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Rating employee performance. Almost all companies utilize some type of year-end performance review for their employees. Human 91影视 (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, 鈥淚f you have this tendency, consider using a normal distribution鈥�10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable.鈥� Suppose you rate an employee鈥檚 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 鈥渆xemplary鈥� employee if you follow the University of Texas HR guidelines?

b. What is the lowest rating you should give to a 鈥渃ompetent鈥� 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$P\left(x>x_0\right)=0.1$

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

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

Since,

role="math" localid="1660755523320" $z=\frac{x_0-渭}{蟽}$

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

Therefore, the lowest rating for an 鈥渆xemplary鈥� 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$P\left(x>x_0\right)=0.4$

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

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

$z=0.25$

Since,

$z=\frac{x_0-渭}{蟽}$

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

Therefore, the lowest rating for a 鈥渃ompetent鈥� employee is 54.

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