2025年CFA《数量分析》模拟试题

2025年CFA《数量分析》模拟试题考试时间：______分钟总分：______分姓名：______试卷内容1.Consideradatasetconsistingofthefollowingsixvalues:5,9,11,3,7,8.a.Calculatethemean,median,andmodeofthisdataset.b.Determinethevarianceandstandarddeviationofthisdataset.c.Calculatethecoefficientofvariationforthisdataset.2.SupposearandomvariableXfollowsastandardnormaldistribution.FindtheprobabilitythatXislessthan1.5.Also,findthevalueofzsuchthattheprobabilityofXbeinggreaterthanzis0.05.3.Aportfoliomanagerclaimsthattheaveragereturnofhisportfoliooverthepastfiveyearsis12%,withastandarddeviationof3%.Assumethereturnsarenormallydistributed.Whatistheprobabilitythattheportfolioreturnwillbelessthan10%inanygivenyear?Whatisthe95%confidenceintervalforthetrueaveragereturnoftheportfolio?4.Youaregiventhefollowingregressionoutputforamodelpredictingstockreturns(Y)basedonmarketreturn(X):Coefficients:Intercept=0.05,Slope(X)=1.2StandardErrors:Intercept=0.03,Slope(X)=0.2t-statistics:Intercept=1.67,Slope(X)=6.0R-squared=0.64a.Interpretthecoefficientofthemarketreturn(X).b.Testthesignificanceofthemarketreturn(X)atthe5%level.c.Calculatethepredictedstockreturnwhenthemarketreturnis10%.5.Acompany'sdailysales(inthousandsofdollars)arebelievedtofollowanAR(1)model.Youaregiventhatthemeansalesare50thousanddollarsandthefirst-orderautoregressivecoefficient(ρ)is0.8.Whatistheexpectedvalueofsalestwodaysfromnow?6.Youareanalyzinganoptiononanon-dividendpayingstock.Thestockpriceis$100,thestrikepriceis$95,therisk-freerateis5%,andthetimetomaturityis6months(0.5years).UsingtheBlack-Scholes-Mertonmodel,calculatethetheoreticalpriceofacalloption.AssumetheoptionpricefollowsageometricBrownianmotionwithavolatilityof20%.7.Afinancialanalystistestingwhetherthemeanannualreturnoftwodifferentmutualfundsissignificantlydifferent.Hecollectsthefollowingdata:FundA:Samplesize=30,Samplemean=8%,Samplestandarddeviation=2%FundB:Samplesize=35,Samplemean=7.5%,Samplestandarddeviation=1.8%Conductatwo-samplet-test(assumingequalvariances)atthe1%significanceleveltodetermineifthereisasignificantdifferencebetweenthemeanreturnsofthetwofunds.8.YouaregivenaprobabilitydistributionforadiscreterandomvariableX:|X|P(X)||-----|------||0|0.2||1|0.3||2|0.4||3|0.1|a.Calculatetheexpectedvalue(mean)andvarianceofX.b.FindtheprobabilitythatXisgreaterthanitsexpectedvalue.9.Considertworandomvariables,XandY,withthefollowingcharacteristics:E(X)=10,E(Y)=15,Var(X)=4,Var(Y)=9,andCov(X,Y)=-2.a.CalculatetheexpectedvalueoftherandomvariableZ=2X-3Y.b.DeterminethevarianceofZ.c.FindthecorrelationcoefficientbetweenXandY.10.Aresearcherwantstotestifanewdrugiseffectiveinloweringbloodpressure.Sheconductsastudywith50patients.Themeansystolicbloodpressurebeforethetreatmentis140mmHgwithastandarddeviationof15mmHg.Afterthetreatment,themeansystolicbloodpressureis135mmHgwithastandarddeviationof14mmHg.Usingasignificancelevelof0.05,performahypothesistesttodetermineifthetreatmenthasastatisticallysignificanteffectonloweringsystolicbloodpressure.