% −3% 1/3

E[S] = 1/3(−0.07) +1/3(0.12) +1/3(0.28) = 0.11

E[B] = 1/3(0.17) +1/3(0.07) +1/3(−0.03) = 0.07

See File Cov_and_Cor.xlsUsing E[S] = 0.11 and E[B] = 0.07, you can now compute COV(S, B) using the formula as follows:

1/3(−0 07 − 0 11)(0 17 − 0 07)= − 0.0117COV(S, B) =

1/3( 0.07 0.11)(0.17 0.07)+ 1/3(0.12 − 0.11)(0.07 − 0.07)+ 1/3(0.28 − 0.11)(− 0.03 − 0.07)

Using [S] = 0.143 and [B] = 0.082, you can now compute COR(S, B) using the formula as follows:

163

COR(S, B) =COV(S, B)[S] [B]

= − 0.01170.143 (0.082)

− 1

So S and B are perfectly negatively correlated: when S returns are up, B returns are down and vice versa.

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Continuous Random Variables• A continuous random variable assumes any

value, including decimals and fractions, in i t l th l liintervals on the real line.

• Example: X = the number of minutes a customer waits in line. Possible values:– Question: If all values are equally likely, what is

P(X = 5.3789)?A P b(X 5 3789) 1/ 0 b h

[0, ).

164

– Answer: Prob(X = 5.3789) = 1/ = 0 because there are an infinite number of possible values for X.

• Conclusion: For a continuous rv, it is not meaningful to specify the likelihood that the variable is equal to one specific value.

Continuous vs. Discrete RVs• Question: Is there any difference between P(X < a)

and P(X a)?• Answer: For a continuous rv the answer is “no”• Answer: For a continuous rv, the answer is no

because:

P(X a) = P(X < a or X = a)= P(X < a) + P(X = a)= P(X < a) + 0

165

P(X a) 0= P(X < a)

• Note: The foregoing step that P(X = a) = 0 is not true for a finite discrete rv, and this is one major difference between a fnite discrete r.v. and a continuous rv.

Using Density Functions• Solution—Use a probability density function to

describe the likelihood that X is in a given interval.

f( ) Probability density function

Area under the graph =Total area = 1

f(x) Probability density function

P(a<X<b)

To find probabilities, wefind areas under the density function.

Possible values where the density function is higher are morelikely to occur than where the density function is lower. 166

a b xAll Possible Values of XMore

likelyLesslikely

Example: Finding Probabilities

f(x)Suppose that the graph of some probability density function is symmetric around 5.

What is P(X < 5)? 0.5What is P(X > 5)? 0.5

0.30.4

0.20.3

167

x54 6If P(X < 4) = 0.3, find:

P(X > 6)P(X < 4 or X > 6)

0.30.3 + 0.3 = 0.6

P(4 < X < 6) 1 0.6 = 0.4

P(4 < X < 5) 0.4/2 = 0.2

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Example: Uniform DistributionFact: In the real world, you can never find the density function of a continuous rv, so what can you do?Ans. Use a density function that mathematicians have created.E l C id X i h ibl l b 0 d 1

1f(x)

?f(x)1/2

Example: Consider a rv X with possible values between 0 and 1.All values are equally likely.

Uniform DistributionX ~ U[0,1]

168

1 x

Find P(0.2 < X < 0.5)0.50.2

0.3 · 1 = 0.3Find P(X > 0.6)

0.6

0.4 · 1 = 0.4E[X] = 0.5 289.012/1][ XSD

2 x

The Normal Distribution• Fact: In the real world, you can never find the density

function of a continuous rv, so what can you do?• Answer: Borrow an existing density function that

mathematicians have created (like the uniform dist.).• The Normal distribution is one such density function

with many desirable properties.• The Normal distribution applies to a continuous rv whose

possible values can be any real number from – to +.

169

• To write the density function, you must know the:· Mean · Standard Deviation

Usually estimated by computing the average and standard dev. from a sample of values for the rv.

The Normal DistributionThe density function f (x) =

f (x) “The Bell Curve”

x

2

2( )

212

x

e

This density function is:C d h h l h l f

170Smaller values of make the bell part thinner and taller.

