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Chapter 5Chapter 5Discrete Probability DistributionsDiscrete Probability Distributions

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0 1 2 3 40 1 2 3 4

■■ Random VariablesRandom Variables

■■ Discrete Probability DistributionsDiscrete Probability Distributions

■■ Expected Value and VarianceExpected Value and Variance

■■ Binomial DistributionBinomial Distribution

■■ Poisson DistributionPoisson Distribution

■■ Hypergeometric DistributionHypergeometric Distribution


A A random variablerandom variable is a numerical description of theis a numerical description of theoutcome of an experiment.outcome of an experiment.

Random VariablesRandom Variables

A A discrete random variablediscrete random variable may assume either amay assume either afinite number of values or an infinite sequence offinite number of values or an infinite sequence ofvalues.values.

A A continuous random variablecontinuous random variable may assume anymay assume anynumerical value in an interval or collection ofnumerical value in an interval or collection ofintervals.intervals.

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Let Let xx = number of TVs sold at the store in one day,= number of TVs sold at the store in one day,

where where xx can take on 5 values (0, 1, 2, 3, 4)can take on 5 values (0, 1, 2, 3, 4)

Example: JSL AppliancesExample: JSL Appliances

■■ Discrete random variable with a Discrete random variable with a finitefinite number number of valuesof values


Let Let xx = number of customers arriving in one day,= number of customers arriving in one day,

where where xx can take on the values 0, 1, 2, . . .can take on the values 0, 1, 2, . . .

Example: JSL AppliancesExample: JSL Appliances

■■ Discrete random variable with an Discrete random variable with an infiniteinfinite sequence sequence of valuesof values

We can count the customers arriving, but there is noWe can count the customers arriving, but there is nofinite upper limit on the number that might arrive.finite upper limit on the number that might arrive.

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Random VariablesRandom Variables

QuestionQuestion Random Variable Random Variable xx TypeType

FamilyFamily

sizesize

xx = Number of dependents= Number of dependents

reported on tax returnreported on tax returnDiscreteDiscrete

Distance fromDistance from

home to storehome to store

xx = Distance in miles from= Distance in miles from

home to the store sitehome to the store site

ContinuousContinuous

Own dogOwn dogor cator cat

xx = 1 if own no pet;= 1 if own no pet;

= 2 if own dog(s) only; = 2 if own dog(s) only;

= 3 if own cat(s) only; = 3 if own cat(s) only;

= 4 if own dog(s) and cat(s)= 4 if own dog(s) and cat(s)

DiscreteDiscrete


The The probability distributionprobability distribution for a random variablefor a random variabledescribes how probabilities are distributed overdescribes how probabilities are distributed overthe values of the random variable.the values of the random variable.

We can describe a discrete probability distributionWe can describe a discrete probability distributionwith a table, graph, or equation.with a table, graph, or equation.

Discrete Probability DistributionsDiscrete Probability Distributions

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The probability distribution is defined by aThe probability distribution is defined by aprobability functionprobability function, denoted by , denoted by ff((xx), which provides), which providesthe probability for each value of the random variable.the probability for each value of the random variable.

The required conditions for a discrete probabilityThe required conditions for a discrete probabilityfunction are:function are:


ff((xx) ) >> 00

ΣΣff((xx) = 1) = 1


■■ a tabular representation of the probabilitya tabular representation of the probabilitydistribution for TV sales was developed.distribution for TV sales was developed.

■■ Using past data on TV sales, …Using past data on TV sales, …

NumberNumber

Units SoldUnits Sold of Daysof Days

00 8080

11 5050

22 4040

33 1010

44 2020

200200

xx ff((xx))

00 .40.40

11 .25.25

22 .20.20

33 .05.05

44 .10.10

1.001.00

80/20080/200


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.50.50

0 1 2 3 40 1 2 3 4Values of Random Variable Values of Random Variable xx (TV sales)(TV sales)Values of Random Variable Values of Random Variable xx (TV sales)(TV sales)

Pro

bab

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rob

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Pro

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yP

rob

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■■ Graphical Representation of Probability DistributionGraphical Representation of Probability Distribution


Discrete Uniform Probability DistributionDiscrete Uniform Probability Distribution

The The discrete uniform probability distributiondiscrete uniform probability distribution is theis thesimplest example of a discrete probabilitysimplest example of a discrete probabilitydistribution given by a formula.distribution given by a formula.

