商务与经济统计习题答案第8版中文版SBE8SM05.docx
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商务与经济统计习题答案第8版中文版SBE8SM05.docx
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商务与经济统计习题答案第8版中文版SBE8SM05
Chapter5
DiscreteProbabilityDistributions
LearningObjectives
1.Understandtheconceptsofarandomvariableandaprobabilitydistribution.
2.Beabletodistinguishbetweendiscreteandcontinuousrandomvariables.
3.Beabletocomputeandinterprettheexpectedvalue,variance,andstandarddeviationforadiscreterandomvariable.
4.Beabletocomputeandworkwithprobabilitiesinvolvingabinomialprobabilitydistribution.
5.BeabletocomputeandworkwithprobabilitiesinvolvingaPoissonprobabilitydistribution.
6.Knowwhenandhowtousethehypergeometricprobabilitydistribution.
Solutions:
1.a.Head,Head(H,H)
Head,Tail(H,T)
Tail,Head(T,H)
Tail,Tail(T,T)
b.x=numberofheadsontwocointosses
c.
Outcome
Valuesofx
(H,H)
2
(H,T)
1
(T,H)
1
(T,T)
0
d.Discrete.Itmayassume3values:
0,1,and2.
2.a.Letx=time(inminutes)toassembletheproduct.
b.Itmayassumeanypositivevalue:
x>0.
c.Continuous
3.LetY=positionisoffered
N=positionisnotoffered
a.S={(Y,Y,Y),(Y,Y,N),(Y,N,Y),(Y,N,N),(N,Y,Y),(N,Y,N),(N,N,Y),(N,N,N)}
b.LetN=numberofoffersmade;Nisadiscreterandomvariable.
c.
ExperimentalOutcome
(Y,Y,Y)
(Y,Y,N)
(Y,N,Y)
(Y,N,N)
(N,Y,Y)
(N,Y,N)
(N,N,Y)
(N,N,N)
ValueofN
3
2
2
1
2
1
1
0
4.x=0,1,2,...,12.
5.a.S={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}
b.
ExperimentalOutcome
(1,1)
(1,2)
(1,3)
(2,1)
(2,2)
(2,3)
NumberofStepsRequired
2
3
4
3
4
5
6.a.values:
0,1,2,...,20
discrete
b.values:
0,1,2,...
discrete
c.values:
0,1,2,...,50
discrete
d.values:
0≤x≤8
continuous
e.values:
x>0
continuous
7.a.f(x)≥0forallvaluesofx.
∑f(x)=1Therefore,itisaproperprobabilitydistribution.
b.Probabilityx=30isf(30)=.25
c.Probabilityx≤25isf(20)+f(25)=.20+.15=.35
d.Probabilityx>30isf(35)=.40
8.a.
x
f(x)
1
3/20=.15
2
5/20=.25
3
8/20=.40
4
4/20=.20
Total1.00
b.
c.f(x)≥0forx=1,2,3,4.
∑f(x)=1
9.a.
x
f(x)
1
15/462
=0.032
2
32/462
=0.069
3
84/462
=0.182
4
300/462
=0.650
5
31/462
=0.067
b.
c.Allf(x)≥0
∑f(x)=0.032+0.069+0.182+0.650+0.067=1.000
10.a.
x
f(x)
1
0.05
2
0.09
3
0.03
4
0.42
5
0.41
1.00
b.
x
f(x)
1
0.04
2
0.10
3
0.12
4
0.46
5
0.28
1.00
c.P(4or5)=f(4)+f(5)=0.42+0.41=0.83
d.Probabilityofverysatisfied:
0.28
e.Seniorexecutivesappeartobemoresatisfiedthanmiddlemanagers.83%ofseniorexecutiveshaveascoreof4or5with41%reportinga5.Only28%ofmiddlemanagersreportbeingverysatisfied.
11.a.
