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Q. 10

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The Practice of Statistics for AP

Found in: Page 137

Book edition 4th

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

Pages 809 pages

ISBN 9781319113339

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Short Answer

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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 $= 18, 000$

Step 2: 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:

$\frac{25 + 475}{2} = \frac{500}{2} = 250$

Step 3: 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" $1.28$ standard deviations below the mean.

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

$1.28 s = 250 - 25$

Divide each side by $1.28 :$

$= 175.78125$ .

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