Question

almost all companies utilize some type of year - end performance review for their employees. human resources (hr) at a universitys health science center provides guidelines for supervisors rating their subordinates. for example, raters are advised to examine their ratings for a tendency to be either too lenient or too harsh. according to hr, \if you have this tendency, consider using a normal distribution—10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable.\ suppose you are rating an employees performance on a scale of 1 (lowest) to 100 (highest). also, assume the ratings follow a normal distribution with a mean of 49 and a standard deviation of 16. complete parts a and b.
a. what is the lowest rating you should give to an \exemplary\ employee if you follow the universitys hr guidelines? (round to two decimal places as needed.)
b. what is the lowest rating you should give to a \competent\ employee if you follow the universitys guidelines? (round to two decimal places as needed.)

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation. We will use the z - scores corresponding to the percentiles to find the cut - off values.

Step2: Find the z - score for the 'exemplary' percentile

The 'exemplary' employees are the top 10% of the distribution. Looking up the z - score in the standard normal distribution table for the area to the left of $z$ corresponding to $1 - 0.10=0.90$. The z - score $z_{0.90}\approx1.28$.
We know that $\mu = 49$ and $\sigma=16$. Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we can solve for $x$ (the rating). Rearranging the formula gives $x=\mu + z\sigma$.
Substituting the values: $x = 49+1.28\times16=49 + 20.48=69.48$.

Step3: Find the z - score for the 'competent' percentile

The 'competent' employees are the middle 40% of the distribution. The lower bound of the 'competent' range has an area to the left of $0.10 + 0.20=0.30$ and the upper bound has an area to the left of $0.10+0.20 + 0.40=0.70$.
The z - score corresponding to an area of $0.30$ is $z_{0.30}\approx - 0.52$ and the z - score corresponding to an area of $0.70$ is $z_{0.70}\approx0.52$. We want the lower bound for 'competent'.
Using the z - score formula $x=\mu+z\sigma$, with $\mu = 49$, $\sigma = 16$ and $z=-0.52$. Then $x=49+( - 0.52)\times16=49-8.32 = 40.68$.

Answer:

a. 69.48
b. 40.68