a)Suppose the distribution of grades on the midterm is approximately normal. The

Question

a)Suppose the distribution of grades on the midterm is approximately normal. The class average is 72 and the standard deviation is 8. Approximately what proportion of students obtained an 80 or higher?

b)What is the z-score of a student who obtained 68 on the test?

c)Find a range of values (i.e. a lower bound and an upper bound for a range of values) that would containg the middle 95% of grades on the midterm

d)You have been told that you scored at the 97.5 percentile. What is your approximate grade?

Explanation / Answer

mean is 72 and s is 8

z is (x-mean)/s

a) P(x>80)=P(z>(80-72)/8)=P(z>1) or 1-P(z<1)=1-0.8413=0.1587

b) z is (68-72)/8=-0.5

c) z for 95% confidence is 1.96

thus lower bound is mean-z*s=72-1.96*8=56.32

upper bound is mean+z*s=72+1.96*8=87.68

d) for 97.5 % the z value from the normal distribution table is 1.96

thus answer is mean+z*s=72+1.96*8=87.68

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