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Q. 10

Expert-verifiedFound in: Page 137

Book edition
4th

Author(s)
David Moore,Daren Starnes,Dan Yates

Pages
809 pages

ISBN
9781319113339

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 $18,000$ managers. Suppose that Ford's performance scores really are Normally distributed. This year, managers with scores less than $25$ received C's, and those with scores above $475$ received A's. What are the mean and standard deviation of the scores? Show your work.

Mean: $250$ points

Standard deviation: $175.78125$ points

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

A Grade$=10\%$

B Grade $=80\%$

C Grade$=10\%$

No. of managers$=18,000$

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:

$\frac{25+475}{2}=\frac{500}{2}\phantom{\rule{0ex}{0ex}}=250$

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" $1.28$ standard deviations below the mean.

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

$1.28s=250-25$

Divide each side by $1.28:$

$=175.78125$.

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