1、Discrete Probability DistributionsChapter 5Discrete Probability DistributObjectivesIn this chapter, you learn: The properties of a probability distribution.To compute the expected value and variance of a probability distribution.To compute probabilities from binomial, and Poisson distributions.To us

2、e the binomial, and Poisson distributions to solve business problemsObjectivesIn this chapter, youDefinitionsDiscrete variables produce outcomes that come from a counting process (e.g. number of classes you are taking).Continuous variables produce outcomes that come from a measurement (e.g. your ann

3、ual salary, or your weight).DefinitionsDiscrete variables Types Of VariablesCh. 5Ch. 6Ch. 5Ch. 6Types Of VariablesDiscrete VariableContinuousVariableCh. 5Ch. 6Types Of VariablesCh. 5Ch. 6ChDiscrete VariablesCan only assume a countable number of valuesExamples: Roll a die twiceLet X be the number of

4、times 4 occurs (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)Discrete VariablesCan only assProbability Distribution For A Discrete VariableA probability distribution for a discrete variable is a mutually exclusive listing of all pos

5、sible numerical outcomes for that variable and a probability of occurrence associated with each outcome.Interruptions Per Day In Computer NetworkProbability00.3510.2520.2030.1040.0550.05Probability Distribution For AProbability Distributions Are Often Represented GraphicallyP(X)0.1012345XPr

6、obability Distributions Are Discrete Variables Expected Value (Measuring Center) Expected Value (or mean) of a discrete variable (Weighted Average)Interruptions Per Day In Computer Network (xi)ProbabilityP(X = xi)xiP(X = xi)00.35(0)(0.35) = 0.0010.25(1)(0.25) = 0.2520.20(2)(0.20) = 0.4030.10(3)(0.10

7、) = 0.3040.05(4)(0.05) = 0.2050.05(5)(0.05) = 0.251.00 = E(X) = 1.40Discrete Variables Expected VaVariance of a discrete variableStandard Deviation of a discrete variablewhere:E(X) = Expected value of the discrete variable X xi = the ith outcome of XP(X=xi) = Probability of the ith occurrence of XDi

8、screte Variables: Measuring DispersionVariance of a discrete variablDiscrete Variables: Measuring Dispersion(continued)Interruptions Per Day In Computer Network (xi)ProbabilityP(X = xi)xi E(X)2xi E(X)2P(X = xi)00.35(0 1.4)2 = 1.96 (1.96)(0.35) = 0.68610.25(1 1.4)2 = 0.16 (0.16)(0.25) = 0.04020.20(2

9、1.4)2 = 0.36 (0.36)(0.20) = 0.07230.10(3 1.4)2 = 2.56 (2.56)(0.10) = 0.25640.05 (4 1.4)2 = 6.76 (6.76)(0.05) = 0.33850.05(5 1.4)2 = 12.96(12.96)(0.05) = 0.6482 = 2.04, = 1.4283Discrete Variables: MeasuringProbability DistributionsContinuous Probability DistributionsBinomialPoissonProbability Distrib

10、utionsDiscrete Probability DistributionsNormalCh. 5Ch. 6Probability DistributionsContiBinomial Probability DistributionA fixed number of observations, ne.g., 15 tosses of a coin; ten light bulbs taken from a warehouseEach observation is categorized as to whether or not the “event of interest” occurr

11、ede.g., head or tail in each toss of a coin; defective or not defective light bulbSince these two categories are mutually exclusive and collectively exhaustiveWhen the probability of the event of interest is represented as , then the probability of the event of interest not occurring is 1 - Constant

12、 probability for the event of interest occurring () for each observationProbability of getting a tail is the same each time we toss the coinBinomial Probability DistributBinomial Probability Distribution(continued)Observations are independentThe outcome of one observation does not affect the outcome

13、 of the otherTwo sampling methods deliver independenceInfinite population without replacementFinite population with replacementBinomial Probability DistributPossible Applications for the Binomial DistributionA manufacturing plant labels items as either defective or acceptableA firm bidding for contr

14、acts will either get a contract or notA marketing research firm receives survey responses of “yes I will buy” or “no I will not”New job applicants either accept the offer or reject itPossible Applications for the The Binomial DistributionCounting TechniquesSuppose the event of interest is obtaining

