Question

A study of long-distance phone calls made from the corporate offices of the Pepsi Bottling Group Inc. showed the calls follow the normal distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minutes

a) What is the probability the calls lasted between 3.5 and 4.1 minutes?

b) What is the probability the calls lasted less than 3.4 minutes?

c) As part of her report to the president, the director of communications would like to report the minimum length of the longest (in duration) 4% of the calls. What is this time?

d) As part of her report to the president, the director of communications would like to report the maximum length of the shortest (in duration) 8% of the calls. What is this time?

Answer 1

We are given the distribution here as:

a) The probability here is computed as:

Converting it to a standard normal variable, we get here:

Getting it from the standard normal tables, we get here:

Therefore 0.3121 is the required probability here.

b) The probability here is computed as:

P(X < 3.4)

Converting it to a standard normal variable, we get here:

Getting it from the standard normal tables, we get here:

Therefore 0.0912 is the required probability here.

c) From standard normal tables, we have:

P(Z > 1.751) = 0.04

Therefore the minimum time here is computed as:
= Mean + 1.751*Std Dev = 4.2 + 1.751*0.6 = 5.2506

Therefore 5.2506 is the required time here.

d) From standard normal tables, we have:
P(Z < -1.405) = 0.08

Therefore the maximum time here is computed as:
= Mean - 1.405*std Dev = 4.2 - 1.405*0.6 = 3.357

Therefore 3.357 is the required time here.