The The discrete uniform probability functiondiscrete uniform probability function isis

ff((xx) = 1/) = 1/nn

where:where:nn = the number of values the random= the number of values the random

variable may assumevariable may assume

the values of thethe values of therandom variablerandom variableare equally likelyare equally likely

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Expected Value and VarianceExpected Value and Variance

The The expected valueexpected value, or mean, of a random variable, or mean, of a random variableis a measure of its central location.is a measure of its central location.

The The variancevariance summarizes the variability in thesummarizes the variability in thevalues of a random variable.values of a random variable.

The The standard deviationstandard deviation, , σσ, is defined as the positive, is defined as the positivesquare root of the variance.square root of the variance.

Var(Var(xx) = ) = σσ 22 = = ΣΣ((xx -- µµ))22ff((xx))Var(Var(xx) = ) = σσ 22 = = ΣΣ((xx -- µµ))22ff((xx))

EE((xx) = ) = µµ = = ΣΣxfxf((xx))EE((xx) = ) = µµ = = ΣΣxfxf((xx))


■■ Expected ValueExpected Value

expected number of expected number of TVs sold in a dayTVs sold in a day

xx ff((xx)) xfxf((xx))

00 .40.40 .00.00

11 .25.25 .25.25

22 .20.20 .40.40

33 .05.05 .15.15

44 .10.10 .40.40

EE((xx) = 1.20) = 1.20


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■■ Variance and Standard DeviationVariance and Standard Deviation

00

11

22

33

44

--1.21.2

--0.20.2

0.80.8

1.81.8

2.82.8

1.441.44

0.040.04

0.640.64

3.243.24

7.847.84

.40.40

.25.25

.20.20

.05.05

.10.10

.576.576

.010.010

.128.128

.162.162

.784.784

x x -- µµ ((x x -- µµ))22 ff((xx)) ((xx -- µµ))22ff((xx))

Variance of daily sales = Variance of daily sales = σ σ 22 = 1.660= 1.660

xx

TVsTVssquaredsquared

Standard deviation of daily sales = 1.2884 TVsStandard deviation of daily sales = 1.2884 TVs



Binomial DistributionBinomial Distribution

■■ Four Properties of a Binomial ExperimentFour Properties of a Binomial Experiment

3. The probability of a success, denoted by 3. The probability of a success, denoted by pp, does, doesnot change from trial to trial.not change from trial to trial.

4. The trials are independent.4. The trials are independent.

2. Two outcomes, 2. Two outcomes, successsuccess and and failurefailure, are possible, are possibleon each trial.on each trial.

1. The experiment consists of a sequence of 1. The experiment consists of a sequence of nnidentical trials.identical trials.

stationaritystationarityassumptionassumption

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Our interest is in the Our interest is in the number of successesnumber of successesoccurring in the occurring in the nn trials.trials.

We let We let xx denote the number of successesdenote the number of successesoccurring in the occurring in the nn trials.trials.


where:where:

ff((xx) = the probability of ) = the probability of xx successes in successes in nn trialstrials

nn = the number of trials= the number of trials

pp = the probability of success on any one trial= the probability of success on any one trial

( )!( ) (1 )

!( )!x n xn

f x p px n x

−= −−


■■ Binomial Probability FunctionBinomial Probability Function

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( )!( ) (1 )

!( )!x n xn

f x p px n x

−= −−


!

!( )!

n

x n x−( )(1 )x n xp p −−

■■ Binomial Probability FunctionBinomial Probability Function

Probability of a particularProbability of a particularsequence of trial outcomessequence of trial outcomeswith x successes in with x successes in nn trialstrials

Number of experimentalNumber of experimentaloutcomes providing exactlyoutcomes providing exactly

xx successes in successes in nn trialstrials



■■ Example: Evans ElectronicsExample: Evans Electronics

Evans is concerned about a low retention rate for Evans is concerned about a low retention rate for employees. In recent years, management has seen a employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the management estimates a probability of 0.1 that the person will not be with the company next year.person will not be with the company next year.

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■■ Using the Binomial Probability FunctionUsing the Binomial Probability Function

Choosing 3 hourly employees at random, what is Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company the probability that 1 of them will leave the company this year?this year?

f xn

x n xp px n x( )

!!( )!

