This exercise uses the normal probability density function and requires the use of either technology or...

This exercise uses the normal probability density function and requires the use of either technology or a table of values of the standard normal distribution.

The cash operating expenses of the regional phone companies during the first half of 1994 were distributed about a mean of $29.88 per access line per month, with a standard deviation of $2.65. Company A's operating expenses were $27.00 per access line per month. Assuming a normal distribution of operating expenses, estimate the percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean. (Round your answer to two decimal places.)

Expert Solution

The difference between the mean operating expenses of all regional phone companies and mean operating expenses of Company A = $29.88 - $27.00 = $2.88

Let X be the random variable denoting the operatind expense of regional phone companies.

X ~ N(29.88, 2.65) i.e. Z = (X - 29.88)/2.65 ~ N(0,1)

The probability that regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean = P[|X - 29.88| 2.88] = P[|X - 29.88|/2.65 2.88/2.65] = P[|Z| 1.0868] = P[-1.0868 Z 1.0868] = (1.0868) - (-1.0868) = 0.8614 - 0.1386 = 0.7228

[(.) is the cdf of N(0,1)]

Hence, the required percentage is 72.28%.

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orchestra answered 2 years ago

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