Question

Yoonie is a personnel manager in a large corporation. Each month she must review $16$of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of $1.2$hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let $\stackrel{-}{x}$be the random variable representing the meantime to complete the $16$reviews. Assume that the $16$reviews represent a random set of reviews.

Find the $95$th percentile for the meantime to complete one month's reviews. Sketch the graph.

a.

b. The $95$th Percentile =____________

Short answer

a) The${95}^{th}$percentile is$4.49$hours and the graph is given below:

b)The value of $95$th percentile for the meantime to complete one month review is $4.49$hours.

Step-by-step solution

Given Information (part a)

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 $16$.

Explanation (part a)

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

Given Information (part b)

The $95$th Percentile =____________

Step 5: Explanation (part b)

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.

Explanation (part b)

The screenshot is given below:

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