•Centered at the mean .

•The std. dev. controls the “thickness”:•Symmetric about . •“Bell shaped.”

Use the Normal when most values of your rv are close to the mean and then become less likely farther from the mean.

The Normal Distribution: Effect of the SD

22 1

1

171

1

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The Normal DistributionArea left of = 0.5

This area =

P(a ≤ Y ≤ b)f (x)

172

a b x

Note: Excel is used to find areas under the Normal density function.

Excel Function NORMDIST• The Excel function NORMDIST is used to find

the area under the normal curve to the left of a i l h i if ( ) hgiven value z, that is, if X ~ N(, ), then

P(X ≤ z) = NORMDIST(z, , , TRUE).

173x z

Practice with ExcelExample: If X ~ N(20, 2), find P(X ≤ 23).

= 0.933NORMDIST(23, 20, 2, TRUE)P(X ≤ 23) = ( , , , )( ≤ )

174

z2320

Practice with ExcelQuestion: What do you do if the area you are interested in is not “all the way to the left”?Answer: Find a way to compute your area using

= 1 NORMDIST(23, 20, 2, TRUE)P(X > 23) = 1 P(X ≤ 23) = 0.067

Answer: Find a way to compute your area using “area all the way to the left”.

X ~N(20 2)

17523z

20

X ~N(20, 2)

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Practice with Excel

If X ~ N(100, 20), find P(80 ≤ X ≤ 120).

Note: This is really the probability that the value of X is within 1 standard deviation of the mean.

176

AnswerIf X ~ N(100, 20), thenP(80 ≤ X ≤ 120) = P(X ≤ 120) P(X ≤ 80)

NORMDIST(120 100 20 TRUE)= NORMDIST(120, 100, 20, TRUE) NORMDIST(80, 100, 20, TRUE)

= 0.841 0.159 = 0.682Try 2 and 3 std. deviations from th d

17712080

the mean and you will discover the empirical rule.

100

The Normal Distribution and the Empirical Rule

955%.

%2.68

178

2 2

3 3 99 7%.

Interpreting the Std. Dev.When your data are bell-shaped (according to a histogram), you can interpret the pop. / sample standard deviation as follows:standard deviation as follows: is a number so that 68% of your data are within one standard deviation of the mean .

= 72 2 Approx.

Of the valuesare in

The Empirical Rule

95%

100%

179

2 Approx. 68%

are in 1

95% 2100% 3

[ ][ ]68%95%

[ ]

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A Historical NoteQuestion: Before Excel and NORMDIST, how did one find areas under the Normal density function?

A U i t bl i hi h ld l kAnswer: Using a table in which you could look up the area, but,…

Question: It is impossible to create a separate table for every combination of and , so what can you do?Answer: Create a single table for a rv Z~N(0, 1),

180

Fact: Any probability question about a rv X~N(, ) can be stated as an equivalent question about a standard normal rv, as you will now see.

which is called a standard normal rv.

The Standard NormalExample: If X~N(, ) then P(X ≤ s) = NORMDIST(s, , , TRUE).

Thus, P(X ≤ s) = P(Z ≤ t)

But P(X ≤ s) = P(X ≤ s ) =

Z t

= NORMDIST(t, 0, 1, TRUE).

N(0, 1) ~

sXP

181

The Standard Normal DistributionNormal with

Mean SD

Standard Normal

SD

2 2

182

Standard Normalwith Mean 0

and SD 1

0 +1 +2-1-2

The Standard NormalExample: If X ~ N(20, 2), find P(X ≤ 23).

Answer 1: P(X ≤ 23) = NORMDIST(23, 20, 2, TRUE)

= 0.933

20 23 20X

= 0.933= NORMDIST(1.5, 0, 1, TRUE)

Answer 2: P(X ≤ 23) = = P(Z ≤ 1.5)20 23 202 2

XP

183z

1.500

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Excel Function NORMINV• For solving some problems, you know the

probability, p, and want to find the value of z so that the area to the left of z is p.

X ~ N(, )

p

P(X ≤ z) = p

184

xAnswer: z = NORMINV(p, , )

z = ?

