Discrete Probability Distributions

Business Statistics, 6th ed.by Ken Black

Chapter 5Discrete

ProbabilityDistributions

Learning Objectives

Distinguish between discrete random variables and continuous random variables.Know how to determine the mean and variance of a discrete distribution.Identify the type of statistical experiments that can be described by the binomial distribution, and know how to work such problems.

Discrete vs. Continuous Distributions

Discrete distributions – constructed from discrete (individually distinct) random variablesContinuous distributions – based on continuous random variablesRandom Variable - a variable which contains the outcomes of a chance experiment

Discrete vs. Continuous Distributions

Categories of Random VariablesDiscrete Random Variable - the set of all possible values is at most a finite or a countable infinite number of possible valuesContinuous Random Variable - takes on values at every point over a given interval

Describing a Discrete Distribution

A discrete distribution can be described by constructing a graph of the distributionMeasures of central tendency and variability can be applied to discrete distributionsDiscrete values of outcomes are used to represent themselves

Mean of discrete distribution – is the long run average

If the process is repeated long enough, the average of the outcomes will approach the long run average (mean)

Requires the process to eventually have a number which is the product of many processes

Mean of a discrete distribution µ = ∑ (X * P(X))

where (X) is the long run average;X = outcome, P = Probability of X

Variance and Standard Deviation of a discrete distribution are solved by using the outcomes (X) and probabilities of outcomes (P(X)) in a manner similar to computing a meanStandard Deviation is computed by taking the square root of the variance

Some Special Distributions

DiscretebinomialPoissonHypergeometric

Continuousnormaluniformexponentialtchi-squareF

Discrete Distribution -- Example

Observe the discrete distribution in the following table. An executive is considering out-of-town business travel for a given Friday. At least one crisis could occur on the day that the executive is gone. The distribution contains the number of crises that could occur during the day the executive is gone and the probability that each number will occur. For example, there is a .37 probability that no crisis will occur, a .31 probability of one crisis, and so on.

012345

0.370.310.180.090.040.01

Number of Crises Probability

Distribution of Daily Crises

0 1 2 3 4 5

Probability

Number of Crises

Discrete Distribution -- Example

2.1)(22 XPX

212 110. .

Variance and Standard Deviation of a Discrete Distribution

X-10123

P(X).1.2.4.2.1

-2-1012

X 41014

2( ) ( )X P X

Requirements for a Discrete Probability Function -- Examples

X P(X)

-10123

X P(X)

-10123

-.1.3.4.3.1

X P(X)

-10123

: YES NO NO

Mean of a Discrete Distribution

E X X P X( )

X-10123

P(X).1.2.4.2.1

-.1.0.4.4.3

X P X ( )

E X X P X( ) .115

Mean of the Crises Data Example

X P(X) XP(X)

0 .37 .00

1 .31 .31

2 .18 .36

3 .09 .27

4 .04 .16

5 .01 .05

0 1 2 3 4 5

Probability

Number of Crises

141 X P X( ) . 2

141 119. .

Variance and Standard Deviationof Crises Data Example

X P(X) (X-) (X-)2 P(X)

0 .37 -1.15 1.32 .49

1 .31 -0.15 0.02 .01

2 .18 0.85 0.72 .13

3 .09 1.85 3.42 .31

4 .04 2.85 8.12 .32

5 .01 3.85 14.82 .15

Binomial Distribution

nXP qp

Probability function

Mean value

Variance and Standard Deviation

According to the U.S. Census Bureau, approximately 6% of all workers in Jackson, Mississippi, are unemployed. In conducting a random telephone survey in Jackson, what is the probability of getting two or fewer unemployed workers in a sample of 20?

Binomial Distribution:Demonstration Problem 5.3

In the following example, 6% are unemployed => pThe sample size is 20 => n94% are employed => qx is the number of successes desiredWhat is the probability of getting 2 or fewer unemployed workers in the sample of 20?The hard part of this problem is identifying p, n, and x – emphasis this when studying the problems.

P X P X P X P X

2 0 1 2

2901 3703 2246 8850

( ) ( ) ( ) ( )

. . . .

