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January 31, 2015

January 31, 2015

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

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 8:32pmA 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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