A set of final examination grades in an introductory statistics course is normal

Question

A set of final examination grades in an introductory statistics course is normally distributed, with a mean of 72 and a standard deviation of 8. Complete parts (a) through (d). a. What is the probability that a student scored below 86 on this exam? The probability that a student scored below 86 is 0.9599 (Round to four decimal places as needed.) b. What is the probability that a student scored between 64 and 95? The probability that a student scored between 64 and 95 is 0.8393 (Round to four decimal places as needed.) c. The probability is 5% that a student taking the test scores higher than what grade? The probability is 5% that a student taking the test scores higher than 85 (Round to the nearest integer as needed.) d. If the professor grades on a curve (for example, the professor could give A's to the top 10% of the class, regardless of the score), is a student better off with a grade of 88 on this exam or a grade of 68 on a different exam, where the mean is 65 and the standard deviation is 3? Show your answer statistically and explain. A student is with a grade of 88 on this exam because the Z value for the grade of 88 is and the Z value for the grade of 68 is (Round to two decimal places as needed.)

Explanation / Answer

Mean = 72
Stdev = 8

a. P(X<86) = P(Z< (86-72)/8) = P(Z< 1.75) = .9599

b. P(64<X<95) = P(-1<Z<2.875) = .9980-.1587 =.8393

c. P(X>c) = 1-.05 = .95, (c-72)/8 = 1.645, c = 72+8*1.645 = 85.16

d. A student with 88 has a Z score of (88-72)/8 =2. The 2nd student with score of 68 has a Z score of (68-65)/3 = 1
So a score of 88 is better than a score of 68 based on the distribution params given

A student is better with a grade of 88 on this examination because the Z value for the grade of 88 is 2 and the Z value for the grade of 68 is 1

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