doane chapter 06

Discrete DistributionsDiscrete DistributionsDiscrete DistributionsDiscrete DistributionsProbability Models

Discrete Distributions

Uniform Distribution

Bernoulli Distribution

Binomial Distribution

Poisson Distribution

Hypergeometric Distribution

Geometric Distribution (optional)

Transformations of Random Variables (optional)

Chapter6666

Probability ModelsProbability ModelsProbability ModelsProbability Models

McGraw-Hill/Irwin © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

• A A randomrandom (or stochastic) (or stochastic) processprocess is a repeatable is a repeatable random experiment.random experiment.

• Probability can be used to analyze random (or stochastic) Probability can be used to analyze random (or stochastic) processes and to understand business processes.processes and to understand business processes.

• For example, each call arriving at the L.L. Bean For example, each call arriving at the L.L. Bean order center is a random experiment in which the order center is a random experiment in which the variable of interest is the amount of the order.variable of interest is the amount of the order.

Probability ModelsProbability Models

• Probability modelsProbability models depict the essential depict the essential characteristics of a stochastic process to guide characteristics of a stochastic process to guide decisions or make predictions.decisions or make predictions.

• L.L. Bean would need to model its call center L.L. Bean would need to model its call center process using a realistic yet simple model.process using a realistic yet simple model.

• For example, can L.L. Bean predict its total order For example, can L.L. Bean predict its total order amount from the next 50 callers?amount from the next 50 callers?

• Probability models with well-known properties can Probability models with well-known properties can be used to describe many stochastic processes.be used to describe many stochastic processes.

Probability ModelsProbability ModelsProbability ModelsProbability Models

Probability ModelsProbability Models

Discrete DistributionsDiscrete Distributions

• A A random variablerandom variable is a function or rule that assigns a numerical value to is a function or rule that assigns a numerical value to each outcome in the sample space of a random experiment.each outcome in the sample space of a random experiment.

Random VariablesRandom Variables

• Nomenclature:Nomenclature:- Capital letters are used to represent- Capital letters are used to represent random variablesrandom variables (e.g., (e.g., XX, , YY).).- Lower case letters are used to represent - Lower case letters are used to represent valuesvalues of the random variable (e.g., of the random variable (e.g., xx, , yy).).

• A A discrete random variablediscrete random variable has a countable has a countable number of distinct values.number of distinct values.

Decision ProblemDiscrete Random Variable

(Range)

On the late morning (9 to 12) work shift, L.L. Bean’s order processing center staff can handle up to 5 orders per minute. The mean arrival rate is 3.5 orders per minute. What is the probability that more than 5 orders will arrive in a given minute?

X = number of phone calls that arrive in a given minute at the L.L. Bean order processing center (X = 0, 1, 2, ...)

For example,


Random VariablesRandom Variables

Probability DistributionsProbability Distributions• A A discrete probability distributiondiscrete probability distribution assigns a probability assigns a probability

to each value of a discrete random variable to each value of a discrete random variable XX..

• To be a valid probability, each probability must be To be a valid probability, each probability must be betweenbetween 0 0 P P((xxii)) 1 1

• and the sum of all the probabilities for the values of and the sum of all the probabilities for the values of XX must be equal to unity. must be equal to unity.

1

( ) 1n

ii

P x


When you flip a coin When you flip a coin three times, the three times, the sample space has sample space has eight equally likely eight equally likely simple events. simple events. They are:They are:

Example: Coin FlipsExample: Coin Flips

11stst Toss Toss 22ndnd Toss 3 Toss 3rdrd Toss TossHH HH HHHH HH TTHH TT HHHH TT TTTT HH HHTT HH TTTT TT HHTT TT TT


If If XX is the number of heads, then is the number of heads, then XX is a random is a random variable whose probability distribution is as follows:variable whose probability distribution is as follows:

Possible EventsPossible Events xx PP((xx))

TTTTTT 00 1/81/8

HTT, THT, TTHHTT, THT, TTH 11 3/83/8

HHT, HTH, THHHHT, HTH, THH 22 3/83/8

HHHHHH 33 1/81/8

TotalTotal 11



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Note that the values of Note that the values of XX need need not be equally likely. However, not be equally likely. However, they must sum to unity.they must sum to unity.

Note also that a discrete probability Note also that a discrete probability distribution is defined only at distribution is defined only at specific points on the specific points on the XX-axis.-axis.



• The The expected valueexpected value E E((XX) of a discrete random variable is the ) of a discrete random variable is the sum of all sum of all XX-values -values weightedweighted by their respective probabilities. by their respective probabilities.

• The The EE((XX) is a measure of ) is a measure of central tendencycentral tendency..

• If there are If there are nn distinct values of distinct values of XX, ,

1

( ) ( )n

i ii

E X x P x


Expected ValueExpected Value

The probability distribution of emergency service calls The probability distribution of emergency service calls on Sunday by Ace Appliance Repair is:on Sunday by Ace Appliance Repair is:

What is the average or What is the average or expectedexpected number of service number of service calls?calls?

xx PP((xx))

00 0.050.05

11 0.100.10

22 0.300.30

33 0.250.25

44 0.200.20

55 0.100.10

TotalTotal 1.001.00


Example: Service CallsExample: Service Calls

The sum of the The sum of the xPxP((xx) column is the expected ) column is the expected value or mean of the discrete distribution.value or mean of the discrete distribution.

x P(x) xP(x)

0 0.05 0.00

1 0.10 0.10

2 0.30 0.60

3 0.25 0.75

4 0.20 0.80

5 0.10 0.50

Total 1.00 2.75

5

1

( ) ( )i ii

E X x P x

First calculate First calculate xxiiPP((xxii):):



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This particular This particular probability distribution probability distribution is not symmetric is not symmetric around the mean around the mean = 2.75. = 2.75.

However, the mean However, the mean is still the balancing is still the balancing point, or fulcrum.point, or fulcrum. = 2.75= 2.75

Because Because EE((XX) is an ) is an averageaverage, it does not have to be an , it does not have to be an observable point.observable point.



• Expected value is the basis of life insurance.Expected value is the basis of life insurance.• For example, what is the probability that a 30-year-old white female will For example, what is the probability that a 30-year-old white female will

die within the next year?die within the next year?

• Based on mortality statistics, the probability is .00059 and the probability Based on mortality statistics, the probability is .00059 and the probability of living another year is 1 - .00059 = .99941.of living another year is 1 - .00059 = .99941.

• What premium should a life insurance company charge to break even on What premium should a life insurance company charge to break even on a $500,000 1-year term policy?a $500,000 1-year term policy?


Application: Life InsuranceApplication: Life Insurance

Let Let XX be the amount paid by the company to settle the policy. be the amount paid by the company to settle the policy.

Event x P(x) xP(x)

Live 0 .99941 0.00

Die 500,000 .00059 295.00

Total 1.00000 295.00Source: Centers for Disease Control and Prevention, National Vital Statistics Reports, 47, no. 28 (1999).

The total expected payout isThe total expected payout is

So, the premium should be $295 plus whatever return the company needs So, the premium should be $295 plus whatever return the company needs to cover administrative overhead and profit.to cover administrative overhead and profit.


Application: Life InsuranceApplication: Life Insurance

• Expected value can be applied to raffles and lotteries.Expected value can be applied to raffles and lotteries.

