商业应用统计英文版ch05

1、AppliedBusinessStatistics,7thed.byKenBlack,Chapter5DiscreteDistributions,LearningObjectives,Distinguishbetweendiscreterandomvariablesandcontinuousrandomvariables.Knowhowtodeterminethemeanandvarianceofadiscretedistribution.Identifythetypeofstatisticalexperimentsthatcanbedescribedbythebinomialdistribu

2、tion,andknowhowtocalculateprobabilitiesbasedonthebinomialdistribution.,Discretevs.ContinuousDistributions,RandomVariable-avariablewhichcontainstheoutcomesofachanceexperimentDiscreteRandomVariableArandomvariablethatonlytakesondistinctvaluesex:Numberofheadson10flips,Numberofdefectiveitemsinarandomsamp

3、leof100,Numberoftimesyoucheckyourwatchduringclass,etc.ContinuousRandomVariableArandomvariablethattakesoninfinitevaluesbyincreasingprecision.Foreachtwovalues,therealwaysexistsavalidvalueinbetweenthem.ex:Timeuntilabulbgoesout,height,etc.,DescribingaDistribution,Adistributioncanbedescribedbyconstructin

4、gagraphofthedistributionMeasuresofcentraltendencyandvariabilitycanbeappliedtodistributions,DescribingaDiscreteDistribution,MeanofdiscretedistributionisthelongrunaverageIftheprocessisrepeatedlongenough,theaverageoftheoutcomeswillapproachthelongrunaverage(mean)Meanofadiscretedistribution=(Xi*P(Xi)wher

5、eisthelongrunaverage,Xi=theithoutcome,DescribingaDiscreteDistribution,Varianceofadiscretedistributionisobtainedinamannersimilartorawdata,summingthesquareddeviationsfromthemeanandweightingthembyP(Xi)(ratherthandividingbyn):Var(Xi)=(Xim)2*P(Xi)StandardDeviationiscomputedbytakingthesquarerootofthevaria

6、nce,DiscreteDistribution-Example,Anexecutiveisconsideringout-of-townbusinesstravelforagivenFriday.Atleastonecrisiscouldoccuronthedaythattheexecutiveisgone.Thedistributiononthefollowingslidecontainsthenumberofcrisesthatcouldoccurduringthedaytheexecutiveisgoneandtheprobabilitythateachnumberwilloccur.F

7、orexample,thereisa.37probabilitythatnocrisiswilloccur,a.31probabilityofonecrisis,andsoon.,DiscreteDistribution-Example,MeanandStandardDeviationofaDiscreteDistribution,RequirementsforaDiscreteProbabilityFunction-Examples,VALID,NOTVALID,Eachprobabilitymustbebetween0and1Thesumofallprobabilitiesmustbeeq

8、ualto1.,NOTVALID,BinomialDistribution,ThebinomialdistributionisadiscretedistributionwhereXisthenumberof“successes”andthefollowingfourconditionsaremet:TherearentrialsThentrialsareindependentofeachotherTheoutcomeisdichotomousonlytwooutcomespossibleTheprobabilityof“success”isconstantExample,10coinflips

9、,X=#ofheadsX=thenumberof“successes”andwesayXfollowsaBinomialdistributionwithntrialandP(success)=pIfthedatafollowabinomialdistribution,thenwecansummarizeP(Xi)forallvaluesofXi=1,nthroughthebinomialprobabilitydistributionformula,BinomialDistribution,ProbabilityfunctionMeanvalueVarianceandStandardDeviat

10、ion,AccordingtotheU.S.CensusBureau,approximately6%ofallworkersinJackson,Mississippi,areunemployed.InconductingarandomtelephonesurveyinJackson,whatistheprobabilityofgettingtwoorfewerunemployedworkersinasampleof20?,BinomialDistribution:DemonstrationProblem5.3,BinomialDistribution:DemonstrationProblem5

11、.3,Inthisexample,6%areunemployed=pThesamplesizeis20=n94%areemployed=qXisthenumberofsuccessesdesiredWhatistheprobabilityofgetting2orfewerunemployedworkersinthesampleof20?=P(X2)Thehardpartofthisproblemisidentifyingp,n,andxemphasizethiswhenstudyingtheproblems.,BinomialDistribution:DemonstrationProblem5

12、.3,BinomialDistributionTable:DemonstrationProblem5.3,BinomialDistribution:DemonstrationProblem5.3,Whatarethemeanandstandarddeviationofthisdistribution?,ExcelsBinomialFunction,XP(X=x)00.00000010.00000020.00000030.00000140.00000650.00003760.00019970.00085880.00305190.009040100.022500110.047273120.0840

13、41130.126420140.160533150.171236160.152209170.111421180.066027190.030890200.010983210.002789220.000451230.000035,Binomialwithn=23andp=0.64,MinitabsBinomialFunction,PoissonDistribution,ThePoissondistributionfocusesonlyonthenumberofdiscreteoccurrencesoversomeintervalorcontinuumPoissondoesnothaveagiven

14、numberoftrials(n)asabinomialexperimentdoesOccurrencesareindependentofotheroccurrencesOccurrencesoccuroveraninterval,PoissonDistribution,IfPoissondistributionisstudiedoveralongperiodoftime,alongrunaveragecanbedeterminedTheaverageisdenotedbylambda()EachPoissondistributioncontainsalambdavaluefromwhicht

15、heprobabilitiesaredeterminedAPoissondistributioncanbedescribedbyalone,PoissonDistribution,Probabilityfunction,PoissonDistribution:DemonstrationProblem5.7,Bankcustomersarriverandomlyonweekdayafternoonsatanaverageof3.2customersevery4minutes.Whatistheprobabilityofhavingmorethan7customersina4-minuteinte

16、rvalonaweekdayafternoon?,ExcelsPoissonFunction,MinitabsPoissonFunction,XP(X=x)00.14956910.28418020.26997130.17098240.08121650.03086260.00977370.00265380.00063090.0001330.000025,Poissonwithmean=1.9,PoissonDistribution:DemonstrationProblem5.7-Solution,=3.2customers/4minutesWewanttocalculateP(X7custome

17、rs/4minutes)Theproblemcaneitherbesolvedas:P(X7)=P(X=8)+P(X=9)+,orP(X7)=1P(X7)=1P(X=7)+P(X=6)+P(X=0)TheanswercanbeobtaineddirectlyorthroughsoftwareTheansweryougetis1.7%ofthetime.Bankofficerscouldusetheseresultstohelpthemmakestaffingdecisions.,HypergeometricDistribution,Samplingwithoutreplacementfroma

18、finitepopulationThenumberofobjectsinthepopulationisdenotedN.Eachtrialhasexactlytwopossibleoutcomes,successandfailure.TrialsarenotindependentXisthenumberofsuccessesinthentrialsThebinomialisanacceptableapproximation,ifn5%N.Otherwiseitisnot.,HypergeometricDistribution,ProbabilityfunctionNispopulationsizenissamplesizeAisnumberofsuccessesinpopulationxisnumberofsuccessesinsampleMeanValueVarianceandstandardde