An instructor who taught two sections of Math 123A, the first with 20 students a

Question

An instructor who taught two sections of Math 123A, the first with 20 students and the second with 30 students, gave a midtem exam. After al the students had rumed in their exam papers, the instructor randomly ordered them before grading. Consider the first 15 graded exam papers (a) Find the probability that exactly 10 of these are from the second section. b) Find the probability that at least 10 of these are from the second section. (c) Find the probability that at least 10 of these are from the same section (d) Find the mean value and standard deviation of the number among these 15 that are from the second section. (e) Find the mean and standard deviation of the number of exam papers not among these first 15 that are from the second section

Explanation / Answer

a) P(exactly 10 from second section) = 30C10 * 20C5/50C15 = 0.207

b) P(at least 10 are from the second section) = ( 30C10 * 20C5 + 30C11 * 20C4 + 30C12 * 20C3 + 30C13 * 20C2 + 30C14 * 20C1 + 30C5)/50C15 = 0.3797

c) P(at least 10 are from the same section) = ( 30C10 * 20C5 + 30C11 * 20C4 + 30C12 * 20C3 + 30C13 * 20C2 + 30C14 * 20C1 + 30C5 + 20C10 * 30C5 + 20C11 * 30C4 + 20C12 * 30C3 + 20C13 * 30C2 + 20C14 * 30C1 + 20C15)/50C15 = 0.3937

d) mean = n * k / N = 15 * 30/50 = 9

variance = n * k * (N - k) * (N - n)/[N2 * (N - 1)]

                              = 15 * 30 * (50 - 30) * (50 - 15)/((50)^2 * 49)

                             = 2.5714

Standard deviation = sqrt(2.5714) = 1.6036

e) mean = n * k / N = 15 * 20/50 = 6

variance = n * k * (N - k) * (N - n)/[N2 * (N - 1)]

                = 15 * 20 * (50 - 20) * (50 - 15)/((50)^2 * 49)

                = 2.5714

standard deviation = sqrt(2.5714) = 1.6036

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