91Ó°ÊÓ

According to the given details, the manager reviews $16$employees each month. Thus, the meantime for each review is $4$hours and the standard deviation is $1.2$hours and the sample size is $16$.

Let's consider$k={95}^{\text{th}}\text{Precentile,}$

The graph for$k={95}^{\text{*}}\text{Precentile is given as below:}$

In the above graph, the shaded area represents probability p(x<k)=0.95

The $95$th Percentile =____________

Let's use Ti-83 calculator to compute the $95$th percentile for the meantime to meet one monthly review.

For this, click on localid="1648811383788" $2^{nd}$,then DISTR, and then scroll down to the inventory option and enter the provided details.

After that we need to click on ENTER button on the calculator to have the desired result.

The screenshot is given below:

Therefore, the value of $95$th percentile for the meantime to complete one month review is$4.49$hours.