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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 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 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

reply for ur query in additional details:
ya i kno table given in link goes upto 3.9 only....
wat v normally do is v find out z using formula (x-mu)/sigma wer v know x, mu and sigma. suppose for eg that value of z comes to 1.34, then in table v refer to 1.3 in leftmost vertical column (let us call it left margin of the page) and 0.04 in topmost row (upper margin of the page) (thus 1.3 + 0.04 makes 1.34). then v see wer this column n row intersect. That value comes to 0.90988 (this value is found in main body of the page). Which means that when value of z is 1.34, probability = 0.90988 = 90.988%

since here we r given % of grades in the above qs (i.e. u can relate them with probability) 10% means 0.10
so now in that table, u have to look for value nearest to 0.10 in the main body of the page. then c that - that value is intersection of which no. at left margin n upper margin. that comes to -1.28
it means that z = -1.28

wen u c for +1.28, u'll get probability at around 90% which is from left to right of the normal curve. and that 90% consists of 10% of grade A and 80% of grade B. At that point (90%) , x=711
so for first part (till grade A covering 10%), v take z as -1.28 and for second part (till grade A + b covering 90%) v take z as +1.28

Hope this wil help :)

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Please help me answer this Stats question?

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