DurationofCall
x
f(x)
1
0.25
2
0.25
3
0.25
4
0.25
1.00
b.
c.f(x)≥0andf
(1)+f
(2)+f(3)+f(4)=0.25+0.25+0.25+0.25=1.00
d.f(3)=0.25
e.P(overtime)=f(3)+f(4)=0.25+0.25=0.50
12.a.Yes;f(x)≥0forallxand∑f(x)=.15+.20+.30+.25+.10=1
b.P(1200orless)=f(1000)+f(1100)+f(1200)
=.15+.20+.30
=.65
13.a.Yes,sincef(x)≥0forx=1,2,3and∑f(x)=f
(1)+f
(2)+f(3)=1/6+2/6+3/6=1
b.f
(2)=2/6=.333
c.f
(2)+f(3)=2/6+3/6=.833
14.a.f(200)=1-f(-100)-f(0)-f(50)-f(100)-f(150)
=1-.95=.05
ThisistheprobabilityMRAwillhavea$200,000profit.
b.P(Profit)=f(50)+f(100)+f(150)+f(200)
=.30+.25+.10+.05=.70
c.P(atleast100)=f(100)+f(150)+f(200)
=.25+.10+.05=.40
15.a.
x
f(x)
xf(x)
3
.25
.75
6
.50
3.00
9
.25
2.25
1.00
6.00
E(x)=μ=6.00
b.
x
x-μ
(x-μ)2
f(x)
(x-μ)2f(x)
3
-3
9
.25
2.25
6
0
0
.50
0.00
9
3
9
.25
2.25
4.50
Var(x)=σ2=4.50
c.σ=
=2.12
16.a.
y
f(y)
yf(y)
2
.20
.40
4
.30
1.20
7
.40
2.80
8
.10
.80
1.00
5.20
E(y)=μ=5.20
b.
y
y-μ
(y-μ)2
f(y)
(y-μ)2f(y)
2
-3.20
10.24
.20
2.048
4
-1.20
1.44
.30
.432
7
1.80
3.24
.40
1.296
8
2.80
7.84
.10
.784
4.560
17.a/b.
x
f(x)
xf(x)
x-μ
(x-μ)2
(x-μ)2f(x)
0
.10
.00
-2.45
6.0025
.600250
1
.15
.15
-1.45
2.1025
.315375
2
.30
.60
-.45
.2025
.060750
3
.20
.60
.55
.3025
.060500
4
.15
.60
1.55
2.4025
.360375
5
.10
.50
2.55
6.5025
.650250
2.45
2.047500
E(x)=μ=2.45
σ2=2.0475
σ=1.4309
18.a/b.
x
f(x)
xf(x)
x-μ
(x-μ)2
(x-μ)2f(x)
0
.01
0
-2.3
5.29
0.0529
1
.23
.23
-1.3
1.69
0.3887
2
.41
.82
-0.3
0.09
0.0369
3
.20
.60
0.7
0.49
0.098
4
.10
.40
1.7
2.89
0.289
5
.05
.25
2.7
7.29
0.3645
2.3
1.23
E(x)=2.3Var(x)=1.23
Theexpectedvalue,E(x)=2.3,oftheprobabilitydistributionisthesameasthatreportedinthe1997StatisticalAbstractoftheUnitedStates.
19.a.E(x)=∑xf(x)=0(.50)+2(.50)=1.00
b.E(x)=∑xf(x)=0(.61)+3(.39)=1.17
c.Theexpectedvalueofa3-pointshotishigher.So,iftheseprobabilitiesholdup,theteamwillmakemorepointsinthelongrunwiththe3-pointshot.
20.a.
x
f(x)
xf(x)
0
.90
0.00
400
.04
16.00
1000
.03
30.00
2000
.01
20.00
4000
.01
40.00
6000
.01
60.00
1.00
166.00
E(x)=166.Ifthecompanychargedapremiumof$166.00theywouldbreakeven.
b.
GaintoPolicyHolder
f(Gain)
(Gain)f(Gain)
-260.00
.90
-234.00
140.00
.04
5.60
740.00
.03
22.20
1,740.00
.01
17.40
3,740.00
.01
37.40
5,740.00
.01
57.40
-94.00
E(gain)=-94.00.Thepolicyholderismoreconcernedthatthebigaccidentwillbreakhimthanwiththeexpectedannuallossof$94.00.
21.a.E(x)=∑xf(x)=0.05
(1)+0.09
(2)+0.03(3)+0.42(4)+0.41(5)
=4.05
b.E(x)=∑xf(x)=0.04
(1)+0.10
(2)+0.12(3)+0.46(4)+0.28(5)
=3.84
c.Executives:
σ2=∑(x-μ)2f(x)=1.2475
MiddleManagers:
σ2=∑(x-μ)2f(x)=1.1344
d.Executives:
σ=1.1169
MiddleManagers:
σ=1.0651
e.Theseniorexecutiveshaveahigheraveragescore:
4.05vs.3.84forthemiddlemanagers.Theexecutivesalsohaveaslightlyhigherstandarddeviation.