2901.)2901)(.1)(1()!020(!0

!20)0( 94.06.

P X( )!( )!

( )(. )(. ) .. .

1 20 120 06 3086 3703

1 20 1

P X( )!( )!

( )(. )(. ) .. .

2 20 2190 0036 3283 2246

2 20 2

Binomial Distribution Table:Demonstration Problem 5.3

P X P X P X P X

2 0 1 2

2901 3703 2246 8850

( ) ( ) ( ) ( )

. . . .

P X P X( ) ( ) . . 2 1 2 1 8850 1150

n p ( )(. ) .20 06 1 202

20 06 94 1 128

1 128 1 062

n p q ( )(. )(. ) .

n = 20 PROBABILITY

X 0.05 0.06 0.07

0 0.3585 0.2901 0.2342

1 0.3774 0.3703 0.3526

2 0.1887 0.2246 0.2521

3 0.0596 0.0860 0.1139

4 0.0133 0.0233 0.0364

5 0.0022 0.0048 0.0088

6 0.0003 0.0008 0.0017

7 0.0000 0.0001 0.0002

8 0.0000 0.0000 0.0000

… … …

20 0.0000 0.0000 0.0000

Excel’s Binomial Function

n = 20

p = 0.06

X P(X)

0 =BINOMDIST(A5,B$1,B$2,FALSE)

X P(X =x)

0 0.0000001 0.0000002 0.0000003 0.0000014 0.0000065 0.0000376 0.0001997 0.0008588 0.0030519 0.00904010 0.02250011 0.04727312 0.08404113 0.12642014 0.16053315 0.17123616 0.15220917 0.11142118 0.06602719 0.03089020 0.01098321 0.00278922 0.00045123 0.000035

Binomial with n = 23 and p = 0.64

Minitab’s Binomial Function

Mean and Std Dev of Binomial Distribution

Binomial distribution has an expected value or a long run average denoted by µ (mu)

If n items are sampled over and over for a long time and if p is the probability of success in one trial, the average long run of successes per sample is expected to be np=> Mean µ = np=> Std Dev = √(npq)

Poisson Distribution

The Poisson distribution focuses only on the number of discrete occurrences over some interval or continuum

Poisson does not have a given number of trials (n)as a binomial experiment doesOccurrences are independent of other occurrencesOccurrences occur over an interval

If Poisson distribution is studied over a long periodof time, a long run average can be determined

The average is denoted by lambda (λ)Each Poisson problem contains a lambda value from which the probabilities are determinedA Poisson distribution can be described by λ alone

)logarithms natural of base (the ...718282.2

,...3,2,1,0for !

averagerunlong

Probability function

Mean value

Standard deviation Variance

Poisson Distribution:Demonstration Problem 5.7

Bank customers arrive randomly on weekday afternoons at an average of 3.2 customers every 4 minutes. What is the probability of having more than 7 customers in a 4-minute interval on a weekday afternoon?

Solutionλ = 3.2 customers>minutes X > 7 customers/4 minutesThe solution requires obtaining the values of x = 8, 9, 10, 11, 12, 13, 14, . . . . Each x value is determined until the values are so far away from λ = 3.2 that the probabilities approach zero. The exact probabilities are summed to find x 7. If the bank has been averaging 3.2 customers every 4 minutes on weekday afternoons, it is unlikely that more than 7 people would randomly arrive in any one 4-minute period. This answer indicates that more than 7 people would randomly arrive in a 4-minute period only 1.69% of the time. Bank officers could use these results to help them make staffing decisions.

0528.0!10

=)10=(

!=P(X)

minutes 8customers/ 4.6=

Adjusted

minutes 8customers/ 10 = X

minutes 4customers/ 2.3

0 15866 4

customers / 4 minutes

X = 6 customers / 8 minutes

Adjusted

= . customers / 8 minutes

P(X) =

( = ) =

0060.0000.0002.0011.0047.

