商务统计学（第7版）英文 课件5 Discrete Probability Distributions、6 The Normal Distribution

DiscreteProbabilityDistributionsChapter5ObjectivesInthischapter,youlearn:

Thepropertiesofaprobabilitydistribution.Tocomputetheexpectedvalueandvarianceofaprobabilitydistribution.Tocomputeprobabilitiesfrombinomial,andPoissondistributions.Tousethebinomial,andPoissondistributionstosolvebusinessproblemsDefinitionsDiscretevariablesproduceoutcomesthatcomefromacountingprocess(e.g.numberofclassesyouaretaking).Continuousvariablesproduceoutcomesthatcomefromameasurement(e.g.yourannualsalary,oryourweight).TypesOfVariablesCh.5Ch.6Ch.5Ch.6TypesOfVariablesDiscreteVariableContinuousVariableCh.5Ch.6DiscreteVariablesCanonlyassumeacountablenumberofvaluesExamples:Rolladietwice LetXbethenumberoftimes4occurs (thenXcouldbe0,1,or2times)Tossacoin5times. LetXbethenumberofheads(thenX=0,1,2,3,4,or5)ProbabilityDistributionForADiscreteVariableAprobabilitydistributionforadiscretevariableisamutuallyexclusivelistingofallpossiblenumericaloutcomesforthatvariableandaprobabilityofoccurrenceassociatedwitheachoutcome.InterruptionsPerDayInComputerNetworkProbability00.3510.2520.2030.1040.0550.05ProbabilityDistributionsAreOftenRepresentedGraphicallyP(X)0.40.30.20.10 1 2 3 4 5 XDiscreteVariablesExpectedValue(MeasuringCenter)

ExpectedValue(ormean)ofadiscretevariable(WeightedAverage)InterruptionsPerDayInComputerNetwork(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)=0.3040.05(4)(0.05)=0.2050.05(5)(0.05)=0.251.00μ=E(X)=1.40VarianceofadiscretevariableStandardDeviationofadiscretevariable where:

E(X)=ExpectedvalueofthediscretevariableX xi=theithoutcomeofX P(X=xi)=ProbabilityoftheithoccurrenceofXDiscreteVariables:

MeasuringDispersionDiscreteVariables:

MeasuringDispersion(continued)InterruptionsPerDayInComputerNetwork(xi)ProbabilityP(X=xi)[xi–E(X)]2[xi–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(2–1.4)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)=0.648σ2=2.04,σ=1.4283ProbabilityDistributionsContinuous

ProbabilityDistributionsBinomialPoissonProbabilityDistributionsDiscrete

ProbabilityDistributionsNormalCh.5Ch.6BinomialProbabilityDistributionAfixednumberofobservations,ne.g.,15tossesofacoin;tenlightbulbstakenfromawarehouseEachobservationiscategorizedastowhetherornotthe“eventofinterest”occurrede.g.,headortailineachtossofacoin;defectiveornotdefectivelightbulbSincethesetwocategoriesaremutuallyexclusiveandcollectivelyexhaustiveWhentheprobabilityoftheeventofinterestisrepresentedasπ,thentheprobabilityoftheeventofinterestnotoccurringis1-πConstantprobabilityfortheeventofinterestoccurring(π)foreachobservationProbabilityofgettingatailisthesameeachtimewetossthecoinBinomialProbabilityDistribution(continued)ObservationsareindependentTheoutcomeofoneobservationdoesnotaffecttheoutcomeoftheotherTwosamplingmethodsdeliverindependenceInfinitepopulationwithoutreplacementFinitepopulationwithreplacementPossibleApplicationsfortheBinomialDistributionAmanufacturingplantlabelsitemsaseitherdefectiveoracceptableAfirmbiddingforcontractswilleithergetacontractornotAmarketingresearchfirmreceivessurveyresponsesof“yesIwillbuy”or“noIwillnot”NewjobapplicantseitheraccepttheofferorrejectitTheBinomialDistribution

