Chapter-4-Discrete-Probability-Distributions-Statistics-统计学-英文教材

ElementaryStatisticsLarsonFarber4x=numberofontimearrivalsx=numberofpointsscoredinagamex=numberofemployeesreachingsalesquotax=numberofcorrectanswers4DiscreteProbabilityDistributions1ProbabilityDistributionsSection4.12Arandomvariable,x

isthenumericaloutcomeofaprobabilityexperiment.

x=Thenumberofpeopleinacar.

x=Thegallonsofgasboughtinaweek.

x=Thetimeittakestodrivefromhometoschool

x=ThenumberoftripstoschoolyoumakeperweekRandomVariables3Arandomvariableisdiscreteifthenumberofpossibleoutcomesisfiniteorcountable.Discreterandomvariablesaredeterminedbyacount.Arandomvariableiscontinuousifitcantakeonanyvaluewithinaninterval.Thepossibleoutcomescannotbelisted.Continuousrandomvariablesaredeterminedbyameasure.TypesofRandomVariables4

x=Thenumberofpeopleinacar.

x=Thegallonsofgasboughtinaweek.

x=Thetimeittakestodrivefromhometoschool

x=ThenumberoftripstoschoolyoumakeperweekIdentifyeachrandomvariableasdiscreteorcontinuous.Discrete-youcountthenumberofpeopleinacar0,1,2,3…Possiblevaluescanbelisted.Continuous-youmeasurethegallonsofgas.Youcannotlistthepossiblevalues.Continuous-youmeasuretheamountoftime.Thepossiblevaluescannotbelisted.Discrete-youcountthenumberoftripsyoumake.Thepossiblenumberscanbelisted.TypesofRandomVariables5Adiscreteprobabilitydistributionlistseachpossiblevalueoftherandomvariable,togetherwithitsprobability.Asurveyasksasampleoffamilieshowmanyvehicleseachowns.numberofvehicles

Propertiesofaprobabilitydistribution

Eachprobabilitymustbebetween0and1,inclusive.

Thesumofallprobabilitiesis1.

DiscreteProbabilityDistributions6

Theheightofeachbarcorrespondstotheprobabilityofx.Whenthewidthofthebaris1,theareaofeachbarcorrespondstotheprobabilitythevalueofxwilloccur.0123ProbabilityHistogram7Mean,VarianceandStandardDeviationThevarianceofadiscreteprobabilitydistributionis:Thestandarddeviationofadiscreteprobabilitydistributionis:Themeanofadiscreteprobabilitydistributionis:8Mean(ExpectedValue)Multiplyeachvaluebyitsprobability.AddtheproductsTheexpectedvalue(themean)is1.763vehicles.Calculatethemean9CalculatetheVarianceandStandardDeviation

Thestandarddeviationis0.775vehicles.Themeanis1.763vehicles.μμμ2variance10BinomialDistributionsSection4.211Thereareafixednumberoftrials.(n)ThentrialsareindependentandrepeatedunderidenticalconditionsEachtrialhas2outcomes,S=SuccessorF=Failure.Theprobabilityofsuccessonasingletrialisp.P(S)=p

Theprobabilityoffailureisq.P(F)=q

where

p+q=1Thecentralproblemistofindtheprobabilityofx

successesoutofntrials.Wherex=0or1or2…n.

BinomialExperimentsCharacteristicsofaBinomialExperimentTherandomvariablex

isacountofthenumberofsuccessesinntrials.121.Whatisthe11thdigitafterthedecimalpointfortheirrationalnumbere?(a)2 (b)7 (c)4 (d)52.WhatwastheDowJonesAverageonFebruary27,1993?(a)3265 (b)3174 (c)3285 (d)33273.HowmanystudentsfromSriLankastudiedatU.S.universitiesfrom1990-91?(a)2320 (b)2350 (c)2360 (d)22404.Howmanykidneytransplantswereperformedin1991?(a)2946 (b)8972 (c)9943 (d)73415.HowmanywordsareintheAmericanHeritageDictionary?(a)60,000 (b)80,000 (c)75,000 (d)83,000Guesstheanswers13QuizResultsCountthenumberofcorrectanswers.Letthenumberofcorrectanswers=x.Whyisthisabinomialexperiment?Whatarethevaluesofn,pandq?Whatarethepossiblevaluesforx?Thecorrectanswerstothequizare:1.d2.a3.b4.c5.b14

Amultiplechoicetesthas8questionseachofwhichhas3choices,oneofwhichiscorrect.Youwanttoknowtheprobabilitythatyouguessexactly5questionscorrectly.Findn,p,q,andx.Adoctortellsyouthat80%ofthetimeacertaintypeofsurgeryissuccessful.Ifthissurgeryisperformed7times,findtheprobabilityexactly6surgerieswillbesuccessful.