• If it costs $2 to buy a ticket in a raffle to win a new car worth $55,000 and If it costs $2 to buy a ticket in a raffle to win a new car worth $55,000 and 29,346 raffle tickets are sold, what is the expected value of a raffle ticket?29,346 raffle tickets are sold, what is the expected value of a raffle ticket?

• If you buy 1 ticket, what is the chance you will If you buy 1 ticket, what is the chance you will

win =win = 1 1 29,34629,346 lose =lose =

29,345 29,345 29,34629,346


Application: Raffle TicketsApplication: Raffle Tickets

• Now, calculate the Now, calculate the EE((XX):):

EE((XX)) = (value if you win)= (value if you win)PP(win) + (value if you lose)(win) + (value if you lose)PP(lose)(lose)

= (55,000)= (55,000) 1 1 29,34629,346

+ (0)+ (0) 29,345 29,345 29,34629,346

= (55,000)(.000034076) + (0)(.999965924) = $1.87= (55,000)(.000034076) + (0)(.999965924) = $1.87

• The raffle ticket is actually worth $1.87. Is it worth spending $2.00 for The raffle ticket is actually worth $1.87. Is it worth spending $2.00 for it?it?


Application: Raffle TicketsApplication: Raffle Tickets

• An An actuarially fairactuarially fair insurance program must collect as much in overall insurance program must collect as much in overall revenue as it pays out in claims.revenue as it pays out in claims.

• Accomplish this by setting the premiums to reflect empirical experience Accomplish this by setting the premiums to reflect empirical experience with the insured group.with the insured group.

• If the pool of insured persons is large enough, the total payout is If the pool of insured persons is large enough, the total payout is predictable.predictable.


Actuarial FairnessActuarial Fairness

• If there are If there are nn distinct values of distinct values of XX, then the , then the variancevariance of a discrete of a discrete random variable is:random variable is:

• The variance is a The variance is a weightedweighted average of the average of the dispersiondispersion about the mean and about the mean and is denoted either as is denoted either as 22 or or VV((XX).).

2 2

1

( ) [ ] ( )n

i ii

V X x P x

• The The standard deviationstandard deviation is the square root of the variance and is denoted is the square root of the variance and is denoted ..

2 ( )V X


Variance and Standard DeviationVariance and Standard Deviation

The probability distribution of The probability distribution of room rentals during February is:room rentals during February is:

The Bay Street Inn is a 7-room bed-and-The Bay Street Inn is a 7-room bed-and-breakfast in Santa Theresa, Ca.breakfast in Santa Theresa, Ca.

x P(x)

0 0.05

1 0.05

2 0.06

3 0.10

4 0.13

5 0.20

6 0.15

7 0.26

Total 1.00


Example: Bed and BreakfastExample: Bed and Breakfast

First find the expected valueFirst find the expected value x P(x)

0 0.05

1 0.05

2 0.06

3 0.10

4 0.13

5 0.20

6 0.15

7 0.26

Total 1.00

7

1

( ) ( )i ii

E X x P x

= 4.71 rooms= 4.71 rooms

x P(x)

0.00

0.05

0.12

0.30

0.52

1.00

0.90

1.82

= 4.71



The The EE((XX) is then used to find ) is then used to find the variance:the variance:

xx PP((xx)) x Px P((xx))

00 0.050.05 0.000.00

11 0.050.05 0.050.05

22 0.060.06 0.120.12

33 0.100.10 0.300.30

44 0.130.13 0.520.52

55 0.200.20 1.001.00

66 0.150.15 0.900.90

77 0.260.26 1.821.82

TotalTotal 1.001.00 = 4.71= 4.71

72 2

1

( ) [ ] ( )i ii

V X x P x

= 4.2259 rooms2

The standard deviation is:The standard deviation is:

= 4.2259= 4.2259 = 2.0577 rooms= 2.0577 rooms

[[xx]]22 P P((xx))

1.1092051.109205

0.6882050.688205

0.4406460.440646

0.2924100.292410

0.0655330.065533

0.0168200.016820

0.2496150.249615

1.3634661.363466

22 = 4.225900 = 4.225900

[[xx]]22

22.184122.1841

13.764113.7641

7.34417.3441

2.92412.9241

0.50410.5041

0.08410.0841

1.66411.6641

5.24415.2441



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Number of Rooms Rented

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The histogram shows that the distribution is skewed to the left and The histogram shows that the distribution is skewed to the left and bimodal.bimodal.

= 2.06 indicates considerable variation around = 2.06 indicates considerable variation around ..

The mode is 7 The mode is 7 rooms rented but rooms rented but

the average is only 4.71 room rentals.the average is only 4.71 room rentals.



• A A probability distribution functionprobability distribution function (PDF) is a mathematical function that (PDF) is a mathematical function that shows the probability of each shows the probability of each XX-value.-value.

• A A cumulative distribution functioncumulative distribution function (CDF) is a mathematical function that shows the cumulative (CDF) is a mathematical function that shows the cumulative sum of probabilities, adding from the smallest to the largest sum of probabilities, adding from the smallest to the largest XX-value, gradually approaching unity.-value, gradually approaching unity.


What is a PDF or CDF?What is a PDF or CDF?

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Illustrative PDFIllustrative PDF(Probability Density Function)(Probability Density Function)

Cumulative CDFCumulative CDF(Cumulative Density Function)(Cumulative Density Function)

Consider the following illustrative histograms:Consider the following illustrative histograms:

The equations for these functions depend on the The equations for these functions depend on the parameter(s)parameter(s) of the distribution. of the distribution.


What is a PDF or CDF?What is a PDF or CDF?

Characteristics of the Uniform DistributionCharacteristics of the Uniform Distribution• The The uniform distributionuniform distribution describes a random variable with a finite number describes a random variable with a finite number

of integer values from of integer values from aa to to bb (the only two parameters). (the only two parameters).

• Each value of the random variable is equally likely to occur.Each value of the random variable is equally likely to occur.

• Consider the following summary of the uniform distribution:Consider the following summary of the uniform distribution:

Uniform DistributionUniform Distribution

ParametersParameters aa = lower limit = lower limitbb = upper limit = upper limit

PDFPDF

RangeRange aa xx bb

MeanMean

Std. Dev.Std. Dev.

Random data generation Random data generation in Excelin Excel =a+INT((b-a+1)*RAND())=a+INT((b-a+1)*RAND())

CommentsComments Used as a benchmark, to generate random Used as a benchmark, to generate random integers, or to create other distributions.integers, or to create other distributions.

2( ) 1 1

12

b a

1( )

1P x

b a

2

a b


• The number of dots on the roll of a die form a uniform random variable The number of dots on the roll of a die form a uniform random variable with six equally likely integer values: 1, 2, 3, 4, 5, 6with six equally likely integer values: 1, 2, 3, 4, 5, 6

0.00

0.02

0.04

0.06

0.08

0.10

0.12

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0.16

0.18

1 2 3 4 5 6

Number of Dots Showing on the Die

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1.00

1 2 3 4 5 6

Number of Dots Showing on the Die

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PDF for one diePDF for one die CDF for one dieCDF for one die

• What is the probability of rolling any of these?What is the probability of rolling any of these?


