Rating employee performance. Almost all companies utilize some type of year-end performance review for their employees. Human Resources (HR) at the University of Texas Health Science Center provide guidelines for supervisors to rate their subordinates. For example, raters are advised to examine their ratings for a tendency to be either too lenient or too harsh. HR says, “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 rate an employee’s performance on a scale of 1 (lowest) to 100 (highest). Also, assume the ratings follow a normal distribution with a mean of 50 and a standard deviation of 15.

a. What is the lowest rating you should give to an “exemplary” employee if you follow the University of Texas HR guidelines?

b. What is the lowest rating you should give to a “competent” employee if you follow the University of Texas HR guidelines?

Step by step solution

01

Given information

Let x be the ratings thatfollow a normal distribution with a mean of 50 and a standard deviation of 15.

02

(a) Calculating the lowest rating for an “exemplary” employee

According to HR, 10% of employees rated exemplary.

To find the lowest rating $x_0$for which the$P\left(x>x_0\right)=0.1$

$$\begin{gathered} P\left(x>x_0\right)=1-P\left(x⩽x_0\right)\backslash \\ P\left(x⩽x_0\right)=0.90 \end{gathered}$$

Hence, the z-score corresponding to the probability value of 0.90 from the table is,$z=1.28$

Since,

role="math" localid="1660755523320" $z=\frac{x_0-\mu }{\sigma }$

$\begin{array}{rcl}1.28& =& \frac{x_0-50}{15}\\ x_0& =& 69.2\\ & \approx & 69\\ & & \end{array}$

Therefore, the lowest rating for an “exemplary” employee is 69

03

(b) Calculating the lowest rating for a “competent” employee

According to HR, 40% of employees rated competent.

To find the lowest rating $x_0$for which the$P\left(x>x_0\right)=0.4$

$$\begin{gathered} P\left(x>x_0\right)=1-P\left(x⩽x_0\right) \\ P\left(x⩽x_0\right)=0.60 \end{gathered}$$

Hence, the z-score corresponding to the probability value of 0.60, from the table is,

$z=0.25$

Since,

$z=\frac{x_0-\mu }{\sigma }$

$\begin{array}{rcl}0.25& =& \frac{x_0-50}{15}\\ x_0& =& 53.75\\ & \approx & 54\\ & & \end{array}$

Therefore, the lowest rating for a “competent” employee is 54.

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