Practice with NORMINV

k0 95 0

185

k0.95 0

ExampleIf X ~ N(50, 8), find j so that P(50 j X 50 + j) = 0.95

50+j50 j 50

0.950.025

186

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Stat Camp for theMBA Program

Daniel Solow

Lecture 4Th N l Di t ib ti d th

187

The Normal Distribution and the Central Limit Theorem

You wrote that a woman is pregnant for 266 days. Who said so? I carried my baby for ten months and five days, and there is no doubt about it because I k th t d t b b i d M

Example 1: Dear Abby

know the exact date my baby was conceived. My husband is in the Navy and it couldn’t possibly have been any other time because I saw him only once for an hour, and I didn’t see him again until the day before the baby was born.

I don’t drink or run around and there is no way

188

I don t drink or run around, and there is no way this baby isn’t his, so please print a retraction about the 266-day carrying time because otherwise I am in a lot of trouble.

San Diego Reader

Dear AbbyStep 1: Identify an appropriate random variable.

Y = number of days of pregnancyAb 230 290?

cont.What are the possible values for Y?What is the density function for Y?

About 230 – 290?

Prob. Density???

189

265 270260 275255… …Days

Idea: Approximate the density of Y with a normal!

Dear Abby• Question: If you are going to use a normal approximation,

what information do you need? • Answer: The mean and standard deviation.

Fact: A di t th ll ti i f ti• Fact: According to the collective experience of generations of pediatricians, pregnancies have a mean of 266 and standard deviation of 16 days, so Y ~ N ( = 266, = 16).

• Question: What are the possible values for Y?• Question: How can the number of days of pregnancy be

< 230?• Answer: Using the normal distribution you have that

– to

190

Answer: Using the normal distribution, you have thatP(Y < 230) = NORMDIST(230, 266, 16, true) 0.01.

• Thus, when using the normal approximation, there is only about 1% chance that a pregnancy lasts less than 230 days.

Models are NOT the real world but hopefully good approximations!

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Dear Abby•Step 2: State what you are looking for as a probability question in terms of the rv.

Y t t fi d P(Y ≥ 10 d 5 d )You want to find P(Y ≥ 10 mo. and 5 days) =P(Y ≥ 310).

•Step 3: Use the probability distribution of the rv to answer the probability question.P(Y ≥ 310) = 1 – P(Y < 310)

191

= 1 – NORMDIST(310, 266, 16, TRUE)= 0.00298 Was she telling the truth?Possibly, but highly unlikely.

Example 2: Problem of GoodTire

GoodTire has a new tire for which, in order to be competitive, they want to offer a warranty of 30,000 miles. Before doing so, h k h f i f

192

the company wants to know what fraction of tires they can expect to be returned under the warranty.

The Problem of GoodTire

•For GoodTire, let X = number of miles such a tire will last

Step 1: Identify an appropriate random variable.

X = number of miles such a tire will last.What are the possible values for X?What is the density function for X?

0 – 90000?

It is unknown, so estimate it using a model, as follows:

From statistical analysis of a random sample, GoodTireb li h il f ll i l l

(cont.)

193

X ~N( = 40000, = 10000)with possible values:

believes the mileage follows approximately a normal distribution with a mean of 40,000 miles and a standard deviation of 10,000 miles, so assume that

– to

The Problem of GoodTire

Step 2: State what you are looking for in t f b bilit ti t i iterms of a probability question pertaining to the random variable.

•GoodTire wants to know theFraction of tires returned =

194

P{X 30000} = ?Likelihood a tire fails =

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The Problem of GoodTireStep 3: Use the probability distribution of the random variable to answer theof the random variable to answer the probability question.•For GoodTire, you have P{X 30000} = ?

X N(40000, 10000)

NORMDIST(30000, 40000, 10000, TRUE) = 0.1587

1954000030000

The Problem of GoodTireQuestion: The CEO finds that a 16% return rate is too high. What warranty mileage s should they offer to get a 5% return rate?gStep 2: Probability Question: What should s be so that P{X s} = 0.05?

Step 3: s = NORMINV(0.05, 40000, 10000) = 23551.47Fact: While you cannot control the

19640000s = ?

