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

49 and a standard deviation of 17.

Complete parts a and b.

a. 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.)

b. What is the lowest rating you should give to a "competent" employee if you follow the university's guidelines?

(Round to two decimal places as needed.)

Answer 1

Mean of 49 and a standard deviation of 17. z=x-49/17

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

Ans:
z = Normsinv(.90) = 1.282 ..................using excel function
15(1.282) + 49 = 70.79 =71 ...lowest rating you should give to an “exemplary” employee

b. What is the lowest rating you should give to a "competent" employee if you follow the university's guidelines?

Ans:

z = Normsinv(.30) = -.524 ..................using excel function
17(-.524) + 49= 44.08,
44, the lowest rating you should give to an “competent” employee

Hope this will be helpful. Thanks and God Bless You :-)