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Reading 9: Common Probability Distributions - LOS h, (Part

Q1. 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)   117 to 183 inches.

B)   110 to 190 inches.

C)   137 to 163 inches.

Q2. 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.

Q3. 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)   3.59%.

B)   1.80%.

C)   96.41%.

答案和详解如下:

Q1. 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)   117 to 183 inches.

B)   110 to 190 inches.

C)   137 to 163 inches.

Correct answer is A)

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

Q2. 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.

Correct answer is C)

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.

Q3. 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)   3.59%.

B)   1.80%.

C)   96.41%.

Correct answer is A)

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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