Findn,p,q,andx.n=8p=1/3q=2/3x=5n=7p=0.80q=0.20x=6BinomialExperiments15Findtheprobabilityofgettingexactly3questionscorrectonthequiz.Writethefirst3correctandthelast2wrongasSSSFFP(SSSFF)=(.25)(.25)(.25)(.75)(.75)=(.25)3(.75)2=0.00879Sinceorderdoesnotmatter,youcouldgetanycombinationofthreecorrectoutoffivequestions.Listthesecombinations.SSSFF

SSFSF

SSFFS

SFFSS

SFSFSFFSSS

FSFSS

FSSFS

SFSSF

FFSSFEachofthese10wayshasaprobabilityof0.00879.P(x=3)=10(0.25)3(0.75)2=10(0.00879)=0.0879BinomialProbabilities16Findtheprobabilityofgettingexactly3questionscorrectonthequiz.Eachofthese10wayshasaprobabilityof0.00879.P(x=3)=10(0.25)3(0.75)2=10(0.00879)=0.0879Combinationofnvalues,choosingx

Thereareways.17BinomialProbabilitiesInabinomialexperiment,theprobabilityofexactlyxsuccessesinntrialsisUsetheformulatocalculatetheprobabilityofgettingnonecorrect,exactlyone,two,three,fourcorrectorall5correctonthequiz.P(3)=0.088P(4)=0.015P(5)=0.00118BinomialDistributionx P(x)0 0.2371 0.3962 0.2643 0.0884 0.0155 0.001BinomialHistogramx19Probabilities1.Whatistheprobabilityofansweringeither2or4questionscorrectly?2.Whatistheprobabilityofansweringatleast3questionscorrectly?3.Whatistheprobabilityofansweringatleastonequestioncorrectly?P(x=2orx=4)=0.264+0.015=0.279P(x

3)=P(x=3orx=4orx=5)=0.088+0.015+0.001=0.104P(x

1)=1-P(x=0)=1-0.237=0.763x P(x)0 0.2371 0.3962 0.2643 0.0884 0.0155 0.00120ParametersforaBinomialExperimentUsethebinomialformulastofindthemean,varianceandstandarddeviationforthedistributionofcorrectanswersonthequiz.Mean:Variance:Standarddeviation:21MoreDiscreteProbabilityDistributionsSection4.322TheGeometricDistributionAmarketingstudyhasfoundthattheprobabilitythatapersonwhoentersaparticularstorewillmakeapurchaseis0.30.Theprobabilitythefirstpurchasewillbemadebythefirstpersonwhoentersthestore0.30.ThatisP(1)=0.30Theprobabilitythefirstpurchasewillbemadebythesecondpersonwhoentersthestoreis(0.70)(0.30).SoP(2)=(0.70)(0.30)=0.21.Theprobabilitythefirstpurchasewillbemadebythethirdpersonwhoentersthestoreis(0.70)(0.70)(0.30).SoP(3)=(0.70)(0.70)(0.30)=0.147.Theprobabilitythefirstpurchasewillbemadebypersonnumberxis23TheGeometricDistributionAgeometricdistributionisadiscreteprobabilitydistributionoftherandomvariablexthatsatisfiesthefollowingconditions.Atrialisrepeateduntilasuccessoccurs2.Therepeatedtrialsareindependentofeachother.3.Theprobabilityofsuccesspisthesameforeachtrial.Theprobabilitythatthefirstsuccesswilloccurontrialnumberxiswhereq=1-p24ApplicationAcerealmakerplacesagamepieceinitsboxes.Theprobabilityofwinningaprizeisoneinfour.FindtheprobabilityyouWinyourfirstprizeonthe4thpurchaseb)

Winyourfirstprizeonyour2ndor3rdpurchasec)Donotwinyourfirstprizeinyourfirst4purchases.P(4)=(.75)3.(.25)=0.1055P(2)=(.75)1(.25)=0.1875andP(3)=(.75)2(.25)=0.1406SoP(2or3)=0.1875+0.1406=0.32811–(P(1)+P(2)+P(3)+P(4))1–(0.25+0.1875+0.140