Question

There are 43 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 4 min. (Round your answers to four decimal places.) (a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins? 8974 (b) f the sports report begins at 11:10, what is the probability that he misses part of the report if he waits until grading is done before turning on the TV? .5793

Answer 1

Say X be the time needed to grade a randomly chosen paper.

X~ N(5, 4)

We have a total of 43 students and all follow a normal distribution. The total time would be the sum of distributions of the individual students. The total time will also follow a normal distribution:
Mean = n*5 = 43*5 = 215
Standard deviation = √n*4 = (√43)*4 = 26.2298

Hence, Total time = T~(215, 26.2298)

a) Probability that he is through grading before 11 PM OR the total time is less than 250 minutes.

P(T < 250) = P(T < (250 - 215)/26.2298) = P(T < 1.33) = 0.9082 (from the z-table)

b) P(T > 260) = P(T > (260 - 215)/26.2298) = P(T > 1.72) = 1 - 0.9573 = 0.0427