Question

An analyst at a management consultancy is studying the price of gasoline for a large U.5, metropolitan area, Suppose that the mean price of a galion of gasoline is 52.34, with a standard deviation of $0.20. a. Compute the probability that for a randomly-chosen gas station, the price of gasoline is within $0.05 from the population mean, Assume that the price of gasoline is normally distributed. Probability 15 (Round to two decimal places.) b. Compthe the probability that the average gasoline price for a sample of 70 gas stations is within 50.05 from the population mean. Probability is (Round to two decimal places.) c. Suppose that the sample size in b. decreases to 50 .

Answer 1

a. The probability that the price of gasoline is within $0.05 from the mean is 1.20. b. The probability that the average gasoline price for a sample of 70 gas stations is within $0.05 from the population mean is 1.96. c. If the sample size decreases to 50, the probability is 1.92.

Step-by-step explanation:

a. To compute the probability that the price of gasoline is within $0.05 from the mean, we need to find the area under the normal curve within that range. We can use the z-score formula to standardize the values. The z-score for $0.05 below the mean is (0.05 - 0) / 0.20 = 0.25. Looking up the z-table, the probability corresponding to a z-score of 0.25 is 0.5987. However, since we are interested in the area on both sides of the mean, we double this probability to get 0.5987 * 2 = 1.1974. Rounding to two decimal places, the probability is 1.20.

b. To compute the probability that the average gasoline price for a sample of 70 gas stations is within $0.05 from the population mean, we can use the same approach as in part a, but with a different standard deviation. The standard deviation of the sample mean can be calculated using the formula: standard deviation of the sample mean = standard deviation of the population / square root of the sample size. In this case, the standard deviation of the sample mean is 0.20 / sqrt(70) = 0.024. The z-score for $0.05 below the mean is (0.05 - 0) / 0.024 = 2.0833. Looking up the z-table, the probability corresponding to a z-score of 2.0833 is 0.9818. However, since we are interested in the area on both sides of the mean, we double this probability to get 0.9818 * 2 = 1.9636. Rounding to two decimal places, the probability is 1.96.

c. If the sample size decreases to 50, the standard deviation of the sample mean would be 0.20 / sqrt(50) = 0.0283. Using the z-score formula, the z-score for $0.05 below the mean would be (0.05 - 0) / 0.0283 = 1.7668. Looking up the z-table, the probability corresponding to a z-score of 1.7668 is 0.9618. However, since we are interested in the area on both sides of the mean, we double this probability to get 0.9618 * 2 = 1.9236. Rounding to two decimal places, the probability is 1.92.