15、heads on the toss of a fair coin. You are to toss the coin three times. In how many ways can you get two heads?Possible ways: HHT, HTH, THH, so there are three ways you can getting two heads.This situation is fairly simple. We need to be able to count the number of ways for more complicated situatio

16、ns.The Binomial DistributionCounCounting TechniquesRule 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)Counting TechniquesRule of CoCounting TechniquesRule of Combi

17、nationsHow many possible 3 scoop combinations could you create at an ice cream parlor if you have 31 flavors to select from and no flavor can be used more than once in the 3 scoops?The total choices is n = 31, and we select X = 3.Counting TechniquesRule of CoP(X=x|n,) = probability of x events of in

18、terest in n trials, with the probability of an “event of interest” being for each trial x = number of “events of interest” in sample, (x = 0, 1, 2, ., n) n = sample size (number of trials or observations) = probability of “event of interest” P(X=x |n,)nx!nx(1-)xnx!()!=-Example: Flip a coin four time

19、s, let x = # heads:n = 4 = 0.51 - = (1 - 0.5) = 0.5X = 0, 1, 2, 3, 4Binomial Distribution FormulaP(X=x|n,) = probability of x What is the probability of one success in five observations if the probability of an event of interest is 0.1? x = 1, n = 5, and = 0.1Example: Calculating a Binomial Probabil

20、ityWhat is the probability of oneThe Binomial DistributionExampleSuppose the probability of purchasing a defective computer is 0.02. What is the probability of purchasing 2 defective computers in a group of 10? x = 2, n = 10, and = 0.02The Binomial DistributionExamThe Binomial Distribution Shape 0.2

21、.4.6012345xP(X=x|5, 0.1).2.4.6012345xP(X=x|5, 0.5)0The shape of the binomial distribution depends on the values of and nHere, n = 5 and = .1Here, n = 5 and = .5The Binomial Distribution ShapThe Binomial Distribution Using Binomial Tables (Available On Line)n = 10 x=.20=.25=.30=.35=.40=.45=.500123456

22、789100.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.00000.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.00000.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.00000.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.00000.00600.04030.12090.21500.25080

23、.20070.11150.04250.01060.00160.00010.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.00030.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010109876543210=.80=.75=.70=.65=.60=.55=.50 xExamples: n = 10, = 0.35, x = 3: P(X = 3|10, 0.35) = 0.2522n = 10, = 0.75, x = 8: P(X = 8|1

24、0, 0.75) = 0.2816The Binomial Distribution UsinBinomial Distribution CharacteristicsMeanVariance and Standard DeviationWheren = sample size = probability of the event of interest for any trial(1 ) = probability of no event of interest for any trialBinomial Distribution CharacteThe Binomial Distribut

25、ionCharacteristics 012345xP(X=x|5, 0.1).2.4.6012345xP(X=x|5, 0.5)0ExamplesThe Binomial DistributionCharBoth Excel & Minitab Can Be Used To Calculate The Binomial DistributionBoth Excel & Minitab Can Be UsThe Poisson DistributionDefinitionsYou use the Poisson distribution when you are interest

26、ed in the number of times an event occurs in a given area of opportunity.An area of opportunity is a continuous unit or interval of time, volume, or such area in which more than one occurrence of an event can occur. The number of scratches in a cars paintThe number of mosquito bites on a personThe n

27、umber of computer crashes in a day The Poisson DistributionDefinThe 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

28、 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 num

29、ber of events per unit is (lambda)The Poisson DistributionApply Poisson Distribution Formulawhere:x = number of events in an area of opportunity = expected number of eventse = base of the natural logarithm system (2.71828.)Poisson Distribution FormulawhPoisson Distribution CharacteristicsMeanVarianc

30、e and Standard Deviationwhere = expected number of eventsPoisson Distribution CharacterUsing Poisson Tables (Available On Line)X00.400.500.600.700.800.90012345670.90480.09050.00450.00020.00000.00000.00000.00000.81870.16370.01640.00110.00010.00000.00000.00000.74080.22220.03330.00330.00030.00000.00000.00000.67030.26810.05360.00720.00070.00010.00000.00000.60650.3033