Example: Rolling a DieExample: Rolling a Die

• The PDF for all The PDF for all xx is: is:

• Calculate the standard deviation as:Calculate the standard deviation as:

• Calculate the mean as:Calculate the mean as:

1 1 1( )

1 6 1 1 6P x

b a

1 63.5

2 2

a b

2 2( ) 1 1 (6 1) 1 1

1.70812 12

b a


Example: Rolling a DieExample: Rolling a Die

On a gas pump, the last two digits (pennies) displayed will be a uniform On a gas pump, the last two digits (pennies) displayed will be a uniform random integer (assuming the pump stops automatically).random integer (assuming the pump stops automatically).

0.000

0.002

0.004

0.006

0.008

0.010

0.012

0 10 20 30 40 50 60 70 80 90

Pennies Digits on Pump

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0 10 20 30 40 50 60 70 80 90

Pennies Digits on Pump

PDFPDF CDFCDFThe parameters are: The parameters are: aa = 00 and = 00 and bb = 99 = 99


Application: Pumping GasApplication: Pumping Gas

• The PDF for all The PDF for all xx is: is:

• Calculate the standard deviation as:Calculate the standard deviation as:

• Calculate the mean as:Calculate the mean as:

1 1 1( ) .010

1 99 0 1 100P x

b a

0 9949.5

2 2

a b

2 2( ) 1 1 (99 0) 1 1

28.8712 12

b a


Application: Pumping GasApplication: Pumping Gas

• To generate random integers from a discrete uniform distribution, use Excel functionTo generate random integers from a discrete uniform distribution, use Excel function=a+INT((b-a+1*RAND()))=a+INT((b-a+1*RAND()))

• To obtain To obtain nn distinct random integers, generate a few extra numbers and distinct random integers, generate a few extra numbers and then delete the duplicate values.then delete the duplicate values.

• To create integers 1 through To create integers 1 through NN, set , set aa = 1 and = 1 and bb = = NN and use Excel function and use Excel function=1+INT(N*RAND())=1+INT(N*RAND())


Uniform Random IntegersUniform Random Integers

• The finance department at Zymurgy, Inc., has a new digital copier that The finance department at Zymurgy, Inc., has a new digital copier that requires a unique user ID code for each of the 37 users.requires a unique user ID code for each of the 37 users.

• Generate unique 4-digit uniform random integers from 1000 to 9999 using Generate unique 4-digit uniform random integers from 1000 to 9999 using the function =1000+INT(9000*RAND()) in an Excel spreadsheet.the function =1000+INT(9000*RAND()) in an Excel spreadsheet.


Application: Copier CodesApplication: Copier Codes

After entering the formula, drag it After entering the formula, drag it down to fill 50 cells with randomly down to fill 50 cells with randomly generated numbers following the generated numbers following the uniform distribution. uniform distribution.



After highlighting and copying After highlighting and copying the cells to the clipboard, the cells to the clipboard, paste only the values (not the paste only the values (not the formulas) to another column formulas) to another column using using Paste Special – ValuesPaste Special – Values. . Now these values can be Now these values can be sorted. sorted.



Sort the random numbers using Sort the random numbers using Data – Sort.Data – Sort.

Use the first 37 random numbers as Use the first 37 random numbers as copier codes for the current copier codes for the current employees and save the remaining employees and save the remaining codes for future employees.codes for future employees.



Here is the uniform distribution Here is the uniform distribution for one die from for one die from LearningStatsLearningStats..


Uniform Model in LearningStatsUniform Model in LearningStats

• A random experiment with only 2 outcomes is a A random experiment with only 2 outcomes is a Bernoulli experimentBernoulli experiment..

• One outcome is arbitrarily labeled a One outcome is arbitrarily labeled a “success” (denoted “success” (denoted XX = 1) and the other a “failure” (denoted = 1) and the other a “failure” (denoted XX = 0). = 0).

is the is the PP(success), 1 – (success), 1 – is the is the PP(failure).(failure).

• Note that Note that PP(0) + (0) + PP(1) = (1 – (1) = (1 – ) + ) + = 1 and = 1 and0 0 << << 1. 1.

• ““Success” is usually defined as the less likely outcome so that Success” is usually defined as the less likely outcome so that < .5 < .5 for convenience.for convenience.

Bernoulli ExperimentsBernoulli Experiments

Bernoulli DistributionBernoulli Distribution

Bernoulli Experiment Possible Outcomes Probability of “Success”

Flip a coin 1 = heads0 = tails

= .50

Consider the following Bernoulli experiments:Consider the following Bernoulli experiments:

Inspect a jet turbine blade 1 = crack found0 = no crack found

= .001

Purchase a tank of gas 1 = pay by credit card0 = do not pay by credit card

= .78

Do a mammogram test 1 = positive test0 = negative test

= .0004



2

2 2 2

1

( ) ( ) ( ) (0 ) (1 ) (1 ) ( ) (1 )i ii

V X x E X P x

• The expected value (mean) of a Bernoulli experiment is calculated as:The expected value (mean) of a Bernoulli experiment is calculated as:

2

1

( ) ( ) (0)(1 ) (1)( )i ii

E X x P x

• The variance of a Bernoulli experiment is calculated as:The variance of a Bernoulli experiment is calculated as:

• The mean and variance are useful in developing the next model.The mean and variance are useful in developing the next model.



• The The binomial distributionbinomial distribution arises when a Bernoulli experiment is arises when a Bernoulli experiment is repeated repeated nn times. times.

• Each Bernoulli trial is independent so the probability of success Each Bernoulli trial is independent so the probability of success remains remains constant on each trial.constant on each trial.

• In a binomial experiment, we are interested in In a binomial experiment, we are interested in XX = number of successes = number of successes in in nn trials. So, trials. So,

XX = = xx11 + + xx22 + ... + + ... + xxnn • The probability of a particular number of successes The probability of a particular number of successes PP((XX) is determined ) is determined

by parameters by parameters nn and and ..

Binomial DistributionBinomial Distribution

Characteristics of the Binomial DistributionCharacteristics of the Binomial Distribution

• The mean of a binomial distribution is found by adding the means for The mean of a binomial distribution is found by adding the means for each of the each of the nn Bernoulli independent events: Bernoulli independent events:

+ + + … + + … + = = nn• The variance of a binomial distribution is found by adding the variances The variance of a binomial distribution is found by adding the variances

for each of the for each of the nn Bernoulli independent events: Bernoulli independent events:

(1-(1-)+ )+ (1-(1-) + … + ) + … + (1-(1-) = ) = nn(1-(1-))• The standard deviation isThe standard deviation is

nn(1-(1-))


Characteristics of the Binomial DistributionCharacteristics of the Binomial Distribution

Parameters n = number of trials = probability of success

PDF

Excel function =BINOMDIST(k,n,,0)

Range X = 0, 1, 2, . . ., n

Mean n

Std. Dev.

Random data generation in Excel

Sum n values of =1+INT(2*RAND()) or use Excel’s Tools | Data Analysis

Comments Skewed right if < .50, skewed left if > .50, and symmetric if = .50.

!( ) (1 )

!( )!x n xn

P xx n x

(1 )n


• It is important to quick oil change shops to ensure that a car’s service It is important to quick oil change shops to ensure that a car’s service time is not considered “late” by the customer.time is not considered “late” by the customer.

• Service times are defined as either Service times are defined as either latelate or or not latenot late..• XX is the number of cars that are late out of the total number of cars is the number of cars that are late out of the total number of cars

serviced.serviced.

