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 x- be the random variable representing the meantime to complete the $16$reviews. Assume that the $16$reviews represent a random set of reviews.

What causes the probabilities in Exercise $7.3$andExercise $7.4$to be different?

Short answer

Here, X and $\stackrel{-}{x}$are two different variables, and the distribution of X and $\stackrel{-}{x}$have the same means but different standard deviations. The distribution of X is more spread than $\stackrel{-}{x}$. This causes the probabilities, P($3.5$<x<$4.25$) and P($3.5$<$\stackrel{-}{x}$<$4.25$) to be different.

Step-by-step solution

Given Information

The required probability in Exercise $7.3$is P($3.5$<x<$4.25$) and in Exercise $7.4$is P($3.5$<$\stackrel{-}{x}$<$4.25$).

Explanation

We have to evaluate the probability that required to be computed in Exercise $7.3$and in Exercise $7.4$.

The required probability in Exercise $7.3$is P($3.5$<x<$4.25$) and in Exercise $7.4$is P($3.5$<$\stackrel{-}{x}$<$4.25$).

According to the information, It is also given that, X follows a normal distribution with mean =$4$and standard deviation =$1.2$.

The sample size is n=$1.6$.

Here, X indicates the time taken to complete one review and $\stackrel{-}{x}$represents the mean time taken to complete the $16$reviews.

Explanation

Since, $X~N({\mu }_x=4,{\sigma }_x=1.2)$, the distribution of the sample mean is

$\overline{X}~N({\mu }_X,\frac{{\sigma }_X}{\sqrt{n}})\text{.Thus,}$

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

$\overline{X}-N(4,0.3)$