CountingTechniquesSupposetheeventofinterestisobtainingheadsonthetossofafaircoin.Youaretotossthecointhreetimes.Inhowmanywayscanyougettwoheads?Possibleways:HHT,HTH,THH,sotherearethreewaysyoucangettingtwoheads.Thissituationisfairlysimple.Weneedtobeabletocountthenumberofwaysformorecomplicatedsituations.CountingTechniques

RuleofCombinationsThenumberofcombinationsofselectingxobjectsoutofnobjectsiswhere: n!=(n)(n-1)(n-2)...(2)(1) x!=(X)(X-1)(X-2)...(2)(1) 0!=1(bydefinition)CountingTechniques

RuleofCombinationsHowmanypossible3scoopcombinationscouldyoucreateatanicecreamparlorifyouhave31flavorstoselectfromandnoflavorcanbeusedmorethanonceinthe3scoops?Thetotalchoicesisn=31,andweselectX=3.P(X=x|n,π)=probabilityofxeventsofinterest inntrials,withtheprobabilityofan “eventofinterest”beingπ

for eachtrialx=numberof“eventsofinterest”insample,(x=0,1,2,...,n)n=samplesize(numberoftrials orobservations)

π=probabilityof“eventofinterest”P(X=x|n,π)nx!nxπ(1-π)xnx!()!=--Example:Flipacoinfourtimes,letx=#heads:n=4π=0.51-π=(1-0.5)=0.5X=0,1,2,3,4BinomialDistributionFormulaWhatistheprobabilityofonesuccessinfiveobservationsiftheprobabilityofaneventofinterestis0.1? x=1,n=5,andπ=0.1Example:

CalculatingaBinomialProbabilityTheBinomialDistribution

ExampleSupposetheprobabilityofpurchasingadefectivecomputeris0.02.Whatistheprobabilityofpurchasing2defectivecomputersinagroupof10? x=2,n=10,andπ=0.02TheBinomialDistributionShape0.2.4.6012345xP(X=x|5,0.1).2.4.6012345xP(X=x|5,0.5)0TheshapeofthebinomialdistributiondependsonthevaluesofπandnHere,n=5andπ=.1Here,n=5andπ=.5TheBinomialDistributionUsingBinomialTables(AvailableOnLine)n=10x…π=.20π=.25π=.30π=.35π=.40π=.45π=.50012345678910……………………………0.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.00000.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.00000.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.00000.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.00000.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.00010.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.00030.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010109876543210…π=.80π=.75π=.70π=.65π=.60π=.55π=.50xExamples:n=10,π=0.35,x=3:P(X=3|10,0.35)=0.2522n=10,π=0.75,x=8:P(X=8|10,0.75)=0.2816BinomialDistributionCharacteristicsMeanVarianceandStandardDeviationWhere n=samplesize

π=probabilityoftheeventofinterestforanytrial (1–π)=probabilityofnoeventofinterestforanytrialTheBinomialDistribution

Characteristics0.2.4.6012345xP(X=x|5,0.1).2.4.6012345xP(X=x|5,0.5)0ExamplesBothExcel&MinitabCanBeUsedToCalculateTheBinomialDistributionThePoissonDistribution

DefinitionsYouusethePoissondistributionwhenyouareinterestedinthenumberoftimesaneventoccursinagivenareaofopportunity.Anareaofopportunityisacontinuousunitorintervaloftime,volume,orsuchareainwhichmorethanoneoccurrenceofaneventcanoccur.Thenumberofscratchesinacar’spaintThenumberofmosquitobitesonapersonThenumberofcomputercrashesinadayThePoissonDistributionApplythePoissonDistributionwhen:YouwishtocountthenumberoftimesaneventoccursinagivenareaofopportunityTheprobabilitythataneventoccursinoneareaofopportunityisthesameforallareasofopportunity

ThenumberofeventsthatoccurinoneareaofopportunityisindependentofthenumberofeventsthatoccurintheotherareasofopportunityTheprobabilitythattwoormoreeventsoccurinanareaofopportunityapproacheszeroastheareaofopportunitybecomessmallerTheaveragenumberofeventsperunitis

(lambda)PoissonDistributionFormulawhere: x=numberofeventsinanareaofopportunity

=expectednumberofevents e=baseofthenaturallogarithmsystem(2.71828...)

