The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. The manager has decided to have a signal system attached to the phone so that after a certain period of time, a sound will occur on her employees' phone if she exceeds the time limit. The manager wants to set the time limit at a level such that it will sound on only 8 percent of all calls. The time limit should be:

We are given that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes i.e.;

Mean, [tex]\mu[/tex] = 8.21 minutes and Standard deviation, [tex]\sigma[/tex] = 2.14 minutes

Also, z score = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

It is also given that a sound will occur on her employees' phone if she exceeds the time limit.

Let X = time limit after which sound which occur

The manager wants to set the time limit at a level such that it will sound on only 8 percent of all calls i.e.; P(X > x) = 0.08

P(X > x) = 0.08 ⇒ P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-8.21}{2.14}[/tex] ) = 0.08

⇒ P(Z > [tex]\frac{x-8.21}{2.14}[/tex] ) = 0.08

In the z table, the critical value which exceeds the area of 0.08 is 1.427 which means;

⇒ [tex]\frac{x-8.21}{2.14}[/tex] = 1.427

⇒ x = (1.427 * 2.14) + 8.21

⇒ x = 11.264 minutes

Therefore, the time limit at a level such that it will sound on only 8 percent of all calls is 11.26 minutes .