Almost all companies utilize some type of year-end performance review for their

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 along 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 48 and a standard deviation of 16. 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.)

Explanation / Answer

(a) If we are following University's HR guidlines.

Mean rating = 48

standard deviation = 16

so We will give only 10% of rated employees exemplary.

so that means if let say X is the rating of a random employeee.

Pr(X>x) = NORM(X>x; 48; 16) = 0.10

or, NORM(X < x; 48; 16) = 0.90

Z - value for p - value = 0.90 is

Z = 1.645

1.645 = (X - 48)/16

X = 48 + 16 * 1.645 = 74.32

so The exempary employee must get a rating of 74.32 to be an "exemplary" employee.

(b) Now there are 40% competent employee.

so percentile of last competernt employee is = 100 - (10 + 20 + 40) = 30%

so p - value = 0.30

so here

NORM(X < x; 48; 16) = 0.30

Z - value for p - value = 0.30 is

Z = - 0.525

(X - 48)/16 = - 0.525

X = 48 - 16 * 0.525

X = 39.60

so lowest rating we should gie to "competent" employee is 39.60 .

1.645 = (X - 48)/16

X = 48 + 16 * 1.645 = 74.32

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