Grading managers. Some 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 this year assigned 10% A grades, 80% B grades, and 10% C grades to the company’s managers. Suppose Ford’s performance scores really are Normally distributed. This year, managers with scores less than 25 received C grades and those with scores above 475 received A grades. What are the mean and standard deviation of the scores?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the grades of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]

For this case we have two conditions given:

[tex] P(X<25) = 0.1[/tex]

[tex] P(X>475) = 0.1[/tex] or equivalently [tex]P(X<475) =0.9[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

So we can find a value from the normal standard distribution that accumulates 0.1 and 0.9 of the area in the left, for this case the two values are:

[tex] z= -1.28, z=-1.28[/tex]

We can verify that P(Z<-1.28) =0.1[/tex] and P(Z<1.28) =0.9[/tex]

And then using the z score we have the following formulas:

[tex] -1.28 = \frac{25 -\mu}{\sigma}[/tex] (1)

[tex] 1.28 = \frac{475 -\mu}{\sigma}[/tex] (2)

If we add equations (1) and (2) we got:

[tex] \frac{25 -\mu}{\sigma} + \frac{475 -\mu}{\sigma} =0[/tex]

We can multiply both sides of the equation by [tex]\sigma[/tex] and we got:

[tex] 25+ 475 -2 \mu = 0[/tex]

[tex] \mu = \frac{500}{2}= 250[/tex]

And then we can find the standard deviation for example from equation (1) and we got:

[tex] \sigma = \frac{25-250}{-1.28}=175.781[/tex]

So then the answer would be:

[tex] \mu = 250, \sigma = 175.781[/tex]