0.05

cannot control the value of a rv, you can control the likelihood of certain events occurring with that RV.

Example 3: Marketing Projections

• From historical data over a number of years, a firm knows that its annual sales average $25firm knows that its annual sales average $25 million. For planning purposes, the CEO wants to know the likelihood that sales next year will:– Exceed $30 million.– Be within $1.5 million of the average.

197

The CEO is willing to issue bonuses if sales are “sufficiently” high. What level should be set so that bonuses are given at most 20% of the time?

Marketing Projections

•LetStep 1: Identify an appropriate random variable.

Let Y = next year’s sales in $ millions.

What are the possible values for Y?What is the density function for Y?

0 – 50????

From statistical analysis over a number of years, they

198

Y ~N( = 25, = 3)

believe that annual sales follows approximately a normal distribution with a mean of $25 mil. and a standard deviation of $3 mil., so assume that

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Marketing ProjectionsStep 2: State what you are looking for in terms of a probability question pertaining t th d i blto the random variable.•You want to know:

•P(sales exceeds $30 mil.) =

•P(sales is within $1.5 of $25 mil.) =

P(Y ≥ 30).

P(23 5 Y 26 5)

199

P(giving a bonus) = 0.20?P(Y ≥ s) = 0.20?

P(23.5 Y 26.5).•What should be the value of sales (s) so that

Marketing ProjectionsStep 3: Use the probability distribution of the random variable to answer the probability question.

•From Excel, using = 25 and = 3:

•P(Y ≥ 30) =

•P(23 5 Y 26 5) =

1 NORMDIST(30, 25, 3, TRUE)

NORMDIST(26 5 25 3 TRUE) –

= 0.045.

200

P(23.5 Y 26.5) NORMDIST(26.5, 25, 3, TRUE) NORMDIST(23.5, 25, 3, TRUE) = 0.383.

• s = = 27.524.NORMINV(0.8, 25, 3)

Example 4: DUI Test• In many states, a driver is legally drunk if the blood

alcohol concentration, as determined by a breath l i 0 10% hi hanalyzer, is 0.10% or higher.

• Suppose that a driver has a true blood alcohol concentration of 0.095%. With the breath analyzer test, what is the probability that the person will be (incorrectly) booked on a DUI charge?

201

Step 1: Identify an appropriate random variable.Let Y = the measurement of the analyzer as a %.Question: What are the possible values for Y? 0 – 0.3?

(cont.)

DUI TestStep 1 (continued).

Question: What is the density function for Y?

Answer: We do not know, but experience indicates that Y follows approximately a normal distribution with mean equal to the person’s true alcohol level and standard deviation equal to 0.004%, so…

Y N( 0 004) h

202

= the person’s true blood alcohol level (%)Y ~N(, = 0.004), where

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DUI TestStep 2: State what you are looking for in terms of a probability question pertaining p y q p gto the random variable.•You want to know the probability that a person with = 0.095 will be (incorrectly) booked on a DUI charge:

203

P(Y ≥ 0.10) P(being booked on a DUI) =

DUI TestStep 3: Use the probability distribution of the random variable to answer thethe random variable to answer the probability question.

•From Excel (using = 0.095 and = 0.004):

P(Y ≥ 0.10) = 0.1056.

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1 NORMDIST(0.10, 0.095, 0.004, true) = •There is about a 10% chance that such a person will be incorrectly charged with a DUI.

An Insurance ProblemGoodHands is considering insuring employees of GoodTire. What

l i h ld thannual premium should the company charge to be sure that there is a likelihood of no more than 1% of losing money on each customer?

This is an example of decision making under

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uncertainty: you have to make a decision today—how much should the annual premium be—

Question: Why is the future uncertain?facing an uncertain future.

Solving the Insurance ProblemStep 1: Identify an appropriate random variable.•Let X = the $ claimed by a customer in one year.Wh t th ibl l f X? [0 100000 (?)]•What are the possible values for X? [0, 100000 (?)]

•Is X continuous or discrete? discrete•What is the density function for X?

It is unknown, so borrow one.