PoissonDistributionCharacteristicsMeanVarianceandStandardDeviationwhere

=expectednumberofeventsUsingPoissonTables(AvailableOnLine)X

0.100.200.300.400.500.600.700.800.90012345670.90480.09050.00450.00020.00000.00000.00000.00000.81870.16370.01640.00110.00010.00000.00000.00000.74080.22220.03330.00330.00030.00000.00000.00000.67030.26810.05360.00720.00070.00010.00000.00000.60650.30330.07580.01260.00160.00020.00000.00000.54880.32930.09880.01980.00300.00040.00000.00000.49660.34760.12170.02840.00500.00070.00010.00000.44930.35950.14380.03830.00770.00120.00020.00000.40660.36590.16470.04940.01110.00200.00030.0000Example:FindP(X=2|=0.50)Excel&MinitabCanBeUsedForThePoissonDistributionGraphofPoissonProbabilitiesX=0.50012345670.60650.30330.07580.01260.00160.00020.00000.0000P(X=2|

=0.50)=0.0758

Graphically:=0.50PoissonDistributionShapeTheshapeofthePoissonDistributiondependsontheparameter:

=0.50

=3.00ChapterSummaryInthischapterwecovered:

Thepropertiesofaprobabilitydistribution.Tocomputetheexpectedvalueandvarianceofaprobabilitydistribution.Tocomputeprobabilitiesfrombinomial,andPoissondistributions.Tousethebinomial,andPoissondistributionstosolvebusinessproblemsTheNormalDistributionChapter6ObjectivesInthischapter,youlearn:

TocomputeprobabilitiesfromthenormaldistributionHowtousethenormaldistributiontosolvebusinessproblemsTousethenormalprobabilityplottodeterminewhetherasetofdataisapproximatelynormallydistributedContinuousProbabilityDistributionsAcontinuousvariableisavariablethatcanassumeanyvalueonacontinuum(canassumeanuncountablenumberofvalues)thicknessofanitemtimerequiredtocompleteatasktemperatureofasolutionheight,ininchesThesecanpotentiallytakeonanyvaluedependingonlyontheabilitytopreciselyandaccuratelymeasureTheNormalDistribution

‘BellShaped’

Symmetrical

Mean,MedianandMode areEqualLocationisdeterminedbythemean,μSpreadisdeterminedbythestandarddeviation,σ

Therandomvariablehasaninfinitetheoreticalrange:+

to

Mean=Median=ModeXf(X)μσTheNormalDistribution

DensityFunctionTheformulaforthenormalprobabilitydensityfunctionisWhere e=themathematicalconstantapproximatedby2.71828

π=themathematicalconstantapproximatedby3.14159

μ=thepopulationmean

σ=thepopulationstandarddeviation X=anyvalueofthecontinuousvariableByvaryingtheparametersμandσ,weobtaindifferentnormaldistributionsABCAandBhavethesamemeanbutdifferentstandarddeviations.BandChavedifferentmeansanddifferentstandarddeviations.TheNormalDistributionShapeXf(X)μσChanging

μshiftsthedistributionleftorright.Changingσincreasesordecreasesthespread.TheStandardizedNormalAnynormaldistribution(withanymeanandstandarddeviationcombination)canbetransformedintothestandardizednormal

distribution(Z)TocomputenormalprobabilitiesneedtotransformXunitsintoZunitsThestandardizednormaldistribution(Z)hasameanof0andastandarddeviationof1TranslationtotheStandardizedNormalDistributionTranslatefromXtothestandardizednormal(the“Z”distribution)bysubtractingthemeanofXanddividingbyitsstandarddeviation:TheZdistributionalwayshasmean=0andstandarddeviation=1TheStandardizedNormalProbabilityDensityFunctionTheformulaforthestandardizednormalprobabilitydensityfunctionisWhere e=themathematicalconstantapproximatedby2.71828

π=themathematicalconstantapproximatedby3.14159 Z=anyvalueofthestandardizednormaldistributionTheStandardized