( ) ( )=−

− −1

1 23!(1) (0.1) (0.9) 3(.1)(.81) .243

1!(3 1)!f = = =

−

LetLet: p: p = .10, = .10, nn = 3, = 3, xx = 1= 1


■■ Tree DiagramTree Diagram


1st Worker1st Worker 2nd Worker2nd Worker 3rd Worker3rd Worker xx Prob.Prob.

Leaves(.1)

Leaves(.1)

Stays(.9)

Stays(.9)

33

22

00

22

22

Leaves (.1)Leaves (.1)

Leaves (.1)Leaves (.1)

S (.9)S (.9)

Stays (.9)Stays (.9)

Stays (.9)Stays (.9)

S (.9)S (.9)

S (.9)S (.9)

S (.9)S (.9)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1) .0010.0010

.0090.0090

.0090.0090

.7290.7290

.0090.0090

11

11

.0810.0810

.0810.0810

.0810.0810

1111

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■■ Using Tables of Binomial ProbabilitiesUsing Tables of Binomial Probabilities

n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50

3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750

2 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750

3 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

p




(1 )np pσ = −

EE((xx) = ) = µµ = = npnp

Var(Var(xx) = ) = σσ 22 = = npnp(1 (1 −− pp))


■■ VarianceVariance

■■ Standard DeviationStandard Deviation

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3(.1)(.9) .52 employeesσ = =

EE((xx) = ) = µµ = 3(.1) = .3 employees out of 3= 3(.1) = .3 employees out of 3

Var(Var(xx) = ) = σσ 22 = 3(.1)(.9) = .27= 3(.1)(.9) = .27



■■ Standard DeviationStandard Deviation


A Poisson distributed random variable is oftenA Poisson distributed random variable is oftenuseful in estimating the number of occurrencesuseful in estimating the number of occurrencesover a over a specified interval of time or spacespecified interval of time or space

It is a discrete random variable that may assumeIt is a discrete random variable that may assumean an infinite sequence of valuesinfinite sequence of values (x = 0, 1, 2, . . . ).(x = 0, 1, 2, . . . ).

Poisson DistributionPoisson Distribution

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Examples of a Poisson distributed random variable:Examples of a Poisson distributed random variable:

the number of knotholes in 14 linear feet ofthe number of knotholes in 14 linear feet ofpine boardpine board

the number of vehicles arriving at athe number of vehicles arriving at atoll booth in one hourtoll booth in one hour




■■ Two Properties of a Poisson ExperimentTwo Properties of a Poisson Experiment

2.2. The occurrence or nonoccurrence in anyThe occurrence or nonoccurrence in anyinterval is independent of the occurrence orinterval is independent of the occurrence ornonoccurrence in any other interval.nonoccurrence in any other interval.

1.1. The probability of an occurrence is the sameThe probability of an occurrence is the samefor any two intervals of equal length.for any two intervals of equal length.

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■■ Poisson Probability FunctionPoisson Probability Function


f xe

x

x( )

!=

−µ µ

where:where:

f(x) f(x) = probability of = probability of xx occurrences in an intervaloccurrences in an interval

µµ = mean number of occurrences in an interval= mean number of occurrences in an interval

ee = 2.71828= 2.71828



MERCYMERCY

■■ Example: Mercy HospitalExample: Mercy Hospital

Patients arrive at the Patients arrive at the

emergency room of Mercyemergency room of Mercy

Hospital at the averageHospital at the average

rate of 6 per hour onrate of 6 per hour on

weekend evenings. weekend evenings.

What is theWhat is the

probability of 4 arrivals inprobability of 4 arrivals in

30 minutes on a weekend evening?30 minutes on a weekend evening?

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■■ Using the Poisson Probability FunctionUsing the Poisson Probability Function

4 33 (2.71828)(4) .1680

4!f

−

= =

MERCYMERCY

µµ = 6/hour = 3/half= 6/hour = 3/half--hour, hour, xx = 4= 4



■■ Using Poisson Probability TablesUsing Poisson Probability Tables

µµµµx 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0

0 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .0498

1 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .1494

2 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .2240

3 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .2240

4 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .1680

5 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .1008

6 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .0504

7 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .0216

8 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081

MERCYMERCY

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■■ Poisson Distribution of ArrivalsPoisson Distribution of Arrivals

Poisson DistributionPoisson DistributionMERCYMERCY

Poisson Probabilities

0.00

0.05

0.10

0.15

0.20

0.25

0 1 2 3 4 5 6 7 8 9 10Number of Arrivals in 30 Minutes

Pro

ba

bil

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actually, actually, the sequencethe sequence

continues:continues:11, 12, …11, 12, …



A property of the Poisson distribution is thatA property of the Poisson distribution is thatthe mean and variance are equal.the mean and variance are equal.