From statistical analysis, the annual claim for these l f ll i t l l di t ib ti ith

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X ~N( = 2500, = 1000)

people follows approximately a normal distribution with a mean of $2500 and a standard deviation of $1000, so:

•Note: It can be OK to approximate a discrete RV with a continuous distribution.

discrete or cont.?

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An Insurance ProblemStep 2: State what you are looking for in terms of a probability question pertaining to the RV.

•For GoodHands, what should the premium s be so that

Probability Question: What should the premium s be so that the

For GoodHands, what should the premium s be so that the likelihood of losing money is no more than 1%.

X N(2500 1000)

X s

Question: When do you lose money on a customer?

P( ) = 0.01?

2072500

X N(2500, 1000) 01.0}{ sXP

s

An Insurance ProblemStep 3: Use the probability distribution of the random variable to answer the probability question.

X N(2500 1000)

= NORMINV(0.99, 2500, 1000)2500

X N(2500, 1000) 01.0}{ sXP

s= $4826 3478solution to the model!

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Fact: While you cannot control the value of a rv (such as the claim of a person), you can control the likelihood of certain events occurring with that RV (such as the likelihood of such a claim exceeding the premium).

$4826.3478solution to the model!= $4826.35solution for the real world!

The Insurance Problem (cont.)Question: GoodHands will insure all 100 employees of GoodTire. What premium should GoodHands charge per employee so that the likelihood of losing money on the average of all these claims is 1%?gStep 1: Identify appropriate random variables.

X = the $ / annual claim of customer~N( = 2500, = 1000)

P b Q ti Wh t h ld b th i th t

(i = 1,…,100)•For this problem, you now have the following rvs:

i i

1 100( ... ) /100X X X (a new random var.)

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Prob. Question: What should be the premium, s, so thatP(X > s) = 0.01? XP( > s) = 0.01?

Fact: To answer this prob. quest. about you need to knowXthe density function of .X ???

Idea: When the rv you are interested in is the AVERAGE of other rvs, try…

The Central Limit Theorem

XThe Central Limit Theorem provides an approximate density function when the r.v. you are interested in is the average of n other rvs say X X X that are:

X

(1) Independent

(2) Identically distributed

(knowing the value of one rv tells younothing about the values of the other rvs).

(have the same densityfunction with mean and standard deviation )

average of n other rvs, say, X1, X2, …, Xn, that are:

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function with mean and standard deviation ),then, for “large” n,

(approx.)Nn

XXX n ~...1 ( , / )n

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The Insurance Problem (cont.)For the insurance problem, you have

Xi = annual $ claimed by person i (i = 1, …, 100)

2500 1000N

2500 , 100 .N1 100...100

X XX

(1) Are X1, X2, …, X100 independent random variables?

Yes, because the amount claimed by one person has

~ N

~ 2500, =1000 .N

2500, 1000 / 100

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no effect on the amount claimed by another person.

(2) Are X1, X2, …, X100 identically distributed? Yes, because

Therefore, by the CLT, is approximately Normal with… X

An Insurance Problem

Step 2: State what you are looking for in terms of a probability question pertaining to the random variableprobability question pertaining to the random variable.

•For GoodHands,What should the premium s be so that theprobability that the average of the 100 claimsexceeds s is 0.01?

Probability Question: What should s be so that

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y Q

?01.0100... 1001

sXXXP

An Insurance Problem (cont.)Probability Question: What should the premium s be so that ?01.0 sXP

2500

N(2500, 100) X

St 3 U th b bilit di t ib ti f th

01.0}{ sXP

s

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Step 3: Use the probability distribution of the random variable to answer the probability question.

s = NORMINV(0.99, 2500, 100)

= $2732.64

Another Example of the CLT• In modeling the performance of a team with 5

people, consider the following five rvs:p p , g

Pi = performance contribution of person ifor (i = 1,…,5)

U[0,1]Possible values: [0, 1] (continuous)Density function:

1

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[ ]yE[Pi] = = 0.5 STDEV[Pi] = = 29.0

121

However, what is of interest is the team performance, so let…

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Another Example of the CLTT = performance of the whole team

54321 PPPPP 5

54321

Possible values: [0, 1] (continuous)Density function: ???You cannot find the true density function, so borrow one

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borrow one.Because the rv T is the average of other RVs, think of using the Central Limit Theorem to approximate the density function of T.

