The final points in a statistical course are normally distri

The final points in a statistical course are normally distribution with a mean of 70 and a standard deviation of 10. The professor wants to grade on a curve, and decides to give out 10% A

Solution

You are playing with z-scores a lot in this one.

z-score is the number of standard deviations away from the mean in either direction.

So we are told the breakdown that he wants:

5% F\'s (5th percentile or 0.05)
15% D\'s (20th percentile or 0.2 -- they add up cumulative with previous grades)
40% C\'s (60th percentile or 0.6)
30% B\'s (90th percentile or 0.9
10% A\'s (whatever is left in the range.

So we have 4 percentiles to find the z-scores on.

Using the table linked below, I found that each percentile corresponds to the following z-scores:

0.05 --> z = -1.645
0.2 --> z = -0.84
0.6 --> z = 0.255
0.9 --> z = 1.28

We are told the mean is 70 with standard deviation of 10. So we can set up this equation:

n = m + sz

Score = mean + standard deviation * z-score.

Do that for all 4 z-scores, and you will get your ranges:

n = 70 + 10(-1.645)
n = 70 - 16.45
n = 53.55

Presuming there aren\'t fractional points, we will call 53 an F, but 54 a D.

Next z-score:

n = 70 + 10(-0.84)
n = 70 - 8.4
n = 61.6

So 61 is a D, 62 is a C. Next z-score:

n = 70 + 10(0.255)
n = 70 + 2.55
n = 72.55

72 is a C, 73 is a B, last z-score:

n = 70 + 10(1.28)
n = 70 + 12.8
n = 82.8

82 is a B, 83 is an A.

So your curved grading scale is:

A: 83+
B: 73 - 82
C: 62 - 72
D: 54 - 71
F : 53 and under