A study of long-distance phone calls made from General Electric Corporate Headqu

Question

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 3.90 minutes and the standard deviation was 0.60 minutes.

What fraction of the calls last between 3.90 and 4.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

What fraction of the calls last more than 4.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

What fraction of the calls last between 4.50 and 5.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

What fraction of the calls last between 3.50 and 5.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 3 percent of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.)

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WorksheetDifficulty: 1 Basic

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 3.90 minutes and the standard deviation was 0.60 minutes.

Explanation / Answer

We know params of the normal distribution:

Mean = 3.90
Stdev = .60

a. P(3.9<X<4.5) = P(-1<Z<1) = .68
b. P(X>4.5) = P(Z>1) = .16
c. P(4.5<X<5.5) = P(1<Z<2.67) = .9962 -.8413 = .1549
d. P(3.5<X<5.5) =.9962-.2514 = .7448
e. P(X>x) =.03, Z=1.88, So, x = 1.88*.6+3.9 =5.028

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