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Discrete Probability DistributionsDiscrete Probability Distributions
Prof. Rushen ChahalProf. Rushen Chahal
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Discrete Probability DistributionsDiscrete Probability Distributions
Random VariablesRandom Variables
Discrete Probability DistributionsDiscrete Probability Distributions
Expected Value and VarianceExpected Value and Variance The Binomial Probability DistributionThe Binomial Probability Distribution
The Poisson Probability DistributionThe Poisson Probability Distribution
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Random VariablesRandom Variables
AA random variablerandom variable is a numericalis a numericaldescription of the outcome of andescription of the outcome of anexperiment.experiment.
AA discrete random variablediscrete random variable maymayassume either a finite number of valuesassume either a finite number of values
or an infinite sequence of values.or an infinite sequence of values.
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Finite and Infinite Values for Discrete RandomFinite and Infinite Values for Discrete RandomVariablesVariables
Discrete random variable with a finite number ofDiscrete random variable with a finite number ofvalues:values:
LetLet xx = number of TV sets sold in one day where= number of TV sets sold in one day wherexx can take on 5 values (0, 1, 2, 3, 4)can take on 5 values (0, 1, 2, 3, 4)
Discrete random variable with an infinite sequence ofDiscrete random variable with an infinite sequence ofvalues:values:
LetLet xx = number of customers arriving in one day= number of customers arriving in one day
wherewhere xx can take on the values 0, 1, 2, . . .can take on the values 0, 1, 2, . . .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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Discrete Random VariablesDiscrete Random Variables
A CPA examination has 4 parts. We can define aA CPA examination has 4 parts. We can define adiscrete random variable asdiscrete random variable as
xx = the number of parts of CPA exam passed= the number of parts of CPA exam passed
This discrete random variable may assume finiteThis discrete random variable may assume finite
number of values of 0, 1, 2, 3, 4.number of values of 0, 1, 2, 3, 4.
We may make a call to sell a product. We can defineWe may make a call to sell a product. We can definea discrete random variablea discrete random variable xx..
xx = 1 if we sell,= 1 if we sell, xx = 0 if we dont.= 0 if we dont.
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Continuous Random VariablesContinuous Random Variables
AA 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.
xx = the time between arrival of two customers into a= the time between arrival of two customers into a
supper market.supper market. xx = the exact weight of an individual= the exact weight of an individual
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PracticePractice
187187--188188 Problem 1Problem 1
Problem 2Problem 2
Problem 3Problem 3
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Probability DistributionsProbability Distributions
TheThe probability distributionprobability distribution for afor arandom variable describes howrandom variable describes howprobabilities are distributed over theprobabilities are distributed over the
values of the random vari
able.values of the random vari
able.
The probability distribution is definedThe probability distribution is defined
by aby a probability functionprobability function, denoted by, denoted byff((xx), which provides the probability for), which provides the probability foreach value of the random variable.each value of the random variable.
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Discrete Probability DistributionsDiscrete Probability Distributions
The required conditions for a discreteThe required conditions for a discreteprobability function are:probability function are:
ff((xx )) >> 0077ff((xx )= 1)= 1
We can describe a discrete probabilityWe can describe a discrete probabilitydistribution with a table, graph, ordistribution with a table, graph, orequation.equation.
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Using Data to Compute ProbabilitiesUsing Data to Compute Probabilities
Using past data on TV sales (below left), a tabularUsing past data on TV sales (below left), a tabularrepresentation of the probability distribution for TVrepresentation of the probability distribution for TVsales (below right) was developed.sales (below right) was developed.
NumberNumber
Units SoldUnits Sold of Daysof Days xx ff((xx ))
00 8080 00 .40.40
11 5050 11 .25.25
22
404022
.2
0.2
033 1010 33 .05.05
44 2020 44 .10.10
200200 1.001.00
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Graphical Representation of A Discrete ProbabilityGraphical Representation of A Discrete ProbabilityDistributionDistribution
A graphical representation of the probabilityA graphical representation of the probabilitydistribution for TV sales in one daydistribution for TV sales in one day
.10.10
.20.20
.30.30
.40.40
.50.50
00 11 22 33 4400 11 22 33 44Values of RandomVariablex (TV sales)Values of RandomVariablex (TV sales)
Probability
Probability
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A Discrete Uniform Probability DistributionA Discrete Uniform Probability Distribution
Discrete Uniform Probability DistributionDiscrete Uniform Probability Distribution
f(x) = 1/n (n is the number ofvalues that x takes)f(x) = 1/n (n is the number ofvalues that x takes)
Rolling a dieRolling a die
x f(x)1 1/6
2 1/6
3 1/6
4 1/65 1/6
6 1/6 1 2 3 4 5 6
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Expected Value and VarianceExpected Value and Variance
TheThe expected valueexpected value, or mean, of a random variable is, or mean, of a random variable isa measure of its central location.a measure of its central location.
