chapter 5 discrete probaility distrinution

Chapter 5

Some Important Discrete Probability Distributions

by Try Sothearith [email protected] [email protected] Tel: 012 585 865 / 016555507Chap 5-*NAA: Basic Statistics Course

NAA: Basic Statistics Course

Learning ObjectivesIn this chapter, you learn: The properties of a probability distributionTo calculate the expected value and variance of a probability distributionTo calculate the covariance and its use in financeTo calculate probabilities from binomial, hypergeometric, and Poisson distributionsHow to use the binomial, hypergeometric, and Poisson distributions to solve business problemsChap 5-*NAA: Basic Statistics Course


Introduction to Probability DistributionsRandom VariableRepresents a possible numerical value from an uncertain event

Random VariablesDiscrete Random VariableContinuousRandom VariableCh. 5Ch. 6Chap 5-*NAA: Basic Statistics Course


Discrete Random VariablesCan only assume a countable number of values

Examples:

Roll a die twiceLet X be the number of times 4 comes up (then X could be 0, 1, or 2 times)

Toss a coin 5 times. Let X be the number of heads (then X = 0, 1, 2, 3, 4, or 5)

Chap 5-*NAA: Basic Statistics Course


Experiment: Toss 2 Coins. Let X = # heads.TTDiscrete Probability Distribution4 possible outcomesTTHHHHProbability Distribution 0 1 2 X X Value Probability 0 1/4 = 0.25 1 2/4 = 0.50 2 1/4 = 0.250.500.25 Probability Chap 5-*NAA: Basic Statistics Course


Discrete Random Variable Summary Measures Expected Value (or mean) of a discrete distribution (Weighted Average)

Example: Toss 2 coins, X = # of heads, compute expected value of X: E(X) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25) = 1.0 X P(X) 0 0.25 1 0.50 2 0.25Chap 5-*NAA: Basic Statistics Course


Variance of a discrete random variable

Standard Deviation of a discrete random variable

where:E(X) = Expected value of the discrete random variable X Xi = the ith outcome of XP(Xi) = Probability of the ith occurrence of XDiscrete Random Variable Summary Measures(continued)Chap 5-*NAA: Basic Statistics Course


Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(X) = 1)Discrete Random Variable Summary Measures(continued)Possible number of heads = 0, 1, or 2Chap 5-*NAA: Basic Statistics Course


The CovarianceThe covariance measures the strength of the linear relationship between two variablesThe covariance:

where:X = discrete variable XXi = the ith outcome of XY = discrete variable YYi = the ith outcome of YP(XiYi) = probability of occurrence of the ith outcome of X and the ith outcome of Y



Computing the Mean for Investment ReturnsReturn per $1,000 for two types of investmentsP(XiYi) Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession- $ 25 - $200 0.5 Stable Economy+ 50 + 60 0.3 Expanding Economy + 100 + 350 InvestmentE(X) = X = (-25)(0.2) +(50)(0.5) + (100)(0.3) = 50E(Y) = Y = (-200)(0.2) +(60)(0.5) + (350)(0.3) = 95Chap 5-*NAA: Basic Statistics Course


Computing the Standard Deviation for Investment ReturnsP(XiYi) Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession- $ 25 - $200 0.5 Stable Economy+ 50 + 60 0.3 Expanding Economy + 100 + 350 InvestmentChap 5-*NAA: Basic Statistics Course


Computing the Covariance for Investment ReturnsP(XiYi) Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession- $ 25 - $200 0.5 Stable Economy+ 50 + 60 0.3 Expanding Economy + 100 + 350 InvestmentChap 5-*NAA: Basic Statistics Course


Interpreting the Results for Investment ReturnsThe aggressive fund has a higher expected return, but much more risk

Y = 95 > X = 50 butY = 193.71 > X = 43.30

The Covariance of 8250 indicates that the two investments are positively related and will vary in the same directionChap 5-*NAA: Basic Statistics Course


The Sum of Two Random VariablesExpected Value of the sum of two random variables:

Variance of the sum of two random variables:

Standard deviation of the sum of two random variables:Chap 5-*NAA: Basic Statistics Course


Portfolio Expected Return and Portfolio RiskPortfolio expected return (weighted average return):

Portfolio risk (weighted variability)

