Statistics- Discrete Probability Distributions

8/3/2019 Statistics- Discrete Probability Distributions

1/36

Discrete Probability DistributionsDiscrete Probability Distributions

Prof. Rushen ChahalProf. Rushen Chahal


2/36

22


Random VariablesRandom Variables


Expected Value and VarianceExpected Value and Variance The Binomial Probability DistributionThe Binomial Probability Distribution

The Poisson Probability DistributionThe Poisson Probability Distribution


3/36

33

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.


4/36

44

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.


5/36

55

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.


6/36

66

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


7/36

77

PracticePractice

187187--188188 Problem 1Problem 1

Problem 2Problem 2

Problem 3Problem 3


8/36

88

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.


9/36

99


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.


10/36

1010

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


11/36

1111

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


12/36

1212

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


13/36

1313

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.


14/36

1414

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


15/36

1515

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


16/36

1616

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


17/36

1717

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


18/36

1818

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..


19/36

1919

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


20/36

2020

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


21/36

2121


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


22/36

2222

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



S (.9)S (.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



23/36

2323

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


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


24/36

2424

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


25/36

2525

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


26/36

2626

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


27/36

2727

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


28/36

2828

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


29/36

2929

The Poisson Probability Distribution Table forThe Poisson Probability Distribution Table for QQ = 5= 5


30/36

3030



31/36

3131



32/36

3232

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


33/36

3333

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


34/36

3434

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.


35/36

3535

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


36/36

3636

The EndThe End

Statistics- Discrete Probability Distributions

Documents

Transcript of Statistics- Discrete Probability Distributions