0 29

The Team ProblemFor the team problem, you have

Pi = performance of person i (i = 1, 2, 3, 4, 5)0 5 d td dU[0 1] ith 0.29.

(1) Are P1, P2, P3, P4, P5 independent random variables?

Yes, assuming that the performance of a person says

(0.5,0.13).N

0.5 and std. dev. =

0.5,

~U[0, 1] with mean =

554321 PPPPPT

)5/29.0,5.0(~ N )13.0,5.0(N

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nothing about the performance of another person.

(2) Are P1, P2, P3, P4, P5 identically distributed?

Therefore, by the CLT, T is approximately Normal with…

Yes, because

The Team ProblemQuestion: What is the probability that the team performance is at least 0.75?

P(T ≥ 0 75) 0 027

T N(0.5, 0.13)

P(T ≥ 0.75) =

1 – NORMDIST(0.75, 0.5, 0.13, TRUE) =

0.027

P(T ≥ 0.75)

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0.5 0.75

Consider a population of N items in which item i has a number, Xi, associated with it and letX = the value of an item selected randomly from the population.

d i bl

Working with a Population

Is X a random variable?

All possible values:

The answer depends on “timing”.If you have already chosen the item, then X is NOT a rv.

(discrete)If you have not yet chosen the item, then X IS a rv.

X1, X2, …, XN

Density function: 1/N 1/N … 1/N

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E[X] = = the population average µ.

STDEV[X] = the population standard deviation .n

XX n ...1

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The Average of a SampleNow suppose you are going to record the numbers X1, X2,…, Xntaken from a sample of size n from a population and then compute:

XXX n

...1 Is a rv? X

All possible values:

G f i

nX

If you have not yet taken the sample, then IS a rv.X

The answer depends on “timing”.If you have already taken the sample, then X is NOT a rv.

The (finite) list of averages of everygroup of size n in the population.

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There is no practical way to list the possible values, so…

G1Groups of size n:

for the group:X A1

G2

A2

G3

A3

Discrete, but…

YOU CANNOT WRITE THE DENSITY FUNCTION.

The Average of a Sample

nXXX n ...1 The rvs X1, X2,…, Xn are iid

from the same population withmean and std dev

) (~ NX

mean = and std. dev. =

Solution: Because is the average of rvs, think of the using the CLT which, if applicable, results in the following density function for

X

:X, n/

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(, +)Possible Values:

Now you can use the Normal Distribution to answer your probability question about .X

,

How Large is Large Enough?• For symmetric but outlier-prone data,

n = 15 samples should be enough to use the normal approximation.pp

• For mild skewness, n = 30 should generally be sufficient to make the normal approximation appropriate.

• For severe skewness, n should be at least 100 to th l i ti

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use the normal approximation.• Generally speaking, the larger n is, the better the

normal approximation is.

A Final Example of the CLT• Historical data collected at a paper mill show that

40% of sheet breaks are due to water drops, resulting from the condensation of steamresulting from the condensation of steam.

• Suppose that the causes of the next 100 sheet breaks are monitored and that the sheet breaks are independent of one another.

• Find the expected value and the standard deviation of the number of sheet breaks that will be caused

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by water drops.• What is the probability that at least 35 of the

breaks will be due to water drops?

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• Success = break due to water drops• P(success) = p =

Exact Answer

0.4• X = number of breaks due to water drops• X is Binomial with n = 100 and p = 0.4• E(X) =

• From ExcelP(X 35) 1 P(X < 35) 1 P(X 34)

np = (100)(0.4) = 40= (100)(0.4)(0.6) = 24 = 4.9SD(X) = n p (1 p)

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P(X 35) = 1 – P(X < 35) = 1 – P(X 34)• = 1 – BINOMDIST(34, 100, 0.4, TRUE)• = 0.8617

Normal Approx. to BinomialFor this problem, let p = P(success) = 0.4, and

1001

on trialsuccessaif ,1i

iXi

In this problem, you are interested in the rvX = number of successes in 100 trials