Expected value of a discrete random variable:Expected value of a discrete random variable:
EE((xx )=)=QQ ==77xfxf((xx ))
TheThe variancevariance summarizes the variability in the valuessummarizes the variability in the valuesof a random variable.of a random variable.
Variance of a discrete random variable:Variance of a discrete random variable:
Var(Var(xx )=)=WW 22 ==77((xx -- QQ ))22ff((xx ))
TheThe standard deviationstandard deviation,, WW, is defined as the positive, is defined as the positivesquare root of the variance.square root of the variance.
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ExampleExample
Expected Value of a Discrete Random VariableExpected Value of a Discrete Random Variable
xx ff((xx ))
00 .40.40
11 .25.2522 .20.20
33 .05.05
44 .10.10
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ExampleExample
Expected Value of a Discrete Random VariableExpected Value of a Discrete Random Variable
xx ff((xx )) xfxf((xx ))
------------------------------------------------------
00 .40.40 .00.0011 .25.25 .25.25
22 .20.20 .40.40
33 .05.05 .15.15
44 .10.10 .40.40
1.20 =1.20 =EE((xx ))
The expected number of TV sets sold in a day is 1.2The expected number of TV sets sold in a day is 1.2
Now Calculate the standard deviation
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Variance and Standard DeviationVariance and Standard Deviationof a Discrete Random Variableof a Discrete Random Variable
xx xx -- QQ ((xx -- QQ ))22 ff((xx ) () (xx -- QQ ))22ff((xx ))_____ _________ ___________ _______ ____________________ _________ ___________ _______ _______________
Example: JSL AppliancesExample: JSL Appliances
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Variance and Standard
DeviationVar
iance and Standard
Deviation
of a Discrete Random Variableof a Discrete Random Variable
xx xx -- QQ ((xx -- QQ ))22 ff((xx ) () (xx -- QQ ))22ff((xx ))_____ _________ ___________ _______ ____________________ _________ ___________ _______ _______________
00--1.2
1.2
1.441.44 .40.40 .576
.576
11 --0.20.2 0.040.04 .25.25 .010.010
22 0.80.8 0.640.64 .20.20 .128.128
33 1.81.8 3.243.24 .05.05 .162.162
44 2.82.8 7.847.84 .10.10 .784.7841.660 =1.660 =WW
The variance of daily sales is 1.66 TV setsThe variance of daily sales is 1.66 TV sets squared.squared.
The standard deviation of sales is 1.29 TV sets.The standard deviation of sales is 1.29 TV sets.
Example: JSL AppliancesExample: JSL Appliances
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The Binomial Probability DistributionThe Binomial Probability Distribution
Properties of a Binomial ExperimentProperties of a Binomial Experiment
The experiment consists of a sequence ofThe experiment consists of a sequence of nnidentical trials.identical trials.
Two outcomes,Two outcomes, successsuccess andand failurefailure, are possible on, are possible oneach trial.each trial.
The probabi
li
ty of a success, denoted byThe probabi
li
ty of a success, denoted bypp, does, doesnot change from trial to trial.not change from trial to trial.
The trials areThe trials are independentindependent..
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ExamplesExamples
Toss a coi
nToss a coi
n Define tail as success and head as failureDefine tail as success and head as failure
f(S) = .5 f(F) = .5f(S) = .5 f(F) = .5
Roll a dieRoll a die
Define 1 as success, any other number as failureDefine 1 as success, any other number as failure f(S) = 1/6f(S) = 1/6 f(F)= 5/6f(F)= 5/6
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Example: Evans ElectronicsExample: Evans Electronics
BinomialBinomial ProbabilityProbability DistributionDistributionEvansEvans isis concernedconcerned aboutabout aa lowlow retentionretention raterate forforemployeesemployees.. OnOn thethe basisbasis ofof pastpast experience,experience, managementmanagementhashas seenseen aa turnoverturnover ofof 1010%% ofof thethe hourlyhourly employeesemployees
annuallyannually.. Thus,Thus, forfor anyany hourlyhourly employeesemployees chosenchosen atatrandom,random, managementmanagement estimatesestimates aa probabilityprobability ofof 00..11 thatthatthethe personperson willwill notnot bebe withwith thethe companycompany nextnext yearyear..