• Assumptions:Assumptions: - cars are independent of each other - cars are independent of each other - probability of a late car is consistent - probability of a late car is consistent


Example: Quick Oil Change ShopExample: Quick Oil Change Shop

• What is the probability that exactly 2 of the next What is the probability that exactly 2 of the next nn = 10 cars serviced = 10 cars serviced are late (are late (PP((X X = 2))?= 2))?

• PP(car is late) = (car is late) = = .10 = .10

• PP(car not late) = 1 - (car not late) = 1 - = .90 = .90

!( ) (1 )

!( )!x n xn

P xx n x

PP((X X = 2) = = 2) = 10!10!

2!(10-2)!2!(10-2)!(.1)(.1)22(1-.10)(1-.10)10-210-2 = .1937= .1937



• Alternatively, we could find Alternatively, we could find PP((X X = 2) using the Excel function = 2) using the Excel function =BINOMDIST(k,n,=BINOMDIST(k,n,,0) where,0) where

kk = the number of “successes” in = the number of “successes” in nn trials trials

nn = the number of independent trials = the number of independent trials = probability of = probability of a “success” a “success”0 means that we 0 means that we want to calculate want to calculate PP((XX = 2) rather = 2) rather than than PP((XX << 2) 2)



• A binomial distribution is A binomial distribution is skewed right if skewed right if < .50, < .50, skewed left if skewed left if > .50, > .50, and symmetric if and symmetric if = .50 = .50

• Skewness decreases as Skewness decreases as nn increases, regardless of the value of increases, regardless of the value of ..

• To illustrate, consider the following graphs:To illustrate, consider the following graphs:


Binomial ShapeBinomial Shape

p = .20Skewed right

p = .50Symmetric

p = .80Skewed left

n = 5

0.00

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0 5 10 15 20

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p = .20Skewed right

p = .50Symmetric

p = .80Skewed left

n = 100.00

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p = .20Skewed right

p = .50Symmetric

p = .80Skewed left

n = 200.00

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0 5 10 15 20

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p = .20Skewed right

p = .50Symmetric

p = .80Skewed left

n = 5

n = 10

n = 200.00

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0 5 10 15 20

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• On average, 20% of the emergency room patients at Greenwood General Hospital lack health insurance.

• In a random sample of 4 patients, what is the probability that at least 2 will be uninsured?

• X = number of uninsured patients (“success”)• P(uninsured) = = 20% or .2020% or .20

• P(insured) = 1 – = 1 – .20 = .801 – .20 = .80

• n =

• The range is X = 0, 1, 2, 3, 4 patients.4 patients4 patients


Application: Uninsured PatientsApplication: Uninsured Patients

• What is the mean and standard deviation of this binomial distribution?What is the mean and standard deviation of this binomial distribution?

Mean = Mean = = = nn = = (4)(.20) = (4)(.20) = 0.8 patients0.8 patients

Standard deviation = Standard deviation = = = (1 )n

= 4(.20(1-.20)= 4(.20(1-.20)

= 0.8 patients= 0.8 patients



Here is the binomial distribution for n = 4, = .20x PDF CDF

0 .4096 = P(X=0) .4096 = P(X<0)=P(0)

1 .4096 = P(X=1) .8192 = P(X<1)=P(0)+P(1)

2 .1536 = P(X=2) .9728 = P(X<2)=P(0)+P(1)+P(2)

3 .0256 = P(X=3) .9984 = P(X<3)=P(0)+P(1)+P(2)+P(3)

4 .0016 = P(X=4) 1.0000 = P(X<4)=P(0)+P(1)+P(2)+P(3)+P(4)

These probabilities can be calculated using a These probabilities can be calculated using a calculator or Excel’s function calculator or Excel’s function =BINOMDIST(x,n,p,cumulative) where =BINOMDIST(x,n,p,cumulative) where cumulativecumulative = 0 for a PDF or = 1 for a CDF = 0 for a PDF or = 1 for a CDF


PDF formula calculations: PDF Excel formula:

=.4096= BINOMDIST(0,4,.2,0)

=.4096= BINOMDIST(1,4,.2,0)

=.1536= BINOMDIST(2,4,.2,0)

=.0256= BINOMDIST(3,4,.2,0)

=.0016= BINOMDIST(4,4,.2,0)4 4 4 4 04!

(4) (.2) (1 .2) 1 .2 .84!(4 4)!

P

3 4 3 3 14!(3) (.2) (1 .2) 4 .2 .8

3!(4 3)!P

2 4 2 2 24!(2) (.2) (1 .2) 4 .2 .8

2!(4 2)!P

1 4 1 1 34!(1) (.2) (1 .2) 4 .2 .8

1!(4 1)!P

0 4 0 0 44!(0) (.2) (1 .2) 1 .2 .8

0!(4 0)!P



Binomial probabilities can also be determined by looking them up in a table Binomial probabilities can also be determined by looking them up in a table ((Appendix AAppendix A) for selected values of ) for selected values of nn (row) and (row) and (column). (column).

n X 0.01 0.02 0.05 0.10 0.15 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.85 0.90 0.95 0.98 0.99

2 0 0.9801 0.9604 0.9025 0.8100 0.7225 0.6400 0.4900 0.3600 0.2500 0.1600 0.0900 0.0400 0.0225 0.0100 0.0025 0.0004 0.00011 0.0198 0.0392 0.0950 0.1800 0.2550 0.3200 0.4200 0.4800 0.5000 0.4800 0.4200 0.3200 0.2550 0.1800 0.0950 0.0392 0.01982 0.0001 0.0004 0.0025 0.0100 0.0225 0.0400 0.0900 0.1600 0.2500 0.3600 0.4900 0.6400 0.7225 0.8100 0.9025 0.9604 0.9801

3 0 0.9703 0.9412 0.8574 0.7290 0.6141 0.5120 0.3430 0.2160 0.1250 0.0640 0.0270 0.0080 0.0034 0.0010 0.0001 -- --1 0.0294 0.0576 0.1354 0.2430 0.3251 0.3840 0.4410 0.4320 0.3750 0.2880 0.1890 0.0960 0.0574 0.0270 0.0071 0.0012 0.00032 0.0003 0.0012 0.0071 0.0270 0.0574 0.0960 0.1890 0.2880 0.3750 0.4320 0.4410 0.3840 0.3251 0.2430 0.1354 0.0576 0.02943 -- -- 0.0001 0.0010 0.0034 0.0080 0.0270 0.0640 0.1250 0.2160 0.3430 0.5120 0.6141 0.7290 0.8574 0.9412 0.9703

4 0 0.9606 0.9224 0.8145 0.6561 0.5220 0.4096 0.2401 0.1296 0.0625 0.0256 0.0081 0.0016 0.0005 0.0001 -- -- --1 0.0388 0.0753 0.1715 0.2916 0.3685 0.4096 0.4116 0.3456 0.2500 0.1536 0.0756 0.0256 0.0115 0.0036 0.0005 -- --2 0.0006 0.0023 0.0135 0.0486 0.0975 0.1536 0.2646 0.3456 0.3750 0.3456 0.2646 0.1536 0.0975 0.0486 0.0135 0.0023 0.00063 -- -- 0.0005 0.0036 0.0115 0.0256 0.0756 0.1536 0.2500 0.3456 0.4116 0.4096 0.3685 0.2916 0.1715 0.0753 0.03884 -- -- -- 0.0001 0.0005 0.0016 0.0081 0.0256 0.0625 0.1296 0.2401 0.4096 0.5220 0.6561 0.8145 0.9224 0.9606



• Individual probabilities can be added to obtain any desired event Individual probabilities can be added to obtain any desired event probability.probability.

