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A second-stage smog alert has been called in a certain

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 74E 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 74E

A second-stage smog alert has been called in a certain area of Los Angeles County in which there are 50 industrial firms. An inspector will visit 10 randomly selected firms to check for violations of regulations. a. ?If 15 of the firms are actually violating at least one regulation, what is the pmf of the number of firms visited by the inspector that are in violation of at least one regulation? b. ?If there are 500 firms in the area, of which 150 are in violation, approximate the pmf of part (a) by a simpler pmf. c. ?For X = the number among the 10 visited that are in violation, compute E(X) and V(X) both for the exact pmf and the approximating pmf in part (b).

Step-by-Step Solution:

Answer : Step 1 : Given a second-stage smog alert has been called in a certain area of Los Angeles County in which there are 50 industrial firms. Here, the population size is N = 50. An inspector will visit 10 randomly selected firms to check for violations of regulations. The sample size is n = 10. a). If 15 of the firms are actually violating at least one regulation,we have to find the pmf of the number of firms visited by the inspector that are in violation of at least one regulation. To begin it is helpful to summarize the information given in the problem: 1).50 industrial firms 2).10 firms randomly selected for inspection 3).15 of the 50 violate at least one regulation Here we know that n=10 ,N=50 and the number of S’s and F’s in the population are M = 15. The Hypergeometric Distribution : The assumptions leading to the hypergeometric distributions are as follows. 1. The population or set to be sampled consists of N individuals, objects, or elements (a finite population). 2. Each individual can be characterized as a success (S) or a failure (F), and there are M successes in the population. 3. A sample of n individuals is drawn in such a way that each subset of size n is equally likely to be chosen. The random variable of interest is X = the number of S’s in the sample. The probability distribution of X depends on the parameters n, M, and N, so we wish to obtain P (X = x) = h (x; n, M, N). Proposition : If X is the number of S’s in a completely random sample of size n drawn from a population consisting of M S’s and (N-M) F’s , then the probability distribution of X, called the hypergeometric distribution, is given by ( x( nx ) P(X = x) = h(x; n, M, N) = N ( n X= the number of successes from a sample size of n, from a population of size N, with M successes in it X~hypergeom( n, M, N ). Where n= number of successes among the population M= sample size N= total population size From this problem, we can define our random variable as: P(X = x) = h(x; n, M, N). Then we substitute n,M and N. Placing these into a hypergeometric function the pmf is h (x; 10, 15, 50). ( )( 5015) P(X = x) = h(x; 10, 15, 50) = x 10x ( 10 Step 2 : b). If there are 500 firms in the area, of which 150 are in violation, approximate the pmf of part (a) by a simpler pmf. Here M=150,N=500. For this problem a method called approximating Hypergeometric Probabilities is used. It’s stated as follows: Let the population size, N, and number of population S’s, M, get large with the ratio M/N remaining fixed at p. Then h(x; n, M, N) approaches b(x; n, p), so for n/N small, the two are approximately equal provided p is not too near either 0 or 1. By using this rationale and knowing N is large relative to n, it yields : h(x; n, M, N) = b(x; n ,MN ) We substitute N and M values. h(x; 10,150, 500) = b(x; 10 , 500 ) So h(x; 10, 150, 500) = b(x, 10, 0.3) As shown in figure , these two distributions are very similar.

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Chapter 3, Problem 74E 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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