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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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.

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Monthly sales of hot water heaters are approximately normally distributed with a mean of 21 and a standard deviation of 5. What is the probabilility of selling 12 hot water heaters or less next month?

A) 1.80%.

B) 3.59%.

C) 96.41%.

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Z = (12 – 21) / 5 = -1.8

From the cumulative z-table, the probability of being more than 1.8 standard deviations below the mean, probability x < -1.8, is 3.59%.

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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.

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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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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.