ChoosingChoosing 33 hourlyhourly employeesemployees aa random,random, whatwhat isis thetheprobabilityprobability thatthat 11 ofof themthem willwill leaveleave thethe companycompany thisthis
year?year?LetLet:: pp == ..1010,, nn == 33,, xx == 11
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Example: Evans ElectronicsExample: Evans Electronics
Using the Tables of Binomial ProbabilitiesUsing the Tables of Binomial Probabilities
p
n x .10 .15 .20 .25 .30 .35 .40 .45 .50
3 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250
1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750
2 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750
3 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250
p
n x .10 .15 .20 .25 .30 .35 .40 .45 .50
3 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250
1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750
2 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750
3 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250
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Using a Tree DiagramUsing a Tree Diagram
FirstWorker
FirstWorker
SecondWorkerSecondWorker
ThirdWorkerThird
Worker
Leaves (.1)Leaves (.1)
Stays (.9)Stays (.9)
Valueof x
Valueof x
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)
Probab.Probab.
.0010.0010
.0090.0090
.0090.0090
.7290.7290
.0090.0090
11
11
11
.0810.0810
.0810.0810
.0810.0810
Example: Evans ElectronicsExample: Evans Electronics
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The Binomial Probability DistributionThe Binomial Probability Distribution
Expected ValueExpected ValueEE((xx )=)=QQ ==npnp
VarianceVariance
Var(Var(xx )=)=WW 22 ==npnp (1(1 --pp ))
Standard DeviationStandard Deviation
Example: Evans ElectronicsExample: Evans Electronics
EE((xx )=)=QQ =3(.1)= .3 employees out of 3=3(.1)= .3 employees out of 3
Var(Var(xx )=)=WW22 =3(.1)(.9)= .27=3(.1)(.9)= .27
SD( ) ( ) x np p! ! W 1SD( ) ( ) x np p! ! W 1
employees52.)9)(.1(.3)(SD !!!Wx employees52.)9)(.1(.3)(SD !!!Wx
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Arrival of customers to a serv
ice stat
ion generally has Po
issonArr
ival of customers to a serv
ice stat
ion generally has Po
issondistribution.distribution.
Arrival of cars to a service station.Arrival of cars to a service station.
Arrival of people to a restaurant.Arrival of people to a restaurant.
Arri
val ai
rplanes to an ai
rport.Arri
val ai
rplanes to an ai
rport.
Properties of a Poisson ExperimentProperties of a Poisson Experiment
The probability of an occurrence is the same for any twoThe probability of an occurrence is the same for any twoi
ntervals of equal length.i
ntervals of equal length. The occurrence or nonoccurrence in any interval isThe occurrence or nonoccurrence in any interval isindependent of the occurrence or nonoccurrence in anyindependent of the occurrence or nonoccurrence in anyother interval.other interval.
The Poisson Probability DistributionThe Poisson Probability Distribution
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Example: Mercy HospitalExample: Mercy Hospital
Patients arrive at the emergency room of Mercy Hospital atPatients arrive at the emergency room of Mercy Hospital atthe average rate of 6 per hour on weekend evenings.the average rate of 6 per hour on weekend evenings.
What is the probability of 4 arrivals in 30 minutes on aWhat is the probability of 4 arrivals in 30 minutes on aweekend evening?weekend evening?
