商务与经济统计习题答案（第8版，中文版）SBE8-SM06

6-1Chapter6ContinuousProbabilityDistributionsLearningObjectives1.Understandthedifferencebetweenhowprobabilitiesarecomputedfordiscreteandcontinuousrandomvariables.2.Knowhowtocomputeprobabilityvaluesforacontinuousuniformprobabilitydistributionandbeabletocomputetheexpectedvalueandvarianceforsuchadistribution.3.Beabletocomputeprobabilitiesusinganormalprobabilitydistribution.Understandtheroleofthestandardnormaldistributioninthisprocess.4.Beabletocomputeprobabilitiesusinganexponentialprobabilitydistribution.5.UnderstandtherelationshipbetweenthePoissonandexponentialprobabilitydistributions.Chapter66-2Solutions:1.a.b.P(x=1.25)=0.Theprobabilityofanysinglepointiszerosincetheareaunderthecurveaboveanysinglepointiszero.c.P(1.0x1.25)=2(.25)=.50d.P(1.20x1.5)=2(.30)=.602.a..15.10.0510203040f(x)x0b.P(x15)=.10(5)=.50c.P(12x18)=.10(6)=.60d.1020()152Exe.2(2010)Var()8.3312xContinuousProbabilityDistributions6-33.a.b.P(x130)=(1/20)(130-120)=0.50c.P(x135)=(1/20)(140-135)=0.25d.120140()1302Exminutes4.a.b.P(.25x.75)=1(.50)=.50c.P(x.30)=1(.30)=.30d.P(x.60)=1(.40)=.405.a.LengthofInterval=261.2-238.9=22.31for238.9261.2()22.30elsewherexfxb.Note:1/22.3=0.045P(x250)=(0.045)(250-238.9)=0.4995Chapter66-4Almosthalfdrivetheballlessthan250yards.c.P(x255)=(0.045)(261.2-255)=0.279d.P(245x260)=(0.045)(260-245)=0.675e.P(x250)=1-P(x250)=1-0.4995=0.5005Theprobabilityofanyonedrivingit250yardsormoreis0.5005.With60players,theexpectednumberdrivingit250yardsormoreis(60)(0.5005)=30.03.Rounding,Iwouldexpect30ofthesewomentodrivetheball250yardsormore.6.a.P(12x12.05)=.05(8)=.40b.P(x12.02)=.08(8)=.64c.(11.98)(12.02).005(8).04.64.08(8)PxPxTherefore,theprobabilityis.04+.64=.687.a.P(10,000x12,000)=2000(1/5000)=.40Theprobabilityyourcompetitorwillbidlowerthanyou,andyougetthebid,is.40.b.P(10,000x14,000)=4000(1/5000)=.80c.Abidof$15,000givesaprobabilityof1ofgettingtheproperty.d.Yes,thebidthatmaximizesexpectedprofitis$13,000.Theprobabilityofgettingthepropertywithabidof$13,000isP(10,000x13,000)=3000(1/5000)=.60.Theprobabilityofnotgettingthepropertywithabidof$13,000is.40.Theprofityouwillmakeifyougetthepropertywithabidof$13,000is$3000=$16,000-13,000.Soyourexpectedprofitwithabidof$13,000isEP($13,000)=.6($3000)+.4(0)=$1800.Ifyoubid$15,000theprobabilityofgettingthebidis1,buttheprofitifyoudogetthebidisonly$1000=$16,000-15,000.Soyourexpectedprofitwithabidof$15,000isEP($15,000)=1($1000)+0(0)=$1,000.ContinuousProbabilityDistributions6-58.9.a.b..6826since45and55arewithinplusorminus1standarddeviationfromthemeanof50.c..9544since40and60arewithinplusorminus2standarddeviationsfromthemeanof50.10.a..3413b..4332c..4772d..493850=5s3540455560650-3-2-1+1+2+3100=10s708090110120130Chapter66-611.a..3413Theseprobabilityvaluesarereaddirectlyfromthetableofareasforthestandardb..4332normalprobabilitydistribution.SeeTable1inAppendixB.c..4772d..4938e..498612.a..2967b..4418c..5000-.1700=.3300d..0910+.5000=.5910e..3849+.5000=.8849f..5000-.2612=.238813.a..6879-.0239=.6640b..8888-.6985=.1903c..9599-.8508=.109114.a.Usingthetableofareasforthestandardnormalprobabilitydistribution,theareaof.4750correspondstoz=1.96.b.Usingthetable,theareaof.2291correspondstoz=.61.c.Lookinthetableforanareaof.5000-.1314=.3686.Thisprovidesz=1.12.d.Lookinthetableforanareaof.6700-.5000=.1700.Thisprovidesz=.44.15.a.Lookinthetableforanareaof.5000-.2119=.2881.Sincethevalueweareseekingisbelowthemean,thezvaluemustbenegative.Thus,foranareaof.2881,z=-.80.b.Lookinthetableforanareaof.9030/2=.4515;z=1.66.c.Lookinthetableforanareaof.2052/2=.1026;z=.26.d.Lookinthetableforanareaof.4948;z=2.56.e.Lookinthetableforanareaof.1915.Sincethevalueweareseekingisbelowthemean,thezvaluemustbenegative.Thus,z=-.50.16.a.Lookinthetableforanareaof.5000-.0100=.4900.Theareavalueinthetableclosestto.4900providesthevaluez=2.33.b.Lookinthetableforanareaof.5000-.0250=.4750.Thiscorrespondstoz=1.96.ContinuousProbabilityDistributions6-7c.Lookinthetableforanareaof.5000-.0500=.4500.Since.4500isexactlyhalfwaybetween.4495(z=1.64)and.4505(z=1.65),weselectz=1.645.However,z=1.64orz=1.65arealsoacceptableanswers.d.Lookinthetableforanareaof.5000-.1000=.4000.Theareavalueinthetableclosestto.4000providesthevaluez=1.28.17.Convertmeantoinches:=69a.Atx=72z=72-693=1P(x72)=0.5000+0.3413=0.8413P(x72)=1-0.8413=0.1587b.Atx=60z=60-693=–3P(x60)=0.5000+0.4986=0.9986P(x60)=1-0.9986=0.0014c.Atx=70z=70-693=0.33P(x70)=0.5000+0.1293=0.6293Atx=66z=66-693=–1P(x66)=0.5000-0.3413=0.1587P(66x70)=P(x70)-P(x66)=0.6293-0.1587=0.4706d.P(x72)=1-P(x72)=1-0.1587=0.841318.a.FindP(x60)Atx=60z=60-4916=0.69P(x60)=0.5000+0.2549=0.7549P(x60)=1-P(x60)=0.2451b.FindP(x30)Atx=30z=30-4916=–1.19P(x30)=0.5000-0.3830=0.1170c.Findz-scoresothatP(zz-score)=0.10z-score=1.28cutsoff10%inuppertailChapter66-8Now,solveforcorrespondingvalueofx.1284916.xx=49+(16)(1.28)=69.48So,10%ofsubscribersspend69.48minutesormorereadingTheWallStreetJournal.19.Wehave=3.5ands=.8.a.5.03.51.88.8zP(x5.0)=P(z1.88)=1-P(z1.88)=1-.9699=.0301Therainfallexceeds5inchesin3.01%oftheApr