22.a.E(x)=∑xf(x)
=300(.20)+400(.30)+500(.35)+600(.15)=445
Themonthlyorderquantityshouldbe445units.
b.Cost:
445@$50=$22,250
Revenue:
300@$70=21,000
$1,250Loss
23.a.Laptop:
E(x)=.47(0)+.45
(1)+.06
(2)+.02(3)=.63
Desktop:
E(x)=.06(0)+.56
(1)+.28
(2)+.10(3)=1.42
b.Laptop:
Var(x)=.47(-.63)2+.45(.37)2+.06(1.37)2+.02(2.37)2=.4731
Desktop:
Var(x)=.06(-1.42)2+.56(-.42)2+.28(.58)2+.10(1.58)2=.5636
c.Fromtheexpectedvaluesinpart(a),itisclearthatthetypicalsubscriberhasmoredesktopcomputersthanlaptops.Thereisnotmuchdifferenceinthevariancesforthetwotypesofcomputers.
24.a.MediumE(x)=∑xf(x)
=50(.20)+150(.50)+200(.30)=145
Large:
E(x)=∑xf(x)
=0(.20)+100(.50)+300(.30)=140
Mediumpreferred.
b.Medium
x
f(x)
x-μ
(x-μ)2
(x-μ)2f(x)
50
.20
-95
9025
1805.0
150
.50
5
25
12.5
200
.30
55
3025
907.5
σ2=2725.0
Large
y
f(y)
y-μ
(y-μ)2
(y-μ)2f(y)
0
.20
-140
19600
3920
100
.50
-40
1600
800
300
.30
160
25600
7680
σ2=12,400
Mediumpreferredduetolessvariance.
25.a.
b.
c.
d.
e.P(x≥1)=f
(1)+f
(2)=.48+.16=.64
f.E(x)=np=2(.4)=.8
Var(x)=np(1-p)=2(.4)(.6)=.48
σ=
=.6928
26.a.f(0)=.3487
b.f
(2)=.1937
c.P(x≤2)=f(0)+f
(1)+f
(2)=.3487+.3874+.1937=.9298
d.P(x≥1)=1-f(0)=1-.3487=.6513
e.E(x)=np=10(.1)=1
f.Var(x)=np(1-p)=10(.1)(.9)=.9
σ=
=.9487
27.a.f(12)=.1144
b.f(16)=.1304
c.P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20)
=.1304+.0716+.0278+.0068+.0008
=.2374
d.P(x≤15)=1-P(x≥16)=1-.2374=.7626
e.E(x)=np=20(.7)=14
f.Var(x)=np(1-p)=20(.7)(.3)=4.2
σ=
=2.0494
28.a.
b.P(atleast2)=1-f(0)-f
(1)
=
=1-.0905-.2673=.6422
c.
29.P(AtLeast5)=1-f(0)-f
(1)-f
(2)-f(3)-f(4)
=1-.0000-.0005-.0031-.0123-.0350
=.9491
30.a.Probabilityofadefectivepartbeingproducedmustbe.03foreachtrial;trialsmustbeindependent.
b.Let:
D=defective
G=notdefective
c.2outcomesresultinexactlyonedefect.
d.P(nodefects)=(.97)(.97)=.9409
P(1defect)=2(.03)(.97)=.0582
P(2defects)=(.03)(.03)=.0009
31.Binomialn=10andp=.05
a.Yes.Sincetheyareselectedrandomly,pisthesamefromtrialtotrialandthetrialsareindependent.
b.f
(2)=.0746
c.f(0)=.5987
d.P(Atleast1)=1-f(0)=1-.5987=.4013
32.a..90
b.P(atleast1)=f
(1)+f
(2)
Alternatively
c.P(atleast1)=1-f(0)
d.Yes;P(atleast1)becomesverycloseto1withmultiplesystemsandtheinabilitytodetectanattackwouldbecatastrophic.
33.a.Usingthebinomialformulaorthetableofbinomialprobabilitieswithp=.5andn=20,wefind:
x
f(x)
12
0.1201
13
0.0739
14
0.0370
15
0.0148
16
0.0046
17
0.0011
18
0.0002
19
0.0000
20
0.0000
0.2517
Theprobability12ormorewillsendrepresentativesis0.25
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