NormalDistributionAlsoknownasthe“Z”distributionMeanis0StandardDeviationis1Zf(Z)01ValuesabovethemeanhavepositiveZ-values.ValuesbelowthemeanhavenegativeZ-values.ExampleIfXisdistributednormallywithmeanof$100andstandarddeviationof$50,theZvalueforX=$200

isThissaysthatX=$200istwostandarddeviations(2incrementsof$50units)abovethemeanof$100.ComparingXandZunitsNotethattheshapeofthedistributionisthesame,onlythescalehaschanged.Wecanexpresstheproblemintheoriginalunits(Xindollars)orinstandardizedunits(Z)Z$1002.00$200$X(μ=$100,σ=$50)(μ=0,σ=1)FindingNormalProbabilities

ProbabilityismeasuredbytheareaunderthecurveabXf(X)PaXb()≤≤PaXb()<<=(Notethattheprobabilityofanyindividualvalueiszero)Probabilityas

AreaUndertheCurveThetotalareaunderthecurveis1.0,andthecurveissymmetric,sohalfisabovethemean,halfisbelowf(X)Xμ0.50.5TheStandardizedNormalTableTheCumulativeStandardizedNormaltableinthetextbook(AppendixtableE.2)givestheprobabilitylessthanadesiredvalueofZ(i.e.,fromnegativeinfinitytoZ)Z02.000.9772Example:

P(Z<2.00)=0.9772TheStandardizedNormalTable

ThevaluewithinthetablegivestheprobabilityfromZ=

uptothedesiredZvalue.97722.0P(Z<2.00)=0.9772

TherowshowsthevalueofZtothefirstdecimalpoint

Thecolumn

givesthevalueofZtotheseconddecimalpoint2.0...(continued)Z0.000.010.02…0.00.1GeneralProcedureforFindingNormalProbabilities

DrawthenormalcurvefortheproblemintermsofXTranslateX-valuestoZ-valuesUsetheStandardizedNormalTableTofindP(a<X<b)whenXisdistributednormally:FindingNormalProbabilitiesLetXrepresentthetimeittakes(inseconds)todownloadanimagefilefromtheinternet.SupposeXisnormalwithameanof18.0secondsandastandarddeviationof5.0seconds.FindP(X<18.6)18.6X18.0LetXrepresentthetimeittakes,insecondstodownloadanimagefilefromtheinternet.SupposeXisnormalwithameanof18.0secondsandastandarddeviationof5.0seconds.FindP(X<18.6)Z0.120X18.618μ=18

σ=5μ

=0σ=1(continued)FindingNormalProbabilitiesP(X<18.6)P(Z<0.12)Z0.12Solution:FindingP(Z<0.12)0.5478StandardizedNormalProbabilityTable(Portion)0.00=P(Z<0.12)P(X<18.6)Z.00.010.0.5000.5040.5080.5398.54380.2.5793.5832.58710.3.6179.6217.6255.020.1.5478

FindingNormal

UpperTailProbabilitiesSupposeXisnormalwithmean18.0andstandarddeviation5.0.NowFindP(X>18.6)X18.618.0NowFindP(X>18.6)…(continued)Z0.120Z0.120.547801.0001.0-0.5478=0.4522P(X>18.6)=P(Z>0.12)=1.0-P(Z≤0.12)=1.0-0.5478=0.4522

FindingNormal

UpperTailProbabilitiesFindingaNormalProbabilityBetweenTwoValuesSupposeXisnormalwithmean18.0andstandarddeviation5.0.FindP(18<X<18.6)P(18<X<18.6)=P(0<Z<0.12)Z0.120X18.618CalculateZ-values:Z0.12Solution:FindingP(0<Z<0.12)0.04780.00=P(0<Z<0.12)P(18<X<18.6)=P(Z<0.12)–P(Z≤0)=0.5478-0.5000=0.04780.5000Z.00.010.0.5000.5040.5080.5398.54380.2.5793.5832.58710.3.6179.6217.6255.020.1.5478StandardizedNormalProbabilityTable(Portion)SupposeXisnormalwithmean18.0andstandarddeviation5.0.NowFindP(17.4<X<18)X17.418.0ProbabilitiesintheLowerTailProbabilitiesintheLowerTailNowFindP(17.4<X<18)…X17.418.0P(17.4<X<18)=P(-0.12<Z<0)=P(Z<0)–P(Z≤-0.12)=0.5000-0.4522=0.0478(continued)0.04780.4522Z-0.120TheNormaldistributionissymmetric,sothisprobabilityisthesameasP(0<Z<0.12)EmpiricalRule

