# An instructor who taught two sections of engineering

Chapter 4, Problem 70E

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QUESTION:

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a.? ?What is the probability that exactly 10 of these are from the second section? b.? ?What is the probability that at least 10 of these are from the second section? c.? ?What is the probability that at least 10 of these are from the same section? d. ?What are the mean value and standard deviation of the number among these 15 that are from the second section? e. ?What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

QUESTION:

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a.? ?What is the probability that exactly 10 of these are from the second section? b.? ?What is the probability that at least 10 of these are from the second section? c.? ?What is the probability that at least 10 of these are from the same section? d. ?What are the mean value and standard deviation of the number among these 15 that are from the second section? e. ?What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

Answer: Step 1 of 4 An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. Taking everything mentioned above into consideration we can see that this is a hypergeometric distribution, where h(x;n, M,N) = h(x;15, 30, 50). a. The probability that exactly 10 of these are from the second section is given below Compute P(X = 10) The Hypergeometric distribution formula is as shown below We have, MC x (N M)C(n x) P(X = x) = NC n P(X = 10) = 30C10 (50 30)(15 10) 50C 15 P(X = 10) = 0.2069 Therefore, P(X = 10) = 0.2069. Hence, the probability that exactly 10 of these are from the second section is 20.69%.