• For example, the probability that the sample of 4 patients will contain For example, the probability that the sample of 4 patients will contain at at leastleast 2 uninsured patients is 2 uninsured patients is

= .1536 + .0256 + .0016 = .1808 = .1536 + .0256 + .0016 = .1808

• HINT: What inequality means “at least?”HINT: What inequality means “at least?”

PP((XX 2) = 2) = PP(2) + (2) + PP(3) + (3) + PP(4)(4)


Compound EventsCompound Events

• What is the probability that What is the probability that fewer thanfewer than 22 patients have insurance?patients have insurance?

= .4096 + .4096 = .8192

• HINT: What inequality means “fewer than?”HINT: What inequality means “fewer than?”

• What is the probability that What is the probability that no more thanno more than 22 patients have insurance?patients have insurance?

= .4096 + .4096 + .1536 = .9728

• HINT: What inequality means “no more than?”HINT: What inequality means “no more than?”

P(X < 2) = P(0) + P(1)

P(X < 2) = P(0) + P(1) + P(2)



0 1 2 3 4

0 1 2 3 4

0 1 2 3 4

P (X 2)

P (X < 2 or X > 2)

"At least two successes."

"Fewer than two or more than two successes."

P (X < 2)"Fewer than two successes."

It is helpful to sketch a diagram:



• Use Excel’s Insert | Function menu to calculate the probability of Use Excel’s Insert | Function menu to calculate the probability of xx = 67 = 67 successes in successes in nn = 1,024 trials with probability = 1,024 trials with probability = .048. = .048.

• Or use =BINOMDIST(67,1024,0.048,0)Or use =BINOMDIST(67,1024,0.048,0)


Binomial Probabilities: ExcelBinomial Probabilities: Excel

• Compute an entire binomial PDF for any Compute an entire binomial PDF for any nn and and (e.g., (e.g., nn= 10, = 10, = .50) in = .50) in MegaStat.MegaStat.


Binomial Probabilities: MegaStatBinomial Probabilities: MegaStat

MegaStat also gives you the option to create a graph of the PDF:

Binomial distribution

10 n0.5 p

cumulativeX p(X) probability0 0.00098 0.000981 0.00977 0.010742 0.04395 0.054693 0.11719 0.171884 0.20508 0.376955 0.24609 0.623056 0.20508 0.828137 0.11719 0.945318 0.04395 0.989269 0.00977 0.99902

10 0.00098 1.000001.00000

5.000 expected value2.500 variance1.581 standard deviation

Binomial distribution (n = 10, p = 0.5)

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p(X

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Binomial Probabilities: MegaStatBinomial Probabilities: MegaStat

• Using Using Visual Statistics Visual Statistics Module 4, here is a binomial distribution for Module 4, here is a binomial distribution for nn = = 10 and 10 and = .50: = .50:

Copy and paste graph Copy and paste graph as a bitmap.as a bitmap.Copy and paste Copy and paste probabilities into Excel. probabilities into Excel.

““Spin” Spin” nn and and and super-impose a and super-impose a normal curve on the binomial distribution.normal curve on the binomial distribution.


Binomial Probabilities: Visual StatisticsBinomial Probabilities: Visual Statistics

Here, Here, nn = 50 and = 50 and = .095. = .095.

Spin buttons Spin buttons let you vary let you vary nn and and ..


Binomial Probabilities: LearningStatsBinomial Probabilities: LearningStats

• Generate a Generate a singlesingle binomial random number in Excel by summing binomial random number in Excel by summing nn Bernoulli random variables Bernoulli random variables (0 or 1) using the function (0 or 1) using the function = 0 + INT(1*RAND()). = 0 + INT(1*RAND()).

• Alternatively, use Excel’s Alternatively, use Excel’s Tools | Data Analysis to get binomial random data.Tools | Data Analysis to get binomial random data.

• This will generate 20 binomial random data values using This will generate 20 binomial random data values using nn = 4 and = 4 and = .20. = .20.


Binomial Random DataBinomial Random Data

• Can you recognize a binomial situation? Look for Can you recognize a binomial situation? Look for nn independent Bernoulli independent Bernoulli trials with constant probability of success.trials with constant probability of success.

In a sample of 20 friends:In a sample of 20 friends:

• How many are left-handed?How many are left-handed?

• How many have ever worked on a factory How many have ever worked on a factory floor?floor?

• How many own a motorcycle?How many own a motorcycle?


Recognizing Binomial ApplicationsRecognizing Binomial Applications

• Can you recognize a binomial situation? Can you recognize a binomial situation? In a sample of 50 cars in a parking lot:In a sample of 50 cars in a parking lot:

• How many are parked end-first?How many are parked end-first?• How many are blue?How many are blue?

• How many have hybrid engines?How many have hybrid engines?In a sample of 10 emergency patients with chest In a sample of 10 emergency patients with chest pain:pain:

• How many will be admitted?How many will be admitted?• How many will need bypass surgery?How many will need bypass surgery?• How many will be uninsured?How many will be uninsured?


Recognizing Binomial ApplicationsRecognizing Binomial Applications

• The The Poisson distributionPoisson distribution was named for French mathematician Sim was named for French mathematician Siméon éon Poisson (1781-1840).Poisson (1781-1840).

• The Poisson distribution describes the number of occurrences within a The Poisson distribution describes the number of occurrences within a randomly chosen unit of time or space.randomly chosen unit of time or space.

• For example, within a minute, hour, For example, within a minute, hour, day, square foot, or linear mile.day, square foot, or linear mile.

Poisson DistributionPoisson Distribution

Poisson ProcessesPoisson Processes

• The events occur randomly and inde-pendently over a continuum of The events occur randomly and inde-pendently over a continuum of time or space:time or space:

One Unit One Unit One Unit

of Time of Time of Time

|---| |---| |---|• • •• • • • •••• • • • •• • • ••• • • •

Flow of Time

• Called the Called the model of arrivalsmodel of arrivals, most Poisson applications model , most Poisson applications model arrivals arrivals per unit of timeper unit of time..

• Each dot (Each dot (••) is an occurrence of the event of interest.) is an occurrence of the event of interest.


• Let Let XX = the number of events per unit of time. = the number of events per unit of time.• XX is a random variable that depends on when the unit of time is is a random variable that depends on when the unit of time is

observed.observed.

• For example, we could get For example, we could get XX = 3 = 3 or or XX = 1 = 1 or or XX = 5 = 5 events, depending on where the randomly chosen unit of time happens to fall. events, depending on where the randomly chosen unit of time happens to fall.

One Unit One Unit One Unit

of Time of Time of Time

|---| |---| |---|• • •• • • • •••• • • • •• • • ••• • • •

Flow of Time


• Arrivals (e.g., customers, defects, accidents) must be independent of each other.

• Some examples of Poisson models in which assumptions are sufficiently met are:

• X = number of customers arriving at a bank ATM in a given minute.