试卷答案1.a.Mean=(5+9+11+3+7+8)/6=7.67.Median=(7+8)/2=7.5.Mode=11.b.Variance=[(5-7.67)²+(9-7.67)²+(11-7.67)²+(3-7.67)²+(7-7.67)²+(8-7.67)²]/6=9.44.StandardDeviation=sqrt(9.44)=3.07.c.CoefficientofVariation=StandardDeviation/Mean=3.07/7.67=0.40.2.a.P(X<1.5)=NORM.DIST(1.5,0,1,TRUE)=0.9332.b.P(X>z)=0.05=>P(X<=z)=0.95.z=NORM.INV(0.95,0,1)=1.645.3.a.Z=(10-12)/3=-0.67.P(X<10)=P(Z<-0.67)=NORM.DIST(-0.67,0,1,TRUE)=0.2514.b.95%CI:Mean±(tcritical*SD/sqrt(n)).Assumingdf=4,tcritical≈2.776.CI:12±(2.776*3/sqrt(5))=[9.72,14.62].4.a.Interpretation:Forevery1%increaseinthemarketreturn,thepredictedstockreturnisexpectedtoincreaseby1.2%.b.H0:Slope=0.H1:Slope≠0.t=Slope/SE=1.2/0.2=6.0.p-value=TDIST(6.0,df=(n-2),TRUE)≈0.0002.p-value<0.05,RejectH0.Themarketreturnissignificantatthe5%level.c.PredictedReturn=Intercept+Slope*X=0.05+1.2*10=12.25.5.E[Sales_t]=ρ*E[Sales_(t-1)]+(1-ρ)*μ=0.8*Sales_(t-1)+0.2*50.Expectedsalestwodaysfromnow(Sales_2)=0.8*E[Sales_1]+0.2*50.First,findE[Sales_1]=0.8*50+0.2*50=50.Then,E[Sales_2]=0.8*50+0.2*50=50.6.d1=[ln(S/K)+(r+σ²/2)T]/(σ*sqrt(T))=[ln(100/95)+(0.05+0.2²/2)*0.5]/(0.2*sqrt(0.5))=[ln(1.0526)+(0.05+0.02)*0.5]/(0.2*0.7071)=[0.0507+0.035]/0.1414=0.0857/0.1414=0.6045.d2=d1-σ*sqrt(T)=0.6045-0.2*0.7071=0.6045-0.1414=0.4631.CallPrice=S*N(d1)-K*exp(-rT)*N(d2)=100*N(0.6045)-95*exp(-0.05*0.5)*N(0.4631).UsingN(0.6045)≈0.7276andN(0.4631)≈0.6788.CallPrice=100*0.7276-95*0.9753*0.6788=72.76-95*0.6623=72.76-62.9385=9.8215.Approximateto9.82.7.H0:μ_A=μ_B.H1:μ_A≠μ_B.Assumptions:Independentsamples,equalvariancesassumed.PooledVarianceSp^2=[(n_A-1)s_A^2+(n_B-1)s_B^2]/(n_A+n_B-2)=[(30-1)2^2+(35-1)1.8^2]/(30+35-2)=[29*4+34*3.24]/63=[116+110.16]/63=226.16/63≈3.58.PooledStandardDeviationSp=sqrt(3.58)≈1.89.StandardError(SE)=Sp*sqrt((1/n_A+1/n_B))=1.89*sqrt((1/30+1/35))=1.89*sqrt(0.0333+0.0286)=1.89*sqrt(0.0619)≈1.89*0.2488≈0.471.t=(Mean_A-Mean_B)/SE=(8-7.5)/0.471=0.5/0.471≈1.06.df=n_A+n_B-2=63.p-value=2*TDIST(1.06,df=63,TRUE)≈2*0.2918=0.5836.p-value(0.5836)>0.01(alpha).FailtorejectH0.Thereisnosignificantdifferenceatthe1%level.8.a.E(X)=Σx*P(x)=0*0.2+1*0.3+2*0.4+3*0.1=0+0.3+0.8+0.3=1.5.Var(X)=E(X^2)-[E(X)]^2.E(X^2)=Σx^2*P(x)=0^2*0.2+1^2*0.3+2^2*0.4+3^2*0.1=0+0.3+1.6+0.9=2.8.Var(X)=2.8-(1.5)^2=2.8-2.25=0.55.b.P(X>E(X))=P(X>1.5).P(X=2)+P(X=3)=0.4+0.1=0.5.9.a.E(Z)=E(2X-3Y)=2E(X)-3E(Y)=2(10)-3(15)=20-45=-25.b.Var(Z)=Var(2X-3Y)=4Var(X)+9Var(Y)-2*2*3*Cov(X,Y)=4(4)+9(9)-12(-2)=16+81+24=121.c.Cov(X,Y)/sqrt(Var(X)Var(Y))=-2/sqrt(4*9)=-2/sqrt(36)=-2/6=-1/3.Correlation=-0.3333.10.H0:μ_before=μ_after.H1:μ_befo