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 4.20 minutes and the standard deviation was 0.40 minutes.

a. What fraction of the calls last between 4.20 and 4.90 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls=?

b. What fraction of the calls last more than 4.90 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls=?

c. What fraction of the calls last between 4.90 and 5.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls =?

d. 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.) Fraction of calls=?

e. As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 6 percent of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.) Duration =?

Explanation / Answer

HERE THE DISTRIBUTION IS NORMAL

THE FORMULA TO BE USED = Z = (X-MEAN)/STANDARD DEVIATION

MEAN = 4.20

STANDARD DEVIATION = 0.4

A) P(4.20<P<4.90) =

For x = 4.20 , z = (4.20 - 4.20) /0.4 = 0 and for x = 4.90, z = (4.90 - 4.20) / 0.4 = 1.75

Hence P(4.20 < x < 4.90) = P(0 < z < 1.75) = [area to the left of z = 1.75] - [area to the left of 0]

= 0.9599 - 0.5 = 0.4599

B)P(X>4.90)

Z =(4.90-4.20)/0.4 = 1.75
Hence P(x > 4.90) = P(z > 1.75) = [total area] - [area to the left of -2.25]

1 - [area to the left of 1.75]

now from the z table we will take the value of z score = 1.75

    = 1 - 0.9599 = 0.0401

C) For x =5.50 , z = (5.50 - 4.20) /0.4 = 3.25 and for x = 4.90, z = (4.90 - 4.20) / 0.4 = 1.75

Hence P(4.90 < x < 5.50) = P(1.75 < z < 3.25) = [area to the left of z = 3.25] - [area to the left of 1.75]

= 0.9994 - 0.9599 = 0.0395

D)For x =5.50 , z = (5.50 - 4.20) /0.4 = 3.25 and for x = 3.50, z = (3.50 - 4.20) / 0.4 = -1.75

Hence P(3.50 < x < 5.50) = P(-1.75 < z < 3.25) = [area to the left of z = 3.25] - [area to the left of -1.75]

= 0.9994 - 0.0401 = 0.9503

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