µµ = = σ σ 22

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Poisson DistributionPoisson DistributionMERCYMERCY

■■ Variance for Number of ArrivalsVariance for Number of Arrivals

During 30During 30--Minute PeriodsMinute Periods

µµ = = σσ 22 = 3 = 3


Hypergeometric DistributionHypergeometric Distribution

The The hypergeometric distributionhypergeometric distribution is closely relatedis closely relatedto the binomial distribution. to the binomial distribution.

However, for the hypergeometric distribution:However, for the hypergeometric distribution:

the trials are not independent, andthe trials are not independent, and

the probability of success changes from trialthe probability of success changes from trialto trial. to trial.

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■■ Hypergeometric Probability FunctionHypergeometric Probability Function


−−

=

n

N

xn

rN

x

r

xf )( for 0 for 0 << xx << rr

where: where: ff((xx) = probability of ) = probability of xx successes in successes in nn trialstrials

nn = number of trials= number of trials

NN = number of elements in the population= number of elements in the population

rr = number of elements in the population= number of elements in the population

labeled successlabeled success


■■ Hypergeometric Probability FunctionHypergeometric Probability Function


( )

r N r

x n xf x

N

n

− −

=

for 0 for 0 << xx << rr

number of waysnumber of waysnn –– x x failures can be selectedfailures can be selectedfrom a total of from a total of NN –– rr failuresfailures

in the populationin the population

number of waysnumber of waysxx successes can be selectedsuccesses can be selectedfrom a total of from a total of rr successessuccesses

in the populationin the populationnumber of waysnumber of ways

a sample of size a sample of size n n can be selectedcan be selectedfrom a population of size from a population of size NN

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■■ Example: NevereadyExample: Neveready

Bob Neveready has removed twoBob Neveready has removed two

dead batteries from a flashlight anddead batteries from a flashlight and

inadvertently mingled them withinadvertently mingled them with

the two good batteries he intendedthe two good batteries he intended

as replacements. as replacements. The four batteries look identical.The four batteries look identical.

Bob now randomly selects two of the four Bob now randomly selects two of the four batteries. What is the probability he selects the two batteries. What is the probability he selects the two good batteries?good batteries?



■■ Using the Hypergeometric FunctionUsing the Hypergeometric Function

2 2 2! 2!

2 0 2!0! 0!2! 1( ) .167

4 4! 6

2 2!2!

r N r

x n xf x

N

n

− − = = = = =

where:where:xx = 2 = number of = 2 = number of goodgood batteries selectedbatteries selectednn = 2 = number of batteries selected= 2 = number of batteries selectedNN = 4 = number of batteries in total= 4 = number of batteries in totalrr = 2 = number of = 2 = number of goodgood batteries in totalbatteries in total

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( )r

E x nN

µ = =

2( ) 11

r r N nVar x n

N N Nσ − = = − −

■■ MeanMean




22 1

4

rn

Nµ = = =

2 2 2 4 2 12 1 .333

4 4 4 1 3σ − = − = = −

■■ MeanMean


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Consider a hypergeometric distribution with Consider a hypergeometric distribution with nn trialstrialsand let and let pp = (= (rr//nn) denote the probability of a success) denote the probability of a successon the first trial.on the first trial.

If the population size is large, the term (If the population size is large, the term (NN –– nn)/()/(NN –– 1)1)

approaches 1.approaches 1.

The expected value and variance can be writtenhe expected value and variance can be written

EE((xx) = ) = npnp and and VarVar((xx) = ) = npnp(1 (1 –– pp).).

Note that these are the expressions for the expectedhe expected

value and variance of a binomial distribution.value and variance of a binomial distribution.

continuedcontinued



When the population size is large, a hypergeometricWhen the population size is large, a hypergeometricdistribution can be approximated by a binomialdistribution can be approximated by a binomialdistribution with distribution with nn trials and a probability of successtrials and a probability of successpp = (= (rr//NN). ).

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End of Chapter 5End of Chapter 5

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