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Eighteen individuals are scheduled to take a driving test

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 68E 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 68E

Eighteen individuals are scheduled to take a driving test at a particular DMV office on a certain day, eight of whom will be taking the test for the first time. Suppose that six of these individuals are randomly assigned to a particular examiner, and let X be the number among the six who are taking the test for the first time. a. What kind of a distribution does X have (name and values of all parameters)? b. Compute c. Calculate the mean value and standard deviation of X.

Step-by-Step Solution:

Answer: Step 1 of 3 Suppose that six of these individuals are randomly assigned to a particular examiner, and let X be the number among the six who are taking the test for the first time. a. Here, X follows hypergeometric distribution. There are N individuals, N = 18 Out of which 18 are first time test takers (we’re going to think of being a first timer as a success) Out of those 18 we need to select 6 people and every subset is equally likely and we want to know how likely is that we selected a certain number of first timers. Taking everything mentioned above into consideration we can see that this is a hypergeometric distribution, where h(x;n, M,N) = h(x;6, 8, 18). Step 2 of 2 b. Compute P(X = 2), P(X 2) and P(X 2) The Hypergeometric distribution formula is as shown below i) We have, MC (N M)C P(X = x) = x (n x) NC n 8C 2 (18 8)(6 2) P(X = 2) = 18C 6 P(X = 2) = 28 × 210 18564 5880 = 18564 Therefore, P(X = 2) = 0.3167 ii) Now, P(X 2) = P(X = 0) + P(X = 1) + P(X = 2) 8C0 (18 8)C(6 0) 8C1 (18 8)C(6 1) 8C2 (18 8)C(6 2) = 18C + 18C + 18C 210 2016 5880 6 6 = 18564 + 18564 + 18564 = 0.0113 + 0.1085 + 0.3167 Hence,P(X 2) = 0.4365 iii) Now, P(X 2) = 1 P(X 2) = 1 P(X = 2) = 1 0.4365 = 0.5635 Therefore, P(X 2) = 0.5635.

Step 3 of 3

Chapter 3, Problem 68E is Solved
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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Eighteen individuals are scheduled to take a driving test

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