(1 point) An instructor has 200 exams to grade. Each exam takes a different amou

Question

(1 point) An instructor has 200 exams to grade. Each exam takes a different amount of time to grade. The times (in minutes) required to grade each exam are independent and all have the same (but unknown) distribution that has mean 18 and variance 10 Approximate the probability that the instructor will grade fewer than 39 exams in the first 713 minutes of work. Hint: If X, is the time to grade the i-th exam, then x = -i X, is the time it takes to grade the first 39 exams. Can you use X to express the event that the instructor will grade fewer than 39 exams in the first 713 minutes of work? 39

Explanation / Answer

Normal distribution

for 39 exams mean = 39 * 18 = 702

and std deviation=(10 * 39)1/2 = 19.75

hence probabilty that it will take more then 713 minutes to grade 39 exams = P(X > 713)

= 1 - P(X > 713) = 1 - P(Z < (713 - 702)/19.75) = 1 - P(Z < 0.56) = 0.2877

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