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　　1、Answer:A

 

　　Solution:E(X)=np=5(0.60)=3.0.Var(X)=np(1-p)=5(0.60)(0.40)=1.2.

 

　　2.Answer:C

 

　　Solution:

 

　　Even though normal curves have different sizes,they all have identical shape characteristics.The kurtosis for all normal distributions is three;an excess kurtosis of three would indicate a leptokurtic distribution.The other choices are true.

 

　　3、Answer:A

 

　　Solution:Use the t-statistic atα/2 and n-1 degrees of freedom when the population variance is unknown,regardless of sample size.

 

　　4、Answer:D

 

　　Solution:Since the probability of default for each bond is independent,∩BB∩CCC)=P(A)×P(BB)×P(CCC)=0.04×0.12×0.30=0.00144=0.14%.

 

　　5、Answer:C

 

　　Solution:Calculating z-values,z1=(110−120)/20=−0.5.z2=(170−120)/20=2.5.Using the z-table,P(−0.5)=(1−0.6915)=0.3085.P(2.5)=0.9938.P(−0.5<X<2.5)=0.9938−0.3085=0.6853.Note that on the exam,you will not have access to z-tables,so you would have to reason this one out using the normal distribution approximations.

 

　　You know that the probability within+/−1 standard deviation of the mean is approximately 68%,meaning that the area within−1 standard deviation of the mean is 34%.Since−0.5 is half of−1,the area under−0.5 to 0 standard deviations under the mean is approximately 34%/2=17%.The probability under+/−2 standard deviations of the mean is approximately 99%.The value$170 is mid way between+2 and+3 standard deviations,so the probability between these values must be(99%/2)=2%.The value from 0 to 2.5 standard deviations must therefore be(99%/2)−(2%/2)=48.5%.Adding these values gives us an approximate probability of(48.5%+17%)=65.5%.

 

　　6、Answer:C

 

　　Solution:Z=($60,000-$47,500)/$12,500=1.0

 

　　From the table of areas under the normal curve,84.13%of observations lie to the left of+1 standard deviation of the mean.So,100%-84.13%=15.87%with incomes of$60,000 or more.

 

　　7、Answer:C

 

　　Solution:Independent random variables have a covariance of zero.Hence,the variance of the sum of two variables will be the sum of the variables’variances.

 

　　8、Answer:A

 

　　Solution:Since we are only concerned with values that are below a 10%return this is a 1 tailed test to the left of the mean on the normal curve.Withμ=19 andσ=4.5,P(X≥10)=P(X≥μ−2σ)therefore looking up-2 on the cumulative Z table gives us a value of 0.0228,meaning that(1−0.0228)=97.72%of the area under the normal curve is above a Z score of-2.Since the Z score of-2 corresponds with the lower level 10%rate of return of the portfolio this means that there is a 97.72%probability that the portfolio will earn at least a 10%rate of return.

 

　　9、Answer:D

 

　　Solution:−10 is 16.5%below the mean return,(16.5/10)=1.65 standard deviations,which leaves approximately 5%of the possible outcomes in the left tail below−10%.

 

　　10、Answer:B

 

　　Solution:The mean is 10%and the standard deviation is 5%.You want to know the probability of a return 5%or better.10%-5%=5%,so 5%is one standard deviation less than the mean.Thirty-four percent of the observations are between the mean and one standard deviation on the down side.Fifty percent of the observations are greater than the mean.So the probability of a return 5%or higher is 34%+50%=84%.