Where w = portion of portfolio value in asset X (1 - w) = portion of portfolio value in asset YChap 5-*NAA: Basic Statistics Course


Portfolio Example Investment X: X = 50 X = 43.30 Investment Y: Y = 95 Y = 193.21 XY = 8250

Suppose 40% of the portfolio is in Investment X and 60% is in Investment Y:

The portfolio return and portfolio variability are between the values for investments X and Y considered individuallyChap 5-*NAA: Basic Statistics Course


Probability DistributionsContinuous Probability DistributionsBinomialHypergeometricPoissonProbability DistributionsDiscrete Probability DistributionsNormalUniformExponentialCh. 6Ch. 7Chap 5-*NAA: Basic Statistics Course


The Binomial DistributionBinomialHypergeometricPoissonProbability DistributionsDiscrete Probability DistributionsChap 5-*NAA: Basic Statistics Course


Binomial Probability DistributionA fixed number of observations, ne.g., 15 tosses of a coin; ten light bulbs taken from a warehouseTwo mutually exclusive and collectively exhaustive categories, generally called success and failureSuccessFailureProbability of success is p, probability of failure is 1 pConstant probability for each Success casee.g., Probability of getting a tail is the same each time we toss the coinChap 5-*NAA: Basic Statistics Course


Binomial Probability Distribution(continued)Observations are independentThe outcome of one observation does not affect the outcome of the otherTwo sampling methodsInfinite population without replacementFinite population with replacement



Possible Binomial Distribution SettingsA manufacturing plant labels items as either defective or acceptableA firm bidding for contracts will either get a contract or notA marketing research firm receives survey responses of yes I will buy or no I will notNew job applicants either accept the offer or reject itChap 5-*NAA: Basic Statistics Course


Rule of CombinationsThe number of combinations of selecting X objects out of n objects is where:n! =(n)(n - 1)(n - 2) . . . (2)(1)X! = (X)(X - 1)(X - 2) . . . (2)(1) 0! = 1 (by definition)Chap 5-*NAA: Basic Statistics Course


P(X) = probability of X successes in n trials, with probability of success p on each trial

X = number of successes in sample, (X = 0, 1, 2, ..., n) n = sample size (number of trials or observations) p = probability of success P(X)nX !nXp(1-p)XnX!()!=--Example: Flip a coin four times, let x = # heads:n = 4p = 0.51 - p = (1 - 0.5) = 0.5X = 0, 1, 2, 3, 4Binomial Distribution FormulaChap 5-*NAA: Basic Statistics Course


Example: Calculating a Binomial ProbabilityWhat is the probability of one success in five observations if the probability of success is .1? X = 1, n = 5, and p = 0.1Chap 5-*NAA: Basic Statistics Course


n = 5 p = 0.1n = 5 p = 0.5Mean 0.2.4.6012345XP(X).2.4.6012345XP(X)0Binomial DistributionThe shape of the binomial distribution depends on the values of p and nHere, n = 5 and p = 0.1Here, n = 5 and p = 0.5Chap 5-*NAA: Basic Statistics Course


Binomial Distribution CharacteristicsMean

Variance and Standard DeviationWheren = sample sizep = probability of success(1 p) = probability of failureChap 5-*NAA: Basic Statistics Course


n = 5 p = 0.1n = 5 p = 0.5Mean 0.2.4.6012345XP(X).2.4.6012345XP(X)0Binomial CharacteristicsExamplesChap 5-*NAA: Basic Statistics Course


Using Binomial TablesExamples: n = 10, p = 0.35, x = 3: P(x = 3|n =10, p = 0.35) = 0.2522n = 10, p = 0.75, x = 2: P(x = 2|n =10, p = 0.75) = 0.0004Chap 5-*NAA: Basic Statistics Course

n = 10xp=.20p=.25p=.30p=.35p=.40p=.45p=.500123456789100.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.00000.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.00000.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.00000.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.00000.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.00010.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.00030.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010109876543210p=.80p=.75p=.70p=.65p=.60p=.55p=.50x


The Poisson DistributionBinomialHypergeometricPoissonProbability DistributionsDiscrete Probability DistributionsChap 5-*NAA: Basic Statistics Course