= X1 + X2 + … +X100

100,...,1 ,

on trialfailureaif ,0i

iXi

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To find P(X ≥ 35) = P(X / 100 ≥ 35 / 100) , you need to know the probability distribution of

which, by the CLT, is approximately normal, so…

,100/XX

Normal Approx. to BinomialEach Xi ~ Binomial(1, p = 0.4), so

E[Xi] = = p = 0 4E[Xi] p 0.449.0)1(][ ppXSD i

Assuming that•The Xi are pairwise independent and•n = 100 is large enough (np > 5 and n(1 – p) > 5),

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)049.0 ,4.0()/,(~100

1001 NnNXXX L

g g ( p ( p) )then by the CLT, the random variable

Normal Approx. to BinomialThen, for X = X1 + …+ X100

,X

= 1 NORMDIST(0 35 0 4 0 049 TRUE)

P(X / 100 ≥ 35 / 100)

)35.0P( XP(X ≥ 35) =

100 100,

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1 NORMDIST(0.35, 0.4, 0.049, TRUE)

= 0.85. (The exact answer was 0.86.)

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A function y = f(x) describes a relationship between the two quantitative variables x and y.

Review of Basic Math

You can represent a function visually as follows:

•y = f(x) = –x + 2 (a linear relationship)

•y = f(x) = x2 – 2x + 1 (a nonlinear relationship)

y y

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x x

Review of FunctionsYou can also think of a function f as transforming an input x into an output y, as follows:

f

x

f(x ) = y

228

Note: A function f can have many input values, instead of just one.

y b

A linear equation y = mx + b, provides a relationship between the two variables, x and y, in which:

Review of Linear Equations

y

x

y = mx + b

y

•b = the y-intercept

•m = the slope of the line

= the value of y when x = 0.

= the change in y per unit of increase in x.

b

x

m1

x + 1

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y

x

•m > 0: as x increases, y increases. m > 0

m < 0

m = 0•m = 0: as x increases, y remains the same.•m < 0: as x increases, y decreases.

An Example of a LineIf

y = the thousands of bushels of wheaty = the thousands of bushels of wheatx = the number of inches of rain

then, for the liney = 80x + 71,

• b = 71 means that there are 71,000 bushels of

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b 71 means that there are 71,000 bushels of wheat when there is no rain.

• m = 80 means that each extra inch of rain results in 80,000 more bushels of wheat.

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Sometimes a line is written in the form:

A Different Equation for a Line

a1x1 + a2x2 = cAssuming that a2 0, you can solve for x2:

x2 = – (a1 / a2) x1 + (c / a2)

+ b

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y = m x + b

Graphing a Line

To draw the graph of the line a1x1 + a2x2 = b:Fi d t diff i t th li ( ll b• Find two different points on the line (usually by setting x1 = 0 and finding x2 and then setting x2 = 0 and finding x1).

• Plotting these two points on a graph.• Drawing the straight line through those two

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points.

Example of Graphing a Line

The line: 2x1 + x2 = 230 x2

When x1 = 0, x2 = 230

When x2 = 0, x1 = 115

Note: Any point on the line gives a value for x1 100

200

300

233

and a value for x2 that satisfies 2x1 + x2 = 230.

x1300200100

100

Solving Two Linear Equations• Objective: Solve the following two equations

for x1 and x2: 1 22x1 + x2 = 230 (a)x1 + 2x2 = 250 (b)

• Solution Procedure:– Solve (a) for x2:

S b tit t 230 2 i (b)x2 = 230 – 2x1 (c)

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– Substitute x2 = 230 – 2x1 in (b):x1 + 2(230 – 2x1) = –3x1 + 460 = 250 (d)

– Solve (d) for x1: – Substitute x1 = 70 in (c):

x1 = 70x2 = 230 – 2x1 = 90.

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• Objective: Solve the following for x1 and x2: (a) 2x1 + x2 = 230(b) x + 2x = 250

(c) 4x1 + 2x2 = 460

Another Approach

–[ ](b) x + 2x = 250(b) x1 + 2x2 250• Alternative Procedure:

– Multiply (a) through by 2.– Subtract (b) from (c).– Solve (d) for x1:

S b i 70 i ( ) d l f

(d) 3x1 = 210

[ ](b) x1 + 2x2 = 250

x1 = 70

235

– Substitute x1 = 70 in (a) and solve for x2:x2 = 230 – 2x1 = 90

• Note: There are computer packages for solving n linear equations in n unknowns.