Using the Poisson Probability FunctionUsing the Poisson Probability Function
QQ =6/hour =3/half=6/hour =3/half--hour,hour, xx =4=4
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Example: Mercy HospitalExample: Mercy Hospital
Using the Tables of Poisson ProbabilitiesUsing the Tables of Poisson Probabilities
Q
x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0
0 .1225 .1 108 .1 003 .0907 .0821 .0743 .0672 .0608 .0550 .0498
1 .2572 .2438 .2306 .2177 .205 2 .1931 .18 15 .1703 .1596 .1494
2 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .2240
3 .1890 .1966 .2033 .2090 .2138 .2176 .22 05 .22 25 .2237 .2240
4 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1 557 .1622 .1680
5 .0417 .0476 .0538 .0602 ..0668 .0735 .0 804 .0872 .0940 .1 008
6 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .0 504
7 .0044 .0055 .0068 .00 83 .0099 .0 1 18 .0 139 .0163 .0 1 88 .0216
8 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0 081
9 .0003 .0004 .0005 .0007 .0009 .0011 .0014 .0 0 18 .0 022 .0027
1 0 .0 001 .0 00 1 .0 001 .0 002 .0002 .0003 .0004 .0005 .0006 .0008
1 1 .0 000 .0 00 0 .0 000 .0 00 0 .0 000 .0 00 1 .0 001 .0 001 .0 002 .0002
12 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 1
Q
x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0
0 .122 5 .1 108 .1 003 .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 .22 05 .22 25 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .148 8 .1 557 .1622 .1680
5 .0417 .0476 .0538 .0602 ..0668 .0735 .0 804 .0872 .0940 .1 008
6 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .045 5 .0 504
7 .0044 .0 0 55 .0 068 .0 083 .0099 .0 1 18 .0 139 .0163 .0 1 88 .0216
8 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0 081
9 .0003 .0004 .0005 .0007 .0009 .0011 .0014 .0 0 18 .0 022 .0027
1 0 .0 001 .0 001 .0 001 .0 002 .0002 .0003 .0004 .0005 .0006 .0008
1 1 .0 000 .0 000 .0 000 .0 000 .0 00 0 .0 001 .0 00 1 .0 001 .0 002 .0002
12 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 0 .0 00 1
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Data was collected for 128 random intervals of 5 minutes inData was collected for 128 random intervals of 5 minutes inweekday mornings over a period of several weeks.weekday mornings over a period of several weeks.
ExampleExample
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Before using Poisson distribution in our example, we need toBefore using Poisson distribution in our example, we need toestimateestimate QQ
The Mean ValueThe Mean Value
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The Poisson Probability Distribution Table forThe Poisson Probability Distribution Table for QQ = 5= 5
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The Poisson Probability Distribution Table forThe Poisson Probability Distribution Table for QQ = 5= 5
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The Poisson Probability Distribution Table forThe Poisson Probability Distribution Table for QQ = 5= 5
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In studyi
ng the need for an addi
ti
onal entrance to a ci
ty parki
ngIn studyi
ng the need for an addi
ti
onal entrance to a ci
ty parki
nggarage, a consultant has recommended an approach that isgarage, a consultant has recommended an approach that isapplicable only in situations where the number of cars enteringapplicable only in situations where the number of cars enteringduring a specified time period follows a Poisson distribution.during a specified time period follows a Poisson distribution.
A random sample of 100 oneA random sample of 100 one--minute time intervals resulted inminute time intervals resulted inthe customer arrivals listed below. A statistical test must bethe customer arrivals listed below. A statistical test must beconducted to see if the assumption of a Poisson distribution isconducted to see if the assumption of a Poisson distribution isreasonable.reasonable.
#Arrivals#Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 120 1 2 3 4 5 6 7 8 9 10 11 12
FrequencyFrequency 0 1 4 10 14 20 12 12 9 8 6 3 10 1 4 10 14 20 12 12 9 8 6 3 1
Example: Troy Parking Garage
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xx ff((xx ))
00 .0025.0025
11 .0149.0149
22 .0446.0446
33 .0892.0892
44 .1339.1339
55 .1620.1620
66 .1606.1606
77 .1389.1389
88 .1041.1041
99 .0694.0694
1010 .0417.0417
1111 .0227.0227
>=12>=12 .0155.0155
TotalTotal 1.00001.0000
Example: Troy Parking Garage
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Poisson Approximation of Binomial DistributionPoisson Approximation of Binomial Distribution
The Poisson probability distribution can be used asThe Poisson probability distribution can be used asan approximation of the binomial probabilityan approximation of the binomial probability
distribution whendistribution whenpp, the probability of success, is small, the probability of success, is small
andand nn, the number of trials, is large., the number of trials, is large.
Approximation is good whenApproximation is good whenpp > 2020
SetSet QQ ==npnp and use the Poisson tables.and use the Poisson tables.
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Poisson Approximation of Binomial DistributionPoisson Approximation of Binomial Distribution
We want to compute the binomial probability of x=3We want to compute the binomial probability of x=3success in 20 trials when p=.01.success in 20 trials when p=.01.
1) Compute it using binomial table1) Compute it using binomial table
2) Compute it using Poisson table2) Compute it using Poisson table
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The EndThe End