Introduction to Statistics MathematicsChapter Discrete Distributions

Important Discrete Probability DistributionsDiscrete Probability DistributionsBinomialHypergeometricPoisson

Binomial Probability Distributionn identical trialse.g.: 15 tosses of a coin; ten light bulbs taken from a warehouseTwo mutually exclusive outcomes on each trialse.g.: Head or tail in each toss of a coin; defective or not defective light bulbTrials are independentThe outcome of one trial does not affect the outcome of the other

Binomial Probability DistributionConstant probability for each triale.g.: Probability of getting a tail is the same each time we toss the coinTwo sampling methodsInfinite population without replacementFinite population with replacement(continued)

Binomial Probability Distribution Function Tails in 2 Tosses of Coin

X P(X) 0 1/4 = .25 1 2/4 = .50 2 1/4 = .25

Proof of the ProbabilityNote that, by the binomial theorem, the probabilities sum to one, that is Proof :

Binomial Distribution CharacteristicsMean E.g. Variance and Standard Deviation

E.g.

Expectation

Variance

Binomial Probabilities Table

Binomial Distribution in PHStatPHStat | probability & prob. Distributions | binomialExample in excel spreadsheet

Histogram

0.59049

0.32805

0.0729

0.0081

0.00045

0.00001

Number of Successes

P(X)

Histogram

Binomial

Binomial Probabilities

Sample size5

Probability of success0.1

Mean0.5

Variance0.45

Standard deviation0.6708203932

Binomial Probabilities Table

XP(X)P(=X)

00.590490.5904900.409511

10.328050.918540.590490.081460.40951

20.07290.991440.918540.008560.08146

30.00810.999540.991440.000460.00856

40.000450.999990.999540.000010.00046

50.0000110.9999900.00001

ExampleS uppose that an airplane engine will fall, when in flight, with prob 1-p independently from engine to engine; suppose that the airplane will make a succesful flight if at least 50 percent of its engines remain operative. For what values of p is a four-engine plane preferable to a two-engine plane ?

SolutionThe probaility that a four-engine plane makes a successful flight is

Whereas the corresponding probability for a two-engine plane is

SolutionHence the four-engine is safer if

Hence, the four-engine plane is safer when the engine success probability is at least as large as 2/3 , whereas the two-engine plane is safer when this probability falls below 2/3

Poisson DistributionPoisson Process:Discrete events in an intervalThe probability of One Success in an interval is stableThe probability of More than One Success in this interval is 0The probability of success is independent from interval to intervale.g.: number of customers arriving in 15 minutese.g.: number of defects per case of light bulbs

Poisson Probability Distribution Functione.g.: Find the probability of 4 customers arriving in 3 minutes when the mean is 3.6.

Poisson Distribution in PHStatPHStat | probability & prob. Distributions | PoissonExample in excel spreadsheet

Histogram

0

0.0273237224

0.0983654008

0.1770577215

0.2124692658

0.1912223392

0.1376800842

0.0826080505

0.0424841403

0.0191178631

0.0076471452

0.0027529723

0.0009009727

0.0002702918

0.00007485

0.0000192472

0.0000046193

0.0000010393

0.0000002201

0.000000044

0.0000000083

0.0000000015

Number of Successes

P(X)

Histogram

Poisson

Poisson Probabilities for Customer Arrivals

Average/Expected number of successes:3.6

Poisson Probabilities Table

XP(X)P(=X)

