cyber21

The mean and standard deviation of four portfolios are listed below in percentage terms. Using Roy's safety first criteria and a threshold of 3%, select the respective mean and standard deviation that corresponds to the optimal portfolio.

A) 5; 3.

B) 14; 20.

C) 19; 28.

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According to the safety-first criterion, the optimal portfolio is the one that has has the largest value for the SFRatio (mean − threshold) / Standard Deviation. A mean = 5 and Standard Deviation = 3 yields the largest SFRatio from the choices given: (5 − 3) / 3 = 0.67.

cyber21

Three portfolios with normally distributed returns are available to an investor who wants to minimize the probability that the portfolio return will be less than 5%. The risk and return characteristics of these portfolios are shown in the following table:

Portfolio	Expected return	Standard deviation
Epps	6%	4%
Flake	7%	9%
Grant	10%	15%

Based on Roy’s safety-first criterion, which portfolio should the investor select?

A) Epps.

B) Grant.

C) Flake.

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Roy’s safety-first ratios for the three portfolios:
Epps = (6 - 5) / 4 = 0.25
Flake = ( 7 - 5) / 9 = 0.222
Grant = (10 - 5) / 15 = 0.33
The portfolio with the largest safety-first ratio has the lowest probability of a return less than 5%. The investor should select the Grant portfolio.

cyber21

The average amount of snow that falls during January in Frostbite Falls is normally distributed with a mean of 35 inches and a standard deviation of 5 inches. The probability that the snowfall amount in January of next year will be between 40 inches and 26.75 inches is closest to:

A) 79%.

B) 87%.

C) 68%.

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To calculate this answer, we will use the properties of the standard normal distribution. First, we will calculate the Z-value for the upper and lower points and then we will determine the approximate probability covering that range.  Note: This question is an example of why it is important to memorize the general properties of the normal distribution.
Z = (observation – population mean) / standard deviation

• Z26.75 = (26.75 – 35) / 5 = -1.65. (1.65 standard deviations to the left of the mean)
• Z40 = (40 – 35) / 5 = 1.0 (1 standard deviation to the right of the mean)

Using the general approximations of the normal distribution:

• 68% of the observations fall within ± one standard deviation of the mean. So, 34% of the area falls between 0 and +1 standard deviation from the mean.

• 90% of the observations fall within ± 1.65 standard deviations of the mean. So, 45% of the area falls between 0 and +1.65 standard deviations from the mean.

Here, we have 34% to the right of the mean and 45% to the left of the mean, for a total of 79%.

cyber21

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) 0.3745.

B) 0.5000.

C) 0.0001.

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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.

cyber21

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) 0.2119.

C) 0.3773.

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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.

cyber21

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) 15.96.

B) 13.34.

C) 41.67.

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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%.

cyber21

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) 15.87%.

C) 5.71%.

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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.

cyber21

Standardizing a normally distributed random variable requires the:

A) mean, variance and skewness.

B) natural logarithm of X.

C) mean and the standard deviation.

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All that is necessary is to know the mean and the variance. Subtracting the mean from the random variable and dividing the difference by the standard deviation standardizes the variable.

cyber21

Which of the following represents the mean, standard deviation, and variance of a standard normal distribution?

A) 0, 1, 1.

B) 1, 1, 1.

C) 1, 2, 4.

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By definition, for the standard normal distribution, the mean, standard deviation, and variance are 0, 1, 1.

cyber21

The owner of a bowling alley determined that the average weight for a bowling ball is 12 pounds with a standard deviation of 1.5 pounds. A ball denoted “heavy” should be one of the top 2% based on weight. Assuming the weights of bowling balls are normally distributed, at what weight (in pounds) should the “heavy” designation be used?

A) 14.22 pounds.

B) 15.08 pounds.

C) 14.00 pounds.

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The first step is to determine the z-score that corresponds to the top 2%. Since we are only concerned with the top 2%, we only consider the right hand of the normal distribution. Looking on the cumulative table for 0.9800 (or close to it) we find a z-score of 2.05. To answer the question, we need to use the normal distribution given: 98 percentile = sample mean + (z-score)(standard deviation) = 12 + 2.05(1.5) = 15.08.