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An instructor who taught two sections of engineering

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 70E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 70E

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?

Step-by-Step Solution:

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%. Step 2 of 4 b. The probability that at least 10 of these are from the second section is given by P(X 10) = P(X = 10) + P(X = 11) + ... + P(X = 15) Using Excel “=HYPGEODIST(x;n, M,N)” x P(x) 10 0.206953879 11 0.117587432 12 0.043807082 13 0.010109327 14 0.00129217 15 6.89157E-05 Sum 0.379818805 Hence, the probability that at least 10 of these are from the second section is 0.3798 = 37.98%.

Step 3 of 4

Chapter 3, Problem 70E is Solved
Step 4 of 4

Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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An instructor who taught two sections of engineering

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