1.mean=85.55k,geometricmean=41.25k,median=30k,mode=20k,harmonicmean=29.2683
Themeanisnotfairbecausethesalarydistributionishighlyskewed.
Themedianwilldoabetterjob.
2.CV=s/µ
𝜇 uk=6.4𝜎 uk=2.43CVuk=0.38
𝜇 us=79𝜎 us=17.81CVus=0.23
3.a)
8
7
6
5
4
3
2
1
0
1
2
3
4
b)Thescoreforthesecondtestisclosesttoanormaldistribution
Quantiles of Input Sample
Quantiles of Input Sample
QQ Plot of Sample Data versus Standard Normal
QQ Plot of Sample Data versus Standard Normal
10
5
8
6
4
3
10
5
0
-5
-4
-2
0
2
Standard Normal Quantiles
4
Quantiles of Input Sample
2
-2
0
2
4
-4
-2
0
2
4
Standard Normal Quantiles
Standard Normal Quantiles
QQ Plot of Sample Data versus Standard Normal
QQ Plot of Sample Data versus Standard Normal
15
6
Quantiles of Input Sample
2
-4
4
4
2
0
-2
-4
-2
0
2
Standard Normal Quantiles
4
c)
Testscrore1:skewness=0.3118,kurtosis=2.4264
Testscrore2:skewness=0.3158,kurtosis=3.1810
Testscrore3:skewness=-0.2721,kurtosis=1.6045
Testscrore4:skewness=-0.1019,kurtosis=1.6639
4.a)
E(X)=sum(Profit.*Probabilities)=1*0.1+1.5*0.2+2*0.4+4*0.2+10*.1=3
Var(X)=sum((sum(Profit.*Probabilities)-Profit).^2.*Probabilities)
=(1-3)^2*0.1+(1.5-3)^2*0.2+(2-3)^2*0.4+(4-3)^2*0.2+(10-3)^2*.1=6.35
b)E(Y)=0.9*E(X)-0.2=2.5
Var(Y)=0.9^2*Var(E)=5.14
c)
m(t)=0.1*et+0.2*e1.5t+0.4*e2t+0.2*e4t+0.1*e10t
1stderivative:m’(t)=0.1*et+0.3*e1.5t+0.8*e2t+0.8*e4t+e10t
2ndderivative:m’(t)=0.1*et+0.45*e1.5t+1.6*e2t+3.2*e4t+10*e10t
E(X)=m’(0)=0.1+0.3+0.8+0.8+1=3
Var(X)=E(X2)-[E(X)]2=m’’(0)-[m’(0)]2=15.35-3^2=6.35
5.
a)P(X=0,Y<=1)=1/6+1/4=5/12
b)MarginalPMFofX
P(X=0)=1/6+1/4+1/8=13/24
P(X=1)=1/8+1/6+1/6=11/24
MarginalPMFofY
P(Y=0)=1/6+1/8=7/24
P(Y=1)=1/4+1/6=5/12
P(Y=2)=1/8+1/6=7/24
c)P(Y=1|X=0)=P(Y=1,X=0)/P(X=0)=(1/4)/(13/24)=6/13
d)Cov(X,Y)=E(XY)–E(X)*E(Y)=1*1*1/6+1*2*1/6–11/24*(5/12+2*7/24)=1/2-11/24=1/24
6.
X=distancebetweenrecharges,
X~N(3000,50^2)Z=(X-3000)/50~N(0,1)
P(X>3100)=P(Z>2)=0.0228
7.
𝑝=1238/1528=81.02%
SE=sqrt(0.8102*(1-0.8102)/1528)=1%
81.02%±1.96*1%
95%ConfidenceInterval79.06–82.98%
8.a)twosamplet-test
b)𝑥' = 120𝑠' - = 457.45
𝑥- = 101𝑠- - = 425.33
s1ands2areclose,sowecanusepooled2-samplettest
𝑠3 =
𝑛' − 1 𝑠' - + 𝑛- − 1 𝑠- = 21.4
𝑛' + 𝑛- − 1
𝑡=
𝑥' − 𝑥-
= 1.87
1
1
𝑠3
+
𝑛' 𝑛p-value=0.039forone-tailedtest
Werejectthenullhypothesisatthesignificantlevelof0.05.
c)
= (
'.:;<='.>: ?@
?AB
) =14.62≈ 15