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 5.10 minutes and the standard deviation was 0.40 minutes. What fraction of the calls last between 5.10 and 5.80 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls What fraction of the calls last more than 5.80 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls What fraction of the calls last between 5.80 and 6.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls What fraction of the calls last between 5.80 and 6.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls What fraction of the calls last between 4.50 and 6.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Fraction of calls 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.) Duration

Explanation / Answer

a)
P(5.1 < x < 5.8)
= P(x < 5.8) - P(x < 5.1)
= P(Z < (5.8 - 5.1) / 0.4) - P(Z < (5.1 - 5.1) / 0.4)
= P(Z < 1.75) - P(Z < 0)
= phi(1.75) - phi(0)
= 0.95994 - 0.5
= 0.45994

Therefore, 0.4599 of calls last between 5.1 minutes and 5.8 minutes

(b)

P(x > 5.8)
=1 - P(x < 5.8)
=1 - P(Z < (5.8 - 5.1) / 0.4)
=1 - P(Z < 1.75)
=1 - phi(1.75)
=1 - 0.95994

= 0. 040006

Therfore, 0.0401 calls last longer then 5.8 mins.

(C) P(5.8 < x < 6.5)
= P(x < 6.5) - P(x < 5.8)
=P(Z < (6.5 - 5.1) / 0.4) P(Z < (5.8 - 5.1) / 0.4)
=P(Z < 3.5) - P(Z < 1.75)
= phi(3.5) - phi(1.75)
= 0.0398

Therefore, 0.0398 of calls last between 5.8 minutes and 6.5 minutes

(D) P(4.5 < x < 6.5)
= P(x < 6.5) - P(x < 4.5)
=P(Z < (6.5 - 5.1) / 0.4) P(Z < (4.5 - 5.1) / 0.4)
=P(Z < 3.5) - P(Z < -1.5)
= phi(3.5) - phi(-1.5)
=0.933

Therefore, 0.933 of calls last between 4.5 minutes and 6.5 minutes

(E) Given P(D > x) = 3% = 0.03
P(D x) = 97% = 0.97

P(Z (x -5.1) / 0.4) = 0.97
(x - 5.1) / 0.4 = inverse phi(0.97)
(x - 5.1) / 0.4 = 1.881
x - 5.1 = 0.4 * 1.881
x = 5.1 + 0.7524

x= 5.8524 minutes

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