Statistics!
posted by Allex on .
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

A score of less than 35 corresponds to a zvalue of 1.28 (lowest 10%), and a score of more than 425 corresponds to a zvalue 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.