How do I find the Mean and Standard Deviation?

Question

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 Companyâ��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 39 received C's and those with scores above 711 received A's. What are the mean and standard deviation of the scores?

mu=__________________points sigma=__________________points

? · Accepted Answer

imagine a normal curve where C grade is represented at the left side of the curve at score of 39 covering 10% area from left and A grade at right side at score of 711 covering 10% area from right.

normally we r given x, mean (x bar) or mu, sigma to find value of z by using formula
z = (x - mu)/sigma
then by that value of z, v refer to table of standard distribution to get area under normal curve.
Here v kno that area. So v have to work backwards.
working reverse, find out in table value nearest to 10%, i.e. 0.10
v get this value at 1.28 which is value of z

so now,
for grade C
-1.28 = (39 - mu)/sigma......(-1.28) coz this is to the left of mu)
so -1.28 sigma = 39- mu
...mu - 1.28 sigma = 39.......eq (1)

for grade A
1.28 = (711 - mu)/sigma......(here + 1.28 coz this is to right of mu)
...1.28 sigma = 711 - mu
...mu + 1.28 sigma = 711....eq (2)

add eq (1) and (2)
...2 mu = 750
...mu = 375

substitute value of mu in eq (2)
...1.28 sigma = 711-375
...1.28 sigma = 336
...sigma = 262.5

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How do I find the Mean and Standard Deviation?

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