6-1. Discrete Distributions Probability Models Probability Models Discrete Distributions Discrete...

Discrete DistributionsDiscrete DistributionsDiscrete DistributionsDiscrete Distributions

Probability Models

Discrete Distributions

Uniform Distribution

Bernoulli Distribution

Binomial Distribution

Poisson Distribution

Chapter6666

McGraw-Hill/Irwin © 2008 The McGraw-Hill Companies, Inc. All rights reserved.

6-3

Probability ModelsProbability ModelsProbability ModelsProbability Models

• 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

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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.

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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


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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


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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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0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0 1 2 3

Number of Heads (X)

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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.



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• 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

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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

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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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0.00

0.05

0.10

0.15

0.20

0.25

0.30

0 1 2 3 4 5

Number of Service Calls

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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.



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• 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

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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

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• 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

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• 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

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• 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

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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

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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



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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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0.00

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0 1 2 3 4 5 6 7

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.



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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

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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


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• 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

0.14

0.16

0.18

1 2 3 4 5 6

Number of Dots Showing on the Die

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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

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• 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

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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

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• 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

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• 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

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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



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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.



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• 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

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• 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

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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


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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

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• 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


Application: Uninsured PatientsApplication: Uninsured Patients

6-1. Discrete Distributions Probability Models Probability Models Discrete Distributions Discrete...

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