• X = number of file server virus infections at a data center during a 24-hour period.

• X = number of blemishes per sheet of white bond paper.


represents the represents the mean number of events per unit of time or spacemean number of events per unit of time or space ..

• The unit of time should be short enough that the mean arrival rate is not The unit of time should be short enough that the mean arrival rate is not large (large (< 20).< 20).

• To make To make smaller, convert to a smaller time unit (e.g., convert hours to smaller, convert to a smaller time unit (e.g., convert hours to minutes).minutes).

• The Poisson model’s only parameter is The Poisson model’s only parameter is (Greek letter “lambda”). (Greek letter “lambda”).



• The number of events that can occur in a given unit of time is not The number of events that can occur in a given unit of time is not bounded, therefore bounded, therefore XX has no obvious limit. has no obvious limit.

• However, Poisson probabilities taper off toward zero as However, Poisson probabilities taper off toward zero as XX increases. increases.

• The Poisson distribution is sometimes called the The Poisson distribution is sometimes called the model of rare eventsmodel of rare events..



Parameters = mean arrivals per unit of time or space

PDF

Range X = 0, 1, 2, ... (no obvious upper limit)

Mean

St. Dev.

Random data Use Excel’s Tools | Data Analysis | Random Number Generation

Comments Always right-skewed, but less so for larger .

( )!

xeP x

x



Here are some Here are some Poisson PDFs. Poisson PDFs.

x = 0.1 = 0.5 = 0.8 = 1.6 = 2.0

0 .9048 .6065 .4493 .2019 .1353

1 .0905 .3033 .3595 .3230 .2707

2 .0045 .0758 .1438 .2584 .2707

3 .0002 .0126 .0383 .1378 .1804

4 -- .0016 .0077 .0551 .0902

5 -- .0002 .0012 .0176 .0361

6 -- -- .0002 .0047 .0120

7 -- -- -- .0011 .0034

8 -- -- -- .0002 .0009

9 -- -- -- -- .0002

Sum 1.0000 1.0000 1.0000 1.0000 1.0000



Poisson distributions are always right-skewed but become less skewed and Poisson distributions are always right-skewed but become less skewed and more bell-shaped as more bell-shaped as increases. increases.

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= 0.8= 0.8

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= 6.4= 6.4



Example: Credit Union CustomersExample: Credit Union Customers

• Find the PDF, mean and standard deviation:Find the PDF, mean and standard deviation:

PDF = PDF = 1.7(1.7)

( )! !

x xe eP x

x x

Mean = Mean = = 1.7 customers per minute. = 1.7 customers per minute.

Standard deviation = Standard deviation = = 1.7= 1.7 = 1.304 cust/min= 1.304 cust/min

On Thursday morning between 9 A.M. and 10 A.M. customers arrive and enter the queue at the On Thursday morning between 9 A.M. and 10 A.M. customers arrive and enter the queue at the Oxnard University Credit Union at a mean rate of 1.7 customers per minute.Oxnard University Credit Union at a mean rate of 1.7 customers per minute.


Example: Credit Union CustomersExample: Credit Union Customers• Here is the Poisson probability distribution for Here is the Poisson probability distribution for

= 1.7 customers per minute on average. = 1.7 customers per minute on average.x

PDFP(X = x)

CDFP(X x)

0 .1827 .1827

1 .3106 .4932

2 .2640 .7572

3 .1496 .9068

4 .0636 .9704

5 .0216 .9920

6 .0061 .9981

7 .0015 .9996

8 .0003 .9999

9 .0001 1.0000

• Note that Note that xx represents the number of represents the number of customers.customers.

• For example, For example, PP((XX=4) is the probability that =4) is the probability that there are exactly 4 customers in the bank.there are exactly 4 customers in the bank.


These probabilities can be These probabilities can be calculated using a calculator or calculated using a calculator or Excel: Excel:

Formula: Excel function:

=POISSON(0,1.7,0)

=POISSON(1,1.7,0)

=POISSON(2,1.7,0)

=POISSON(3,1.7,0)

=POISSON(4,1.7,0)

3 1.71.7(3) .1496

3!

eP

4 1.71.7(4) .0636

4!

eP

2 1.71.7(2) .2640

2!

eP

1 1.71.7(1) .3106

1!

eP

0 1.71.7(0) .1827

0!

eP


Using the Poisson FormulaUsing the Poisson Formula

Formula: Excel function:

=POISSON(5,1.7,0)

=POISSON(6,1.7,0)

=POISSON(7,1.7,0)

=POISSON(8,1.7,0)

=POISSON(9,1.7,0)

9 1.71.7(9) .0001

9!

eP

8 1.71.7(8) .0003

8!

eP

7 1.71.7(7) .0015

7!

eP

6 1.71.7(6) .0061

6!

eP

5 1.71.7(5) .0216

5!

eP

Beyond Beyond XX = 9, the probabilities = 9, the probabilities are below .0001are below .0001


Using the Poisson FormulaUsing the Poisson Formula

• Here are the graphs of the distributions:Here are the graphs of the distributions:

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Pro

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Poisson PDF for Poisson PDF for = 1.7 = 1.7 Poisson CDF for Poisson CDF for = 1.7 = 1.7

• The most likely event is 1 arrival (The most likely event is 1 arrival (PP(1)=.3106 or 31.1% chance).(1)=.3106 or 31.1% chance).

• This will help the credit union schedule tellers.This will help the credit union schedule tellers.


• Cumulative probabilities can be evaluated by summing individual X probabilities.

• What is the probability that two or fewer customers will arrive in a given minute?

= .1827 + .3106 + .2640 = .7573

P(X < 2) = P(0) + P(1) + P(2)



• What is the probability of at least three customers (the complimentary What is the probability of at least three customers (the complimentary event)?event)?

= 1 - .7573 =.2427 = 1 - .7573 =.2427 PP((XX >> 3) = 1 - 3) = 1 - PP((X X << 2) 2)

• You can also use Excel’s function =POISSON(2,1.7,1) to obtain this You can also use Excel’s function =POISSON(2,1.7,1) to obtain this probability.probability.



Appendix B facilitates Poisson Appendix B facilitates Poisson calculations but doesn’t go beyond calculations but doesn’t go beyond = 20. = 20.

X 1.6 1.7 1.8 1.9 2.0 2.1

0 0.2019 0.1827 0.1653 0.1496 0.1353 0.12251 0.3230 0.3106 0.2975 0.2842 0.2707 0.25722 0.2584 0.2640 0.2678 0.2700 0.2707 0.27003 0.1378 0.1496 0.1607 0.1710 0.1804 0.18904 0.0551 0.0636 0.0723 0.0812 0.0902 0.09925 0.0176 0.0216 0.0260 0.0309 0.0361 0.04176 0.0047 0.0061 0.0078 0.0098 0.0120 0.01467 0.0011 0.0015 0.0020 0.0027 0.0034 0.00448 0.0002 0.0003 0.0005 0.0006 0.0009 0.00119 -- 0.0001 0.0001 0.0001 0.0002 0.000310 -- -- -- -- -- 0.000111 -- -- -- -- -- --


Poisson Probabilities: Tables (Appendix B)Poisson Probabilities: Tables (Appendix B)

Excel’s menus for calculating Poisson Excel’s menus for calculating Poisson probabilitiesprobabilities

The resulting probabilities are more accurate than those from Appendix B.The resulting probabilities are more accurate than those from Appendix B.


