Q1. If a stock's return is normally distributed with a mean of 16% and a standard deviation of 50%, what is the probability of a negative return in a given year?

A)   0.5000.

B)   0.0001.

C)   0.3745.

Q2. John Cupp, CFA, has several hundred clients. The values of the portfolios Cupp manages are approximately normally distributed with a mean of $800,000 and a standard deviation of $250,000. The probability of a randomly selected portfolio being in excess of $1,000,000 is:

A)   0.1057.

B)   0.3773.

C)   0.2119.

Q3. A food retailer has determined that the mean household income of her customers is $47,500 with a standard deviation of $12,500. She is trying to justify carrying a line of luxury food items that would appeal to households with incomes greater than $60,000. Based on her information and assuming that household incomes are normally distributed, what percentage of households in her customer base has incomes of $60,000 or more?

A)   2.50%.

B)   15.87%.

C)   5.00%.

Q4. Given a normally distributed population with a mean income of $40,000 and standard deviation of $7,500, what percentage of the population makes between $30,000 and $35,000?

A)   13.34.

B)   41.67.

C)   15.96.

Q5. A grant writer for a local school district is trying to justify an application for funding an after-school program for low-income families. Census information for the school district shows an average household income of $26,200 with a standard deviation of $8,960. Assuming that the household income is normally distributed, what is the percentage of households in the school district with incomes of less than $12,000?

A)   9.92%.

B)   5.71%.

C)   15.87%.

答案和详解如下：

Q1. If a stock's return is normally distributed with a mean of 16% and a standard deviation of 50%, what is the probability of a negative return in a given year?

A)   0.5000.

B)   0.0001.

C)   0.3745.

Correct answer is C)

The selected random value is standardized (its z-value is calculated) by subtracting the mean from the selected value and dividing by the standard deviation. This results in a z-value of (0 − 16) / 50 = -0.32. Changing the sign and looking up +0.32 in the z-value table yields 0.6255 as the probability that a random variable is to the right of the standardized value (i.e. more than zero). Accordingly, the probability of a random variable being to the left of the standardized value (i.e. less than zero) is 1 − 0.6255 = 0.3745.

Q2. John Cupp, CFA, has several hundred clients. The values of the portfolios Cupp manages are approximately normally distributed with a mean of $800,000 and a standard deviation of $250,000. The probability of a randomly selected portfolio being in excess of $1,000,000 is:

A)   0.1057.

B)   0.3773.

C)   0.2119.

Correct answer is C)

Although the number of clients is discrete, since there are several hundred of them, we can treat them as continuous. The selected random value is standardized (its z-value is calculated) by subtracting the mean from the selected value and dividing by the standard deviation. This results in a z-value of (1,000,000 – 800,000) / 250,000 = 0.8. Looking up 0.8 in the z-value table yields 0.7881 as the probability that a random variable is to the left of the standardized value (i.e.less than $1,000,000). Accordingly, the probability of a random variable being to the right of the standardized value (i.e. greater than $1,000,000) is 1 – 0.7881 = 0.2119.

Q3. A food retailer has determined that the mean household income of her customers is $47,500 with a standard deviation of $12,500. She is trying to justify carrying a line of luxury food items that would appeal to households with incomes greater than $60,000. Based on her information and assuming that household incomes are normally distributed, what percentage of households in her customer base has incomes of $60,000 or more?

A)   2.50%.

B)   15.87%.

C)   5.00%.

Correct answer is B)

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.

Q4. Given a normally distributed population with a mean income of $40,000 and standard deviation of $7,500, what percentage of the population makes between $30,000 and $35,000?

A)   13.34.

B)   41.67.

C)   15.96.

Correct answer is C)

The z-score for $30,000 = ($30,000 – $40,000) / $7,500 or –1.3333, which corresponds with 0.0918. The z-score for $35,000 = ($35,000 – $40,000) / $7,500 or –0.6667, which corresponds with 0.2514. The difference is 0.1596 or 15.96%.

Q5. A grant writer for a local school district is trying to justify an application for funding an after-school program for low-income families. Census information for the school district shows an average household income of $26,200 with a standard deviation of $8,960. Assuming that the household income is normally distributed, what is the percentage of households in the school district with incomes of less than $12,000?

A)   9.92%.

B)   5.71%.

C)   15.87%.

Correct answer is B)

Z = ($12,000 – $26,200) / $8,960 = –1.58.

From the table of areas under the standard normal curve, 5.71% of observations are more than 1.58 standard deviations below the mean.

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