00.0273240.0273240.0000000.9726761.000000

10.0983650.1256890.0273240.8743110.972676

20.1770580.3027470.1256890.6972530.874311

30.2124690.5152160.3027470.4847840.697253

40.1912220.7064380.5152160.2935620.484784

50.1376800.8441190.7064380.1558810.293562

60.0826080.9267270.8441190.0732730.155881

70.0424840.9692110.9267270.0307890.073273

80.0191180.9883290.9692110.0116710.030789

90.0076470.9959760.9883290.0040240.011671

100.0027530.9987290.9959760.0012710.004024

110.0009010.9996300.9987290.0003700.001271

120.0002700.9999000.9996300.0001000.000370

130.0000750.9999750.9999000.0000250.000100

140.0000190.9999940.9999750.0000060.000025

150.0000050.9999990.9999940.0000010.000006

160.0000011.0000000.9999990.0000000.000001

170.0000001.0000001.0000000.0000000.000000

180.0000001.0000001.0000000.0000000.000000

190.0000001.0000001.0000000.0000000.000000

200.0000001.0000001.0000000.0000000.000000

&A

Page &P

Poisson Probabilities for Customer Arrivals

Poisson Distribution CharacteristicsMean

Standard Deviation and Variance = 0.5= 6 0.2.4.6012345XP(X) 0.2.4.60246810XP(X)

Approximate a binomial to poissonAn important property of the poisson random variable is that it may be used to approximate a binomial random variabel when the binomial parameter n is large and p is small. To see this, suppose that X is a binomial r.v. with parameters (n,p), and let = np. Then

Proof

Expectation

Variance

Hypergeometric Distributionn trials in a sample taken from a finite population of size NSample taken without replacementTrials are dependentConcerned with finding the probability of X successes in the sample where there are A successes in the population

Hypergeometric Distribution FunctionE.g. 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective?

Hypergeometric Distribution CharacteristicsMean

Variance and Standard Deviation Finite Population Correction Factor

Hypergeometric Distribution in PHStatPHStat | probability & prob. Distributions | Hypergeometric Example in excel spreadsheet

Histogram

0.1666666667

0.5

0.3

0.0333333333

0

Number of Successes

P(X)

Histogram

Hypergeometric

Hypergeometric Probabilities

Sample size3

No. of successes in population4

Population size10

Hypergeometric Probabilities Table

XP(x)

00.1666666667

10.5

20.3

30.0333333333

40

PHStat User Note:If the #NUM! Message appears in the Probabilities Table, the number of successes is larger than the sample size--an impossibility.

Alter the value(s) of the sample size (cell B3) and/or number of successes (B4) before continuing.

Select this note and then select Edit | Cut to delete this note from the worksheet.

Jointly Distributed Random VariablesThe joint probability mass function of X and Y is p(x,y)=P(X=x,Y=y)The probability mass function of X

The probability mass function of Y

Expectation

ExampleAs another example of the usefulness of equation above, let us use it to obtain the expectation of a binomial r.v.

ExampleAt a party N men throw their hats into the center of a room. The hats are mixed up and each man randomly selects one. Find the expected number of men that select their own hats

SolutionLetting X denote the number of men that select their own hats, we can best compute E(X) by noting that X = X1++XN ; where Xi is indicator function if the ith man select his own hat. So P(Xi = 1) = 1/N. And so E(Xi) = 1/N. Hence We obtain that E(X) = 1. So, no matter how many people are at the party, on the average exactly one of the men will select his own hat.

Independent R.VX and Y are independent if

Covariance and Variance of Sums of Random VariablesThe covariance of any two random variables X and Y, denoted by Cov(X,Y), is defined byCov(X,Y)= E[(X-E[X])(Y-E[Y])] = E(XY)-E(X)E(Y) If X and Y are independent Cov(X,Y) = 0

Properties of CovarianceCov(x,X) = Var(X)Cov(cX,Y) = c Cov(X,Y)Cov(X,Y+Z)= E[X(Y+Z)]-E[X]E[Y+Z] = E[XY]-E[X]E[Y] + E[XZ]-E[X]E[Z]= (Cov(X,Y) + Cov(X,Z)The last property easily generalizes to give

Variance of Sum Variabel

Proposition Suppose that X1,,Xn are independent and identically distributed with expected value and variance 2. Then

ExampleSums of independent Poisson Random Variables : Let A and Y be independent Poisson random variables wirh respective means 1 and 2 . Calculate the distribution of X + Y.Solution : Since the event {X+Y = n} may be written as the union of the disjoint events {X=k,Y=n-k}, 0kn, we have

9999999999999999