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 3.6 minutes and the standard deviation was 0.40 minute. a) What is the probability that calls last between 3.6 and 4.2 minutes? b) What is the porbability that the calls last more than 4.2 minutes? c) What is prob calls last beteen 4.2 and 5 minutes/ d) What is prob call last between 3 and 5 minutes? ? e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time? ROUND your z score to 2 decimals and final answer to 4 decimals

Answer 1

Mean = 3.6 minutes

Standard deviation = 0.4 minutes

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(call last between 3.6 and 4.2 minutes) = P(X < 4.2) - P(X < 3.6)

= P(Z < (4.2 - 3.6)/0.4) - P(X < 3.6)

= P(Z < 1.5) - 0.5

= 0.9332 - 0.5

= 0.4332

b) P(call last more than 4.2 minutes) = P(X > 4.2)

= 1 - P(X < 4.2)

= 1 - 0.9332

= 0.0668

c) P(4.2 < X < 5) = P(X < 5) - P(X < 4.2)

= P(Z < (5-3.6)/0.4) - P(Z < 1.5)

= P(Z < 3.5) - P(Z < 1.5)

= 0.9998 - 0.9332

= 0.0666

d) P(3 < X < 5) = P(Z < (5-3.6)/0.4) - P(Z < (3 - 3.6)/0.4)

= P(Z < 3.5) - P(Z < -1.5)

= 0.9998 - 0.0668

= 0.9330

e) The the time be T

P(X > T) = 0.04

P(X < T) = 1-0.04 = 0.96

P(Z < (T - 3.6)/0.4) = 0.96

(T - 3.6)/0.4 = 1.75 (From standard normal distribution table)

T = 4.3 minutes