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-hour search 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

Complete the distributions.

a. X~ _____(_____,_____)

b. $\stackrel{-}{x}$~ _____(_____,_____)

Short answer

From the information given in the question,

a)$X~N(4,1.2)$

b)localid="1648811887752" $\overline{X}~N(4,0.3)$

Step-by-step solution

Given Information (part a)

From the question, we have that

-the time for each review (X) is normally distributed

- the meantime ($\mu$) for each review is $4$hours,

- standard deviation ($\sigma$) is $1.2$hours

- sample size (n) is $16$employees.

Explanation (part a)

We know that the normal distribution with mean ($\mu$)and standard deviation($\sigma$) is given as:

$\mathbf{X}~\mathbf{N}(\mu ,\sigma )$

From our example,we can complete the distribution

$X~N(4,1.2)$

Given Information (part b)

$\stackrel{-}{x}$ ~ _____(_____,_____)

Explanation (part b)

$\text{The distribution for the sample mean}\left(\overline{X}\right)\text{is given as}$

$\overline{\mathbf{X}}~\mathbf{N}(\mu \mathbf{x},\frac{\sigma \mathbf{x}}{\sqrt{\mathbf{n}}})$

From our example, we can complete the distribution

$\overline{X}~N(4,\frac{1.2}{\sqrt{16}})$

localid="1648811912306" $\overline{X}~N(4,0.3)$