Chapter 8.5, Problem 29E

To determine

To find the percentage of calls answered between 0 to 4 minutes.

Answer to Problem 29E

The percentage of calls answered between 0 to 4 minutes is 68.26 %

Explanation of Solution

Given information: Mean $\mu =2$ and standard deviation $\sigma =2$ of the normal distribution.

Formula Used: The probability density function for a normal distribution with mean $\mu$

and standard deviation $\sigma$ is: $f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{(x-\mu )}^2}{2{\sigma }^2}}=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-z^2}{2}};z=\frac{x-\mu }{\sigma }$

Calculation :

Here, $\mu =2$ and $\sigma =2$

When $x=0$ , $z=\frac{0-2}{2}=-1$

When $x=4$ , $z=\frac{4-2}{2}=1$

Between $z=-1\text{to}z=1$ , the area under the normal curve is 68.26 %

Hence, the percentage of calls answered between 0 to 4 minutes is 68.26 %