aianjie

 

The correct answer is D

In stratified random sampling, a researcher classifies a population into smaller groups based on one or more characteristics, takes a simple random sample from each subgroup based on the size of the subgroup, and pools the results.

aianjie

 

15、The probability that an observation lies within three standard deviations of the mean for any probability distribution is at least:

A) 75%.

B) 54%.

C) 89%.

D) 99%.

aianjie

 

The correct answer is C

One can use Chebyshev's Inequality to calculate this proportion. 1 ? (1/32) = 89%.

aianjie

 

14、The probability of returns less than ?10%, assuming a normal distribution with expected return of 6.5% and standard deviation of 10%, is:

A) less than 2.5%.

B) approximately 10%.

C) not defined with only this information.

D) approximately 5%.

aianjie

 

The correct answer is D

?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%.

aianjie

 

12、The distribution of annual returns for a bond portfolio is approximately normal with an expected value of $120 million and a standard deviation of $20 million. Which of the following is closest to the probability that the value of the portfolio one year from today will be between $110 million and $170 million?

A) 74%.

B) 58%.

C) 42%.

D) 66%.

aianjie

 

The correct answer is D

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%.

aianjie

 

13、The annual returns for a portfolio are normally distributed with an expected value of ￡50 million and a standard deviation of ￡25 million. What is the probability that the value of the portfolio one year from today will be between ￡91.13 million and ￡108.25 million?

A) 0.025.

B) 0.040.

C) 0.075.

D) 0.090.

aianjie

 

The correct answer is B

Calculate the standardized variable corresponding to the outcomes. Z1 = (91.13 ? 50) / 25 = 1.645, and Z2 = (108.25 ? 50) / 25 = 2.33. The cumulative normal distribution gives cumulative probabilities of F(1.645) = 0.95 and F(2.33) = 0.99. The probability that the outcome will lie between Z1 and Z2 is the difference: 0.99 ? 0.95 = 0.04. Note that even though you will not have a z-table on the exam, the probability values for 1.645 and 2.33 are commonly used values that you should have memorized.

aianjie

 

The correct answer is A

Note that if you memorize the basic intervals for a normal distribution, you do not need a normal distribution table to answer this question. -27.3% represents a loss of -35.3% from the mean return (-27.3 – 8.0), which is (-35.3/18) = -1.96 standard deviations to the left of the mean. For a normal distribution, we know that approximately 95 percent of all observations lie with +/- 1.96 standard deviations of the mean, so the probability that the return is between -27.3% and 8.0% must be (95%/2) = 47.5%. A return of 37.7 percent represents a gain of (37.7 - 8.0) = 29.7% from the mean return, which is (29.7/18) = 1.65 standard deviations to the right of the mean. For a normal distribution, we know that approximately 90 percent of all observations lie with +/- 1.65 standard deviations of the mean, so the probability that the return is between 8.0% and 37.7% must be (90%/2) = 45%. Therefore the probability that the return is between -27.3 percent and 37.7 percent = (47.5% + 45%) = 92.5%.