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Chapter 2: Q. 10 (page 137)

R2.10 Grading managers Many companies "grade on a bell curve" to compare the performance of their managers and professional workers. This forces the use of some low-performance ratings, so that not all workers are listed as "above average." Ford Motor Company's "performance management process" for a time assigned 10%A grades, 80%B grades, and 10%C grades to the company's 18,000managers. Suppose that Ford's performance scores really are Normally distributed. This year, managers with scores less than 25received C's, and those with scores above 475received A's. What are the mean and standard deviation of the scores? Show your work.

Short Answer

Expert verified

Mean: 250points

Standard deviation:175.78125points

Step by step solution

01

Given Information 

Ford Motor Company's "performance management process" for a time assigned,

A Grade=10%

B Grade =80%

C Grade=10%

No. of managers=18,000

02

Explanation 

Assuming: the performance scores are normally distributed.

Normal distributions are symmetric about the mean, so the mean must lie between the lowest and highest 10%boundaries.

The mean essentially equals the average between these two boundaries:

25+4752=5002=250

03

Explanation 

Determine the z-score corresponding with 10%(or 0.10) in table A:

z=-1.28

As such, the lowest 10%boundary lies localid="1649407327981" 1.28standard deviations below the mean.

As a consequence, the difference between the mean and the boundaries is 1.28 standard deviations:

1.28s=250鈭�25

Divide each side by 1.28:

=175.78125.

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