The Poisson DistributionApply the Poisson Distribution when:You wish to count the number of times an event occurs in a given area of opportunityThe probability that an event occurs in one area of opportunity is the same for all areas of opportunity The number of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunityThe probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smallerThe average number of events per unit is (lambda)Chap 5-*NAA: Basic Statistics Course


Poisson Distribution Formulawhere:X = number of events in an area of opportunity = expected number of eventse = base of the natural logarithm system (2.71828...)Chap 5-*NAA: Basic Statistics Course


Poisson Distribution CharacteristicsMean

Variance and Standard Deviationwhere = expected number of events



Using Poisson TablesExample: Find P(X = 2) if = 0.50Chap 5-*NAA: Basic Statistics Course

X0.100.200.300.400.500.600.700.800.90012345670.90480.09050.00450.00020.00000.00000.00000.00000.81870.16370.01640.00110.00010.00000.00000.00000.74080.22220.03330.00330.00030.00000.00000.00000.67030.26810.05360.00720.00070.00010.00000.00000.60650.30330.07580.01260.00160.00020.00000.00000.54880.32930.09880.01980.00300.00040.00000.00000.49660.34760.12170.02840.00500.00070.00010.00000.44930.35950.14380.03830.00770.00120.00020.00000.40660.36590.16470.04940.01110.00200.00030.0000


Graph of Poisson ProbabilitiesP(X = 2) = 0.0758 Graphically: = 0.50 Chap 5-*NAA: Basic Statistics Course

X =0.50012345670.60650.30330.07580.01260.00160.00020.00000.0000


Chart2

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

x

P(x)

Histogram

0

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

0.0000000588

0.0000000033

0.0000000002

0

0

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Number of Successes

P(X)

Histogram

Poisson2

Poisson Probabilities for Customer Arrivals

Data

Average/Expected number of successes:0.5

Poisson Probabilities Table

XP(X)P(=X)

00.6065310.6065310.0000000.3934691.000000

10.3032650.9097960.6065310.0902040.393469

20.0758160.9856120.9097960.0143880.090204

30.0126360.9982480.9856120.0017520.014388

40.0015800.9998280.9982480.0001720.001752

50.0001580.9999860.9998280.0000140.000172

60.0000130.9999990.9999860.0000010.000014

70.0000011.0000000.9999990.0000000.000001

80.0000001.0000001.0000000.0000000.000000

90.0000001.0000001.0000000.0000000.000000

100.0000001.0000001.0000000.0000000.000000

110.0000001.0000001.0000000.0000000.000000

120.0000001.0000001.0000000.0000000.000000

130.0000001.0000001.0000000.0000000.000000

140.0000001.0000001.0000000.0000000.000000

150.0000001.0000001.0000000.0000000.000000

160.0000001.0000001.0000000.0000000.000000

170.0000001.0000001.0000000.0000000.000000

180.0000001.0000001.0000000.0000000.000000

190.0000001.0000001.0000000.0000000.000000

200.0000001.0000001.0000000.0000000.000000

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Poisson2

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Data



XP(X)P(=X)

00.90480.9048370.0000000.0951631.000000

10.09050.9953210.9048370.0046790.095163

20.00450.9998450.9953210.0001550.004679

30.00020.9999960.9998450.0000040.000155

40.00001.0000000.9999960.0000000.000004

50.00001.0000001.0000000.0000000.000000

60.00001.0000001.0000000.0000000.000000

70.00001.0000001.0000000.0000000.000000

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Poisson Distribution ShapeThe shape of the Poisson Distribution depends on the parameter : = 0.50 = 3.00Chap 5-*NAA: Basic Statistics Course


Chart3

0.0497870684

0.1493612051

0.2240418077

0.2240418077

0.1680313557

0.1008188134

0.0504094067

0.0216040315

0.0081015118

0.0027005039

0.0008101512

0.0002209503

x

P(x)

Histogram

0

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

0.0000000588

0.0000000033

0.0000000002

0

0

0

0

0

0

0

0

0

0

Number of Successes

P(X)

Histogram

Poisson2


Data

Average/Expected number of successes:3


XP(X)P(=X)

