Reading 9: Common Probability Distributions-LOS l 习题精选 2011, average, standard, interpret

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Session 3: Quantitative Methods: Application
Reading 9: Common Probability Distributions

LOS l: Define the standard normal distribution, explain how to standardize a random variable, and calculate and interpret probabilities using the standard normal distribution.

 

 

The average annual rainfall amount in Yucutat, Alaska, is normally distributed with a mean of 150 inches and a standard deviation of 20 inches. The 90% confidence interval for the annual rainfall in Yucutat is closest to:

A) 110 to 190 inches.

B) 137 to 163 inches.

C) 117 to 183 inches.

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The 90% confidence interval is μ ± 1.65 standard deviations. 150 ? 1.65(20) = 117 and 150 + 1.65(20) = 183.

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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%.

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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.

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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) 0.2119.

B) 0.1057.

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.

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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%.

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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.

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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) 15.96.

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%.

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

A) 1, 1, 1.

B) 1, 2, 4.

C) 0, 1, 1.

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

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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.

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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) 87%.

B) 79%.

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

• Z_26.75 = (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%.

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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.

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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.

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The standard normal distribution is most completely described as a:

A) distribution that exhibits zero skewness and no excess kurtosis.

B) symmetrical distribution with a mean equal to its median.

C) normal distribution with a mean of zero and a standard deviation of one.

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The standard normal distribution is defined as a normal distribution that has a mean of zero and a standard deviation of one. The other choices apply to any normal distribution.