Question

Assume the hold time of callers to a cable company is normally distributed with a mean of 4.0 minutes and a standard deviation of 0.4 minute. Determine the percent of callers who are on hold between 3.4 minutes and 4.5 minutes. % (Round to two decimal places as needed.)

Answer 1

According to the problem, we have

`\begin{gathered} \mu=4.0\min \\ \sigma=0.4\min \end{gathered}`

We have to find the percent of callers who are on hold between 3.4 minutes and 4.5 minutes.

First, we find the z-score

`z=(x-\mu)/(\sigma)`

For x = 3.4

`z=(3.4-4.0)/(0.4)=(-0.6)/(0.4)=-1.5`

For x = 4.5

`z=(4.5-4.0)/(0.4)=(0.5)/(0.4)=1.25`

The probability we have to find is

Hence, the probability is 23.51%.