Poisson Probabilities: ExcelPoisson Probabilities: Excel

““Spin” Spin” and and overlay a normal overlay a normal curve. curve.

Module 4 (Module 4 ( = 1.7) = 1.7)

Copy and paste Copy and paste the graph as a the graph as a bitmap; copy and bitmap; copy and paste the paste the probabilities into probabilities into Excel. Excel.


Poisson Probabilities: Visual StatisticsPoisson Probabilities: Visual Statistics

• Can you recognize a Poisson situation? Can you recognize a Poisson situation? • Look for arrivals of “rare” independent events with Look for arrivals of “rare” independent events with no obvious upper limitno obvious upper limit..

• In the last week, how many credit card In the last week, how many credit card applications did you receive by mail?applications did you receive by mail?

• In the last week, how many checks did you write?In the last week, how many checks did you write?

• In the last week, how many e-mail viruses did In the last week, how many e-mail viruses did your firewall detect?your firewall detect?


Recognizing Poisson ApplicationsRecognizing Poisson Applications

• The Poisson distribution may be used to approximate a binomial by The Poisson distribution may be used to approximate a binomial by setting setting = n = n..

• This approximation is helpful when This approximation is helpful when nn is large and Excel is not available. is large and Excel is not available.

• For example, suppose For example, suppose nn = 1,000 women are screened for a rare type of = 1,000 women are screened for a rare type of cancer.cancer.

• This cancer has a nationwide incidence of 6 cases per 10,000. What is This cancer has a nationwide incidence of 6 cases per 10,000. What is ?? = 6/10,000 = .0006= 6/10,000 = .0006

• This is a binomial distribution withThis is a binomial distribution with n n = 1,000 and = 1,000 and =.0006.=.0006.


Poisson Approximation to BinomialPoisson Approximation to Binomial

• Set Set = n = n

• Since the binomial formula involves factorials (which are cumbersome as Since the binomial formula involves factorials (which are cumbersome as nn increases), use the Poisson distribution as an approximation:increases), use the Poisson distribution as an approximation:

• Now use Appendix B or the Poisson PDF to calculate the probability of Now use Appendix B or the Poisson PDF to calculate the probability of xx successes. For example:successes. For example:

= (1000)(.0006)= (1000)(.0006) = .6= .6

= .5488 + .3293 + .0988 = .9769= .5488 + .3293 + .0988 = .9769

PP((XX << 2) = 2) = PP(0) + (0) + PP(1) + (1) + PP(2)(2)


Poisson Approximation to BinomialPoisson Approximation to Binomial

Here is a comparison of Binomial probabilities and the respective Poisson Here is a comparison of Binomial probabilities and the respective Poisson approximations.approximations.

Poisson approximation: Actual Binomial probability:

P(0) = .60 e-0.6 / 0! = .5488 = .5487

P(1) = .61 e-0.6 / 1! = .3293 = .3294

P(2) = .62 e-0.6 / 2! = .0988 = .0988

2 1000 21000!(2) .0006 (1 .0006)

2!(1000 2)!P

1 1000 11000!(1) .0006 (1 .0006)

1!(1000 1)!P

0 1000 01000!(0) .0006 (1 .0006)

0!(1000 0)!P

Rule of thumb: the approximation is adequate if Rule of thumb: the approximation is adequate if nn >> 20 and 20 and << .05. .05.


• The The hypergeometric distributionhypergeometric distribution is similar to the binomial distribution. is similar to the binomial distribution.

• However, unlike the binomial, sampling is However, unlike the binomial, sampling is without replacementwithout replacement from a from a finite population of finite population of NN items. items.

• The hypergeometric distribution may be skewed right or left and is The hypergeometric distribution may be skewed right or left and is symmetric only if the proportion of successes in the population is 50%.symmetric only if the proportion of successes in the population is 50%.

Hypergeometric DistributionHypergeometric Distribution

Characteristics of the Hypergeometric Dist.Characteristics of the Hypergeometric Dist.

ParametersParameters NN = number of items in the population = number of items in the populationnn = sample size = sample sizess = number of “successes” in population = number of “successes” in population

PDFPDF

RangeRange XX = max(0, = max(0, nn--NN++ss) ) xx min( min(ss, , nn))

MeanMean nn where where = = ss//NN

St. Dev.St. Dev.

CommentsComments Similar to binomial, but sampling is without replacement from Similar to binomial, but sampling is without replacement from a finite population. Can be approximated by binomial with a finite population. Can be approximated by binomial with = = ss//NN if if nn//NN < 0.05 (i.e., less than 5% sample). < 0.05 (i.e., less than 5% sample).

(1 )1

N nn

N

( )

s N s

x n xP x

N

n


wherewhere

==

the number of ways to choose the number of ways to choose xx successes from successes from ss successes in successes in the populationthe population

==

the number of ways to choose the number of ways to choose nn––xx failures from failures from NN--ss failures in the population failures in the population

==

the number of ways to choose the number of ways to choose nn items from items from NN items in the items in the populationpopulation

( )

s N s

x n xP x

N

n

N

n

N s

n x

s

x


Characteristics of the Hypergeometric Dist.Characteristics of the Hypergeometric Dist.The hypergeometric PDF uses the formula for combinations:

• In a shipment of 10 iPods, 2 were damaged and 8 are good.In a shipment of 10 iPods, 2 were damaged and 8 are good.

• The receiving department at BestBuy tests a sample of 3 iPods at The receiving department at BestBuy tests a sample of 3 iPods at random to see if they are defective.random to see if they are defective.

• Let the random variable Let the random variable XX be the number of damaged iPods in the be the number of damaged iPods in the sample.sample.


Example: Damaged iPodsExample: Damaged iPods

Now describe the problem:Now describe the problem:

N N = 10 (number of iPods in the shipment)= 10 (number of iPods in the shipment)n n = 3 (sample size drawn from the shipment)= 3 (sample size drawn from the shipment)ss = 2 (number of damaged iPods in the = 2 (number of damaged iPods in the shipment (“successes” in population)) shipment (“successes” in population))NN––s s = 8 (number of non-damaged iPods in= 8 (number of non-damaged iPods in the shipment) the shipment)xx = number of damaged iPods in the sample = number of damaged iPods in the sample (“successes” in sample) (“successes” in sample)nn––x x = number of non-damaged iPods in the sample= number of non-damaged iPods in the sample



• This is not a binomial problem because This is not a binomial problem because is not constant. is not constant.

• What is the probability of getting a damaged iPod on the first draw from What is the probability of getting a damaged iPod on the first draw from the sample?the sample?

11 = 2/10 = 2/10

• Now, what is the probability of getting a damaged iPod on the second Now, what is the probability of getting a damaged iPod on the second draw?draw?

22 = 1/9 (if the first iPod was damaged) or = 1/9 (if the first iPod was damaged) or

= 2/9 (if the first iPod was undamaged) = 2/9 (if the first iPod was undamaged)



• What about on the third draw?What about on the third draw?33 = 0/8 or = 0/8 or

= 1/8 or = 1/8 or = 2/8 = 2/8 depending on what happened in the first two draws.depending on what happened in the first two draws.



