SOLUTION: The marks of students, in a course of Probability and Statistics, at a university, are normally distributed with mean 68% and variance 70%. If passing percentage of marks is at lea

Question 1150437: The marks of students, in a course of Probability and Statistics, at a university, are normally distributed with mean 68% and variance 70%. If passing percentage of marks is at least 40, what proportion of class will fail the course? The teacher decides to award grade ‘A’ to the highest 7% marks, grade ‘B+’ to next 15% marks, grade ‘B’ to next 25% marks , grade ‘C+’ to next 20% marks, grade ‘C’ to next 13% marks, grade ‘D’ to next 11% marks find the range of marks for grade ‘A’,’B+’, ’B’, ’C+’, ’C’, ’D’, ’F’.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!

mean is 68 and sd is sqrt of 70 or 8.4. Note: use mean of 68 and var of 70, not 0.70, because that would require %^2 as units.
z=(x-mean)/sd
z<40 is failing and it is (40-68)/8.4=-2.86
z<-2.86 is probability 0.0021
z(0.93) is 1.48
1.48=(x-68)/8.4
12.43=x-40
x=80.4
B+ has a bottom z-score at the 78th percentile or z=0.77
0.77=(x-68)/8.4
6.47=x-68
x=74.47 or 74 rounded so (74.5, 80.4) is B+
B has bottom z-score at 53rd %ile or z=0.075
0.075=(x-68)/8.4
B=68.63 or range (68.6, 74.5)
C+ goes down to 33rd percentile to z=-0.44
similarly, C+ (64.3, 68.6)
C goes down to 20th percentile or z=-0.84
so the range is 7.1 below the mean or (60.9, 64.3)
D goes down to the 9th percentile or z=-1.34
the range is (56.7, 60.9)
below 56.7 would be failing by the teacher's rule.

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