00.0497870.0497870.0000000.9502131.000000

10.1493610.1991480.0497870.8008520.950213

20.2240420.4231900.1991480.5768100.800852

30.2240420.6472320.4231900.3527680.576810

40.1680310.8152630.6472320.1847370.352768

50.1008190.9160820.8152630.0839180.184737

60.0504090.9664910.9160820.0335090.083918

70.0216040.9880950.9664910.0119050.033509

80.0081020.9961970.9880950.0038030.011905

90.0027010.9988980.9961970.0011020.003803

100.0008100.9997080.9988980.0002920.001102

110.0002210.9999290.9997080.0000710.000292

120.0000550.9999840.9999290.0000160.000071

130.0000130.9999970.9999840.0000030.000016

140.0000030.9999990.9999970.0000010.000003

150.0000011.0000000.9999990.0000000.000001

160.0000001.0000001.0000000.0000000.000000

170.0000001.0000001.0000000.0000000.000000

180.0000001.0000001.0000000.0000000.000000

190.0000001.0000001.0000000.0000000.000000

200.0000001.0000001.0000000.0000000.000000

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Data



XP(X)P(=X)

00.90480.9048370.0000000.0951631.000000

10.09050.9953210.9048370.0046790.095163

20.00450.9998450.9953210.0001550.004679

30.00020.9999960.9998450.0000040.000155

40.00001.0000000.9999960.0000000.000004

50.00001.0000001.0000000.0000000.000000

60.00001.0000001.0000000.0000000.000000

70.00001.0000001.0000000.0000000.000000

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Chart2

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

x

P(x)

Histogram

0

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

0.0000000588

0.0000000033

0.0000000002

0

0

0

0

0

0

0

0

0

0

Number of Successes

P(X)

Histogram

Poisson2


Data



XP(X)P(=X)

00.6065310.6065310.0000000.3934691.000000

10.3032650.9097960.6065310.0902040.393469

20.0758160.9856120.9097960.0143880.090204

30.0126360.9982480.9856120.0017520.014388

40.0015800.9998280.9982480.0001720.001752

50.0001580.9999860.9998280.0000140.000172

60.0000130.9999990.9999860.0000010.000014

70.0000011.0000000.9999990.0000000.000001

80.0000001.0000001.0000000.0000000.000000

90.0000001.0000001.0000000.0000000.000000

100.0000001.0000001.0000000.0000000.000000

110.0000001.0000001.0000000.0000000.000000

120.0000001.0000001.0000000.0000000.000000

130.0000001.0000001.0000000.0000000.000000

140.0000001.0000001.0000000.0000000.000000

150.0000001.0000001.0000000.0000000.000000

160.0000001.0000001.0000000.0000000.000000

170.0000001.0000001.0000000.0000000.000000

180.0000001.0000001.0000000.0000000.000000

190.0000001.0000001.0000000.0000000.000000

200.0000001.0000001.0000000.0000000.000000

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Data



XP(X)P(=X)

00.90480.9048370.0000000.0951631.000000

10.09050.9953210.9048370.0046790.095163

20.00450.9998450.9953210.0001550.004679

30.00020.9999960.9998450.0000040.000155

40.00001.0000000.9999960.0000000.000004

50.00001.0000001.0000000.0000000.000000

60.00001.0000001.0000000.0000000.000000

70.00001.0000001.0000000.0000000.000000

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The Hypergeometric DistributionBinomialPoissonProbability DistributionsDiscrete Probability DistributionsHypergeometricChap 5-*NAA: Basic Statistics Course


The Hypergeometric Distributionn trials in a sample taken from a finite population of size NSample taken without replacementOutcomes of trials are dependentConcerned with finding the probability of X successes in the sample where there are A successes in the populationChap 5-*NAA: Basic Statistics Course


Hypergeometric Distribution FormulaWhereN = population sizeA = number of successes in the population N A = number of failures in the populationn = sample sizeX = number of successes in the sample n X = number of failures in the sampleChap 5-*NAA: Basic Statistics Course


Properties of the Hypergeometric DistributionThe mean of the hypergeometric distribution is

The standard deviation is

Where is called the Finite Population Correction Factor from sampling without replacement from a finite populationChap 5-*NAA: Basic Statistics Course


Using the Hypergeometric DistributionExample: 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded?

N = 10n = 3 A = 4 X = 2The probability that 2 of the 3 selected computers have illegal software loaded is 0.30, or 30%.Chap 5-*NAA: Basic Statistics Course


chapter 5 discrete probaility distrinution

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