Since there are only 2 damaged iPods in the population, the only possible values of Since there are only 2 damaged iPods in the population, the only possible values of xx are are 0, 1, and 2. Here are the probabilities:0, 1, and 2. Here are the probabilities:

2 8 2! 8!0 3 56 70! 2! 3! 5!

(0)10 10! 120 15

3! 7!3

P

= .4667= .4667

2 8 2! 8!1 2 56 71! 1! 2! 6!

(1)10 10! 120 15

3! 7!3

P

= .4667= .4667

2 8 2! 8!2 1 8 12! 0! 1! 7!

(2)10 10! 120 15

3! 7!3

P

= .0667= .0667

=HYPGEOMDIST(0,3,2,10)=HYPGEOMDIST(0,3,2,10)



PDF FormulaPDF Formula Excel FunctionExcel Function

Hypergeometric DistributionHypergeometric DistributionUsing the Hypergeometric FormulaUsing the Hypergeometric Formula

Since the hypergeometric formula and tables are tedious and impractical, Since the hypergeometric formula and tables are tedious and impractical, use Excel’s hypergeometric function to find probabilities.use Excel’s hypergeometric function to find probabilities.


Using the Hypergeometric FormulaUsing the Hypergeometric Formula

Module 4, the probabilities are given below the graph. Module 4, the probabilities are given below the graph.

Copy and paste Copy and paste graph as a bitmap; graph as a bitmap; copy and paste copy and paste probabilities into probabilities into Excel. Excel.

““Spin” Spin” NN and and nn; ; overlay a normal overlay a normal or binomial curve. or binomial curve.


Hypergeometric Probabilities: Visual StatisticsHypergeometric Probabilities: Visual Statistics

LearningStatsLearningStats allows you to spin the allows you to spin the values of values of NN, , nn, and , and ss to get the desired to get the desired probability. probability.


Hypergeometric Probabilities: LearningStatsHypergeometric Probabilities: LearningStats

• Look for a finite populationLook for a finite population ((NN) containing a known number of successes ) containing a known number of successes ((ss) and ) and sampling without replacementsampling without replacement ((nn items in the sample). items in the sample).

• Out of 40 cars are inspected for California Out of 40 cars are inspected for California emissions compliance, 32 are compliant but 8 emissions compliance, 32 are compliant but 8 are not. A sample of 7 cars is chosen at are not. A sample of 7 cars is chosen at random. What is the probability that all are random. What is the probability that all are compliant? At least 5?compliant? At least 5?



• Out of 500 background checks for firearms Out of 500 background checks for firearms purchasers, 50 applicants are convicted felons. purchasers, 50 applicants are convicted felons. Through a computer error, 10 applicants are Through a computer error, 10 applicants are approved without a background check. What is approved without a background check. What is the probability that none is a felon? At least 2?the probability that none is a felon? At least 2?

• Out of 40 blood specimens checked for HIV, 8 Out of 40 blood specimens checked for HIV, 8 actually contain HIV. A worker is accidentally actually contain HIV. A worker is accidentally exposed to 5 specimens. What is the probability exposed to 5 specimens. What is the probability that none contained HIV?that none contained HIV?



• Both the binomial and hypergeometric involve samples of size Both the binomial and hypergeometric involve samples of size nn and treat and treat XX as the number of successes. as the number of successes.

• The binomial samples The binomial samples withwith replacement while the hypergeometric replacement while the hypergeometric samples samples withoutwithout replacement. replacement.

Rule of ThumbRule of ThumbIf n/N < 0.05 it is safe to use the binomial

approximation to the hypergeometric, using sample size n and success probability = s/N.


Binomial Approximation to the HypergeometricBinomial Approximation to the Hypergeometric

• For example, suppose we want For example, suppose we want PP((XX=6) for =6) for a hypergeometric with a hypergeometric with NN = 400, = 400, nn = 10, = 10, ss = 200. = 200.

nn//NN = 10/400 = 0.025 < .05 so the binomial approximation is = 10/400 = 0.025 < .05 so the binomial approximation is acceptable.acceptable.

Set Set = = ss//NN = 200/400 = .50 and use Appendix A to obtain the = 200/400 = .50 and use Appendix A to obtain the probability.probability.

PP((XX=6) = .2051=6) = .2051


Binomial Approximation to the HypergeometricBinomial Approximation to the Hypergeometric

• The The geometric distributiongeometric distribution describes the number of Bernoulli trials until describes the number of Bernoulli trials until the first success.the first success.

• XX is the number of trials until the first success. is the number of trials until the first success.

is the constant probability of a success on each trial.is the constant probability of a success on each trial.

• XX ranges from {1, 2, . . .} since we must have at least one trial to obtain ranges from {1, 2, . . .} since we must have at least one trial to obtain the first success. However, the number of trials is not fixed.the first success. However, the number of trials is not fixed.

Geometric DistributionGeometric Distribution

Characteristics of the Geometric DistributionCharacteristics of the Geometric Distribution

The geometric distribution is always skewed to the right.The geometric distribution is always skewed to the right.Parameters = probability of success

PDF P(x) = (1)x1

Range X = 1, 2, ...

Mean 1/

St. Dev.

Comments Describes the number of trials before the first success. Highly skewed.

2

1

The mean and standard deviation are nearly the same when The mean and standard deviation are nearly the same when is small. is small.


Characteristics of the Geometric DistributionCharacteristics of the Geometric Distribution

• At Faber University, 15% of the alumni make a donation or pledge At Faber University, 15% of the alumni make a donation or pledge during the annual telefund. during the annual telefund.

• What is What is ??

• What is the probability that the first donation will not come until the 7What is the probability that the first donation will not come until the 7 thth call? call?

= .15= .15

• The PDF is: The PDF is: PP((xx) = ) = (1–(1– ))xx11

PP((77) = .15(1–.15)) = .15(1–.15)7711 = .15(.85)= .15(.85)66 = .0566 = .0566


Example: Telefund CallingExample: Telefund Calling

• What are the mean and standard deviation of this distribution? What are the mean and standard deviation of this distribution?

= 1/= 1/• So, we would expect to call between 6 and 7 alumni until the first So, we would expect to call between 6 and 7 alumni until the first

donation. donation.

= 1/(.15) = 6.67 calls= 1/(.15) = 6.67 calls

==2

1

= 6.15= 6.15= = 1-.151-.15

(.15)(.15)22

• The large standard deviation indicates that we should not regard the The large standard deviation indicates that we should not regard the mean as a good prediction of how many trials are needed.mean as a good prediction of how many trials are needed.


Example: Telefund CallingExample: Telefund Calling

LearningStatsLearningStats gives both the graph and gives both the graph and numeric probabilities of the distribution.numeric probabilities of the distribution.


Using LearningStatsUsing LearningStats

• A A linear transformationlinear transformation of a random variable of a random variable XX is performed by adding a is performed by adding a constant or multiplying by a constant.constant or multiplying by a constant.

Rule 1: Rule 1: aX+baX+b = a = aXX + b + b (mean of a transformed (mean of a transformed

variable) variable)Rule 2: Rule 2: aX+baX+b = a = aXX (standard deviation of a (standard deviation of a

transformed variable) transformed variable)

Transformations of Random Transformations of Random VariablesVariables

Linear TransformationsLinear Transformations

Applied Statistics in Business and Economics

End of Chapter 6End of Chapter 6

doane chapter 06

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