商务统计学英文版教学课件第5章.ppt

1、Discrete Probability DistributionsChapter 5ObjectivesIn this chapter,you learn:nThe properties of a probability distribution.nTo compute the expected value and variance of a probability distribution.nTo compute probabilities from binomial,and Poisson distributions.nTo use the binomial,and Poisson di

2、stributions to solve business problemsDefinitionsnDiscrete variables produce outcomes that come from a counting process(e.g.number of classes you are taking).nContinuous variables produce outcomes that come from a measurement(e.g.your annual salary,or your weight).Types Of VariablesCh.5Ch.6Ch.5Ch.6T

3、ypes Of VariablesDiscrete VariableContinuousVariableCh.5Ch.6Discrete VariablesnCan only assume a countable number of valuesExamples:nRoll a die twiceLet X be the number of times 4 occurs(then X could be 0,1,or 2 times)nToss a coin 5 times.Let X be the number of heads (then X =0,1,2,3,4,or 5)Probabil

4、ity Distribution For A Discrete VariablenA probability distribution for a discrete variable is a mutually exclusive listing of all possible numerical outcomes for that variable and a probability of occurrence associated with each outcome.Interruptions Per Day In Computer NetworkProbability00.3510.25

5、20.2030.1040.0550.05Probability Distributions Are Often Represented GraphicallyP(X)0.40.30.20.1012345XDiscrete Variables Expected Value(Measuring Center)n Expected Value(or mean)of a discrete variable(Weighted Average)NiiixXPx1)(E(X)Interruptions Per Day In Computer Network(xi)ProbabilityP(X=xi)xiP(

6、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)=0.3040.05(4)(0.05)=0.2050.05(5)(0.05)=0.251.00=E(X)=1.40nVariance of a discrete variablenStandard Deviation of a discrete variablewhere:E(X)=Expected value of the discrete variable X xi =the ith outcome of XP(X=xi)=Probabil

7、ity of the ith occurrence of XDiscrete Variables:Measuring Dispersion N1ii2i)xP(XE(X)x2 N1ii2i)xP(XE(X)x2Discrete 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 (

8、0.16)(0.25)=0.04020.20(2 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,=1.4283 N1ii2i)xP(XE(X)xProbability DistributionsContinuous Probability DistributionsBinomialPoissonProbability Distributio

9、nsDiscrete Probability DistributionsNormalCh.5Ch.6Binomial Probability DistributionnA fixed number of observations,nne.g.,15 tosses of a coin;ten light bulbs taken from a warehousenEach observation is categorized as to whether or not the“event of interest”occurredne.g.,head or tail in each toss of a

10、 coin;defective or not defective light bulbnSince these two categories are mutually exclusive and collectively exhaustivenWhen the probability of the event of interest is represented as,then the probability of the event of interest not occurring is 1-nConstant probability for the event of interest o

11、ccurring()for each observationnProbability of getting a tail is the same each time we toss the coinBinomial Probability Distribution(continued)nObservations are independentnThe outcome of one observation does not affect the outcome of the othernTwo sampling methods deliver independencenInfinite popu

12、lation without replacementnFinite population with replacementPossible Applications for the Binomial DistributionnA manufacturing plant labels items as either defective or acceptablenA firm bidding for contracts will either get a contract or notnA marketing research firm receives survey responses of“

13、yes I will buy”or“no I will not”nNew job applicants either accept the offer or reject itThe Binomial DistributionCounting TechniquesnSuppose the event of interest is obtaining 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?nPossible ways:

14、HHT,HTH,THH,so there are three ways you can getting two heads.nThis situation is fairly simple.We need to be able to count the number of ways for more complicated situations.Counting TechniquesRule of CombinationsnThe number of combinations of selecting x objects out of n objects isx)!(nx!n!Cxn wher

15、e:n!=(n)(n-1)(n-2).(2)(1)x!=(X)(X-1)(X-2).(2)(1)0!=1 (by definition)Counting TechniquesRule of CombinationsnHow 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?nThe total choices

16、 is n=31,and we select X=3.4,4952953128!12328!2930313!28!31!3)!(313!31!C331 P(X=x|n,)=probability of x events of interest 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 observatio

17、ns)=probability of“event of interest”P(X=x|n,)nx!nx(1-)xnx!()!Example:Flip a coin four times,let x=#heads:n=4=0.51-=(1-0.5)=0.5X=0,1,2,3,4Binomial Distribution FormulaWhat 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

18、:Calculating a Binomial Probability0.32805.9)(5)(0.1)(00.1)(1(0.1)1)!(51!5!)(1x)!(nx!n!5,0.1)|1P(X4151xnxThe 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 =0.02.