)9()8()7()6()5(

XPXPXPXPXP

Poisson Distribution:Using the Poisson Tables

X 0.5 1.5 1.6 3.00 0.6065 0.2231 0.2019 0.04981 0.3033 0.3347 0.3230 0.14942 0.0758 0.2510 0.2584 0.22403 0.0126 0.1255 0.1378 0.22404 0.0016 0.0471 0.0551 0.16805 0.0002 0.0141 0.0176 0.10086 0.0000 0.0035 0.0047 0.05047 0.0000 0.0008 0.0011 0.02168 0.0000 0.0001 0.0002 0.00819 0.0000 0.0000 0.0000 0.002710 0.0000 0.0000 0.0000 0.000811 0.0000 0.0000 0.0000 0.000212 0.0000 0.0000 0.0000 0.0001

4751.3230.2019.1

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

XPXPXPXP

Poisson Distribution:Using the Poisson Tables

X 0.5 1.5 1.6 3.00 0.6065 0.2231 0.2019 0.04981 0.3033 0.3347 0.3230 0.14942 0.0758 0.2510 0.2584 0.22403 0.0126 0.1255 0.1378 0.22404 0.0016 0.0471 0.0551 0.16805 0.0002 0.0141 0.0176 0.10086 0.0000 0.0035 0.0047 0.05047 0.0000 0.0008 0.0011 0.02168 0.0000 0.0001 0.0002 0.00819 0.0000 0.0000 0.0000 0.002710 0.0000 0.0000 0.0000 0.000811 0.0000 0.0000 0.0000 0.000212 0.0000 0.0000 0.0000 0.0001

Excel’s Poisson Function

X P(X)

0 =POISSON(D5,E$1,FALSE)

Minitab’s Poisson Function

X P(X =x)

0 0.1495691 0.2841802 0.2699713 0.1709824 0.0812165 0.0308626 0.0097737 0.0026538 0.0006309 0.00013310 0.000025

Poisson with mean = 1.9

Mean and Std Dev of a Poisson Distribution

Mean of a Poisson Distribution is λUnderstanding the mean of a Poisson distribution gives a feel for the actual occurrences that are likely to happenVariance of a Poisson distribution is also λStd Dev = Square root of λ

Poisson Approximation of the Binomial Distribution

Binomial problems with large sample sizes and small values of p, which then generate rare events, are potential candidates for use of the Poisson DistributionRule of thumb, if n > 20 and np < 7, the approximation is close enough to use the Poisson distribution for binomial problems

Procedure for Approximating binomial with PoissonBegin with the computation of the binomial mean distribution µ = npBecause µ is the expected value of the binomial, it becomes λ for Poisson distributionUse µ as the λ, and using the x from the binomial problem allows for the approximation of the probabilities from the Poisson table or Poisson formula

If and the approximation is acceptable.n n p 20 7,

Use n p.

Binomial probabilities are difficult to calculate when n is large.Under certain conditions binomial probabilities may be approximated by Poisson probabilities.

Poisson approximation

Hypergeometric Distribution

Sampling without replacement from a finite populationThe number of objects in the population is denoted N.Each trial has exactly two possible outcomes, success and failure.Trials are not independentX is the number of successes in the n trialsThe binomial is an acceptable approximation,if n < 5% N. Otherwise it is not.

Hypergeometric Distribution

A N A n N n

NN( ) ( )

A x N A n x

N n( )

Probability functionN is population sizen is sample sizeA is number of successes in populationx is number of successes in sample

MeanValue

Variance and standarddeviation

N = 24X = 8n = 5

x0 0.10281 0.34262 0.36893 0.15814 0.02645 0.0013

P xC C

A x N A n x

N n( )

56 120

42 504

8 3 24 8 5 3

Hypergeometric Distribution:Probability Computations

Excel’s Hypergeometric Function

N = 24

X P(X)

0 =HYPGEOMDIST(A6,B$3,B$2,B$1)

=SUM(B6:B11)

Minitab’s Hypergeometric Function

X P(X =x)

0 0.1027671 0.3425562 0.3689063 0.1581034 0.0263505 0.001318

Hypergeometric with N = 24, A = 8, n = 5

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