Exponentials• An exponent is the power to which a number

(called the base) is raised.• Example: 25 (base = 2; exponent = 5)• Question: How much will $1000 be worth after

5 years at 6% compound interest?Year 1 Year 2 Year 3 Year 4 Year 5

Principal $1,000.00 $1,060.00 $1,123.60 $1,191.02 $1,262.48$60 00 $63 60 $67 42 $71 46 $75 75

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Interest $60.00 $63.60 $67.42 $71.46 $75.75Total $1,060.00 $1,123.60 $1,191.02 $1,262.48 $1,338.23

Answer: Total = f (P, r, n) = P(1 + r )n

= 1000 (1 + 0.06)5 = 1338.23

Properties of Exponents• Laws of Exponents:

– xa + b = xb + a = xa xb (example: 23 + 2 = 23 22)( )b ( b) b ( l (23)2 26)– (xa)b = (xb)a = xab (example: (23)2 = 26)

– x–a = 1 / xa (example: 2–3 = 1 / 23 = 1 / 8)– x0 = 1

• Exponential Functions Increase and Decrease Rapidly:

y = 2 x̂

1200000

y = 2 (̂-x)

0 6

237

0

200000

400000

600000

800000

1000000

1200000

0 5 10 15 20 25

y = 2 x̂

0

0.1

0.2

0.3

0.4

0.5

0.6

0 5 10 15 20 25

y = 2 (̂-x)

Scientific Notation

• Scientific Notation: a 10b (also written as a E ±b) means move the decimal point of a:a E ±b) means move the decimal point of a:– b positions to the right, if b > 0.– b positions to the left, if b < 0.

• Example: 4.000 103 = 4.000 E+3 =• Example: 4 10–3 = 4 E3 =

4000.0.004.

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Logarithms• The log base b of x [written logb(x)] is the

power to which you must raise b to get x.• Examples: log10(100) =• Logs are only defined for positive numbers.• If the base is omitted, the default is 10.• The base e = 2.718… is used in some financial

applications (such as continuous compounding)

2, 5log2(32) =

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applications (such as continuous compounding), in which case, loge(x) is written as ln(x) (the “natural log” of x).

Laws of Logarithms• Logs convert products to sums, that is,

logb(xy) = logb(x) + logb(y).– Ex: log2(64) = log2(416) = log2(4) + log2(16) = 2+4 = 6

• logb(x / y) = logb(x) – logb(y)– Ex: log10(1000 / 100) =

• Logs bring down exponents, that is, logb(xy) = y logb(x). – Example: log2(45) =

• Logs undo exponentiation that is

log10(1000) – log10(100) = 32 = 1

5(2) = 105 log2(4) =

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• Logs undo exponentiation, that is,logb(by) = y logb(b) = y.– Example: log2(25) =

• loga(x) = k logb(x), where k = loga(b) – Example: log2(x) = 3.322 log10(x)

5

Problem Solving with Logs• Question: How many years will it take to

double an investment at i % interest compounded annually?compounded annually?

• Answer: LetP = the initial investmentr = interest rate as a fraction = i / 100n = the number of years of compounding

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n = the number of years of compoundingThen, after n years, you will have

P(1 + r )n.

Problem Solving with Logs• Answer (continued):

Thus, you want to find n so thatP(1 + r )n = 2P (1 + r )n = 2 (a)P(1 + r )n = 2P

To solve (a) for n, take the log of both sides to bring the exponent n down:

log[(1 + r )n] = log(2) n log[(1 + r )] = log(2)

(1 + r )n = 2 (a)

Qn: Log base what?Ans: Log base 10 (but

b ill k)

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g[( )] g( )n = log(2) / log[(1 + r )]

• Example: At 6% (r = 0.06), it will take n = log(2) / log(1.06) = 0.301 / 0.025 = 11.9 years.

any base will work).