19、01531)(.8508)(45)(.0004.02)(1(.02)2)!(102!10!)(1x)!(nx!n!0.02)10,|2P(X2102xnx The Binomial Distribution Shape 0.2.4.6012345xP(X=x|5,0.1).2.4.6012345xP(X=x|5,0.5)0nThe shape of the binomial distribution depends on the values of and nnHere,n=5 and =.1nHere,n=5 and =.5The Binomial Distribution Using Bi

20、nomial Tables(Available On Line)n=10 x=.20=.25=.30=.35=.40=.45=.500123456789100.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.00000.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.00000.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.00000.01350.07250.17570.2522

21、0.23770.15360.06890.02120.00430.00050.00000.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.00010.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.00030.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010109876543210=.80=.75=.70=.65=.60=.55=.50 xExamples:n=10,=0.3

22、5,x=3:P(X=3|10,0.35)=0.2522n=10,=0.75,x=8:P(X=8|10,0.75)=0.2816Binomial Distribution CharacteristicsnMeannVariance and Standard DeviationnE(X)-(1n2 )-(1n Wheren=sample size=probability of the event of interest for any trial(1 )=probability of no event of interest for any trialThe Binomial Distributi

23、onCharacteristics 0.2.4.6012345xP(X=x|5,0.1).2.4.6012345xP(X=x|5,0.5)00.5(5)(.1)n 0.6708.1)(5)(.1)(1)-(1n2.5(5)(.5)n 1.118.5)(5)(.5)(1)-(1nExamplesBoth Excel&Minitab Can Be Used To Calculate The Binomial DistributionThe Poisson DistributionDefinitionsnYou use the Poisson distribution when you are in

24、terested in the number of times an event occurs in a given area of opportunity.nAn 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.nThe number of scratches in a cars paintnThe number of mosquito bites on a perso

25、nnThe number of computer crashes in a day The Poisson DistributionnApply the Poisson Distribution when:nYou wish to count the number of times an event occurs in a given area of opportunitynThe probability that an event occurs in one area of opportunity is the same for all areas of opportunity nThe n

26、umber of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunitynThe probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smallernThe average number of events pe

27、r unit is (lambda)Poisson Distribution Formulawhere:x=number of events in an area of opportunity=expected number of eventse=base of the natural logarithm system(2.71828.)!)|(xexXPx Poisson Distribution CharacteristicsnMeannVariance and Standard Deviation 2 where =expected number of eventsUsing Poiss

28、on Tables(Available On Line)X 0.100.200.300.400.500.600.700.800.90012345670.90480.09050.00450.00020.00000.00000.00000.00000.81870.16370.01640.00110.00010.00000.00000.00000.74080.22220.03330.00330.00030.00000.00000.00000.67030.26810.05360.00720.00070.00010.00000.00000.60650.30330.07580.01260.00160.00

29、020.00000.00000.54880.32930.09880.01980.00300.00040.00000.00000.49660.34760.12170.02840.00500.00070.00010.00000.44930.35950.14380.03830.00770.00120.00020.00000.40660.36590.16470.04940.01110.00200.00030.0000Example:Find P(X=2|=0.50)0.07582!(0.50)ex!e0.50)|2P(X20.50 x Excel&Minitab Can Be Used For The

30、 Poisson DistributionGraph of Poisson ProbabilitiesX =0.50012345670.60650.30330.07580.01260.00160.00020.00000.0000P(X=2|=0.50)=0.0758 Graphically:=0.50 Poisson Distribution ShapenThe shape of the Poisson Distribution depends on the parameter :=0.50=3.00Chapter SummaryIn this chapter we covered:nThe properties of a probability distribution.nTo compute the expected value and variance of a probability distribution.nTo compute probabilities from binomial,and Poisson distributions.nTo use the binomial,and Poisson distributions to solve business problems