μ±1σenclosesabout68.26%ofX’s

f(X)Xμμ+1σμ-1σWhatcanwesayaboutthedistributionofvaluesaroundthemean?Foranynormaldistribution:σσ68.26%TheEmpiricalRule

μ±2σ

coversabout95.44%ofX’s

μ±3σ

coversabout99.73%ofX’sxμ2σ2σxμ3σ3σ95.44%99.73%(continued)StepstofindtheXvalueforaknownprobability: 1.FindtheZvaluefortheknownprobability 2.ConverttoXunitsusingtheformula:GivenaNormalProbability

FindtheXValueFindingtheXvalueforaKnownProbabilityExample:LetXrepresentthetimeittakes(inseconds)todownloadanimagefilefromtheinternet.SupposeXisnormalwithmean18.0andstandarddeviation5.0FindXsuchthat20%ofdownloadtimesarelessthanX.X?18.00.2000Z?0(continued)FindtheZvaluefor

20%intheLowerTail20%areainthelowertailisconsistentwithaZvalueof-0.84Z.03-0.9.1762.1736.2033-0.7.2327.2296.04-0.8.2005StandardizedNormalProbabilityTable(Portion).05.1711.1977.2266…………X?18.00.2000Z-0.8401.FindtheZvaluefortheknownprobability 2.ConverttoXunitsusingtheformula:FindingtheXvalue

So20%ofthevaluesfromadistributionwithmean18.0andstandarddeviation5.0arelessthan13.80BothMinitab&ExcelCanBeUsedToFindNormalProbabilitiesFindP(X<9)whereXisnormalwithameanof7andastandarddeviationof2Excel MinitabEvaluatingNormalityNotallcontinuousdistributionsarenormalItisimportanttoevaluatehowwellthedatasetisapproximatedbyanormaldistribution.Normallydistributeddatashouldapproximatethetheoreticalnormaldistribution:Thenormaldistributionisbellshaped(symmetrical)wherethemeanisequaltothemedian.Theempiricalruleappliestothenormaldistribution.Theinterquartilerangeofanormaldistributionis1.33standarddeviations.EvaluatingNormalityComparingdatacharacteristicstotheoreticalpropertiesConstructchartsorgraphsForsmall-ormoderate-sizeddatasets,constructastem-and-leafdisplayoraboxplottocheckforsymmetryForlargedatasets,doesthehistogramorpolygonappearbell-shaped?ComputedescriptivesummarymeasuresDothemean,medianandmodehavesimilarvalues?Istheinterquartilerangeapproximately1.33σ?Istherangeapproximately6σ?(continued)EvaluatingNormalityComparingdatacharacteristicstotheoreticalpropertiesObservethedistributionofthedatasetDoapproximately2/3oftheobservationsliewithinmean±1standarddeviation?Doapproximately80%oftheobservationsliewithinmean±1.28standarddeviations?Doapproximately95%oftheobservationsliewithinmean±2standarddeviations?EvaluatenormalprobabilityplotIsthenormalprobabilityplotapproximatelylinear(i.e.astraightline)withpositiveslope?(continued)Constructing

ANormalProbabilityPlotNormalprobabilityplotArrangedataintoorderedarrayFindcorrespondingstandardizednormalquantilevalues(Z)Plotthepairsofpointswithobserveddatavalues(X)ontheverticalaxisandthestandardizednormalquantilevalues(Z)onthehorizontalaxisEvaluatetheplotforevidenceoflinearityAnormalprobabilityplotfordatafromanormaldistributionwillbeapproximatelylinear:306090-2-1012ZXTheNormalProbabilityPlot

InterpretationNormalProbabilityPlot

InterpretationLeft-SkewedRight-SkewedRectangular306090-2-1012ZX(continued)306090-2-1012ZX306090-2-1012ZXNonlinearplotsindicateadeviationfromnormalityEvaluatingNormality

AnExamp