Question

Almost all companies utilize some type of​ year-end performance review for their employees. Human Resources​ (HR) at a​ university's 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 distributionlong dash—​10% of employees​ (rated) exemplary,​ 20% distinguished,​ 40% competent,​ 20% marginal, and​ 10% unacceptable." Suppose you are rating 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 51 and a standard deviation of 17.

.What is the lowest rating you should give to an​ "exemplary" employee if you follow the​ university's HR​ guidelines?

_________​(Round to two decimal places as​ needed.)

What is the lowest rating you should give for "distinquished" employee?

What is the lowest rating you should give for "competent" employee?

What is the lowest rating you should give fo "marginal" employee?

What is the lowest rating you should give fo "unacceptable" employee?

Answer 1

a)

Top 10% will have a p-value = 0.9

Using the z table we find z score where probability = 0.9

From table,

Z Score = 1.28

So, (X-51)/17 = 1.28

Which gives X = 51 + 1.28*17 = 72.76

So, the lowest rating for an​ "exemplary" employee = 72.76

e)

Similarly Bottom 10% , p-value = 0.1

Z Score = -1.28

(X-51)/17 = -1.28

X = 51 – 1.28*17 = 29.24

So, the highest rating for an​ "unacceptable" employee = 29.24

So, the lowest rating for an​ "unacceptable" employee = 0

b)

For distinguished, p-value = 0.7

Z Score = 0.525

X = 51 + 0.525*17 = 59.925

So, the lowest rating for an​ " distinguished " employee = 59.925

d)

the lowest rating for an​ " marginal " employee = the highest rating for an​ "unacceptable" employee = 29.25