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 distribution—​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 52 and a standard deviation of 13.

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

(A) Exemplary employee are those who are in top 10% or above bottom 90%

using z and percentile table, z score for 90 percentile is 1.282

it is given that mean = 52 and standard deviation = 13

So, x = (z*standard deviation) + mean, where x is the required lowest rating

this implies

x = (1.282*13) + 52

= 16.666 + 52

= 68.67 (rounded to 2 decimals)

(B)

comptent employee are those who are in top 70% or above bottom 30% because there are exemplary and distinguished above competent.

using z and percentile table, z score for 30 percentile is -0.524

it is given that mean = 52 and standard deviation = 13

So, x = (z*standard deviation) + mean, where x is the required lowest rating

this implies

x = (-0.524*13) + 52

= -6.812 + 52

= 45.19(rounded to 2 decimals)