Posted by **Allex** on Friday, January 18, 2013 at 4:44pm.

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 35 received C's and those with scores above 425 received A's. What are the mean and standard deviation of the scores?

A. μ = 230.0 and σ = 152.3

B. μ = 230.0 and σ = 167.5

C. μ = 230.0 and σ = 140.3

D. μ = 225.0 and σ = 152.3

E. μ = 250.0 and σ = 175.8

F. μ = 250.0 and σ = 168.6

- Statistics! -
**MathGuru**, Friday, January 18, 2013 at 8:32pm
A score of less than 35 corresponds to a z-value of -1.28 (lowest 10%), and a score of more than 425 corresponds to a z-value of +1.28 (upper 10%).

Therefore:

(35 - mean)/sd = -1.28

(425 - mean)/sd = 1.28

Next step:

35 - mean = -1.282(sd)

425 - mean = 1.282(sd)

Adding the equations, we have:

460 - 2(mean) = 0

Solving for mean:

mean = -460/-2 = 230

Now, substitute the mean into either one of the two original equations and solve for sd:

(35 - 230)/sd = -1.28

sd = 152.3

(425 - 230)/sd = 1.28

sd = 152.3

Mean = 230 and sd = 152.3

If I haven't missed anything, this should be it.

I hope this helps.

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