Question

In a large section of a Chemistry class, the points for the final exam are normally distributed, with a mean of 72, and a variance of 81. Grades are to be assigned according to the following rule: the top 10% receive As, the next 20% receive Bs, the middle 40% receive Cs, the next 20% receive Ds, and the bottom 10% receive Fs. Find the lowest score on the final exam that would qualify a student for an A, a B, a C, and a D.

Answer 1

P((x-72)/9)=90%
(x-72)/9=1.28 [normal distribution table]

x=83.52
83.52 at least for getting A

P((x-72)/81)=10%
(x-72)/81=-1.28 [normal distribution table]

x=60.48
60.48 maximum for getting D

for B, score is between 72 and 83.52

for C, score is between 60.48 and 72