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A reservation service employs five information operators

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 109E 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 109E

A reservation service employs five information operators who receive requests for information independently of one another, each according to a Poisson process with rate ? = 2 per minute. a. ?What is the probability that during a given 1-min period, the first operator receives no requests? b. ?What is the probability that during a given 1-min period, exactly four of the five operators receive no requests? c. ?Write an expression for the probability that during a given 1-min period, all of the operators receive exactly the same number of requests.

Step-by-Step Solution:

Answer : Step 1 of 3 : Given, A reservation service employs five information operators who receive requests for information independently of one another, each according to a Poisson process with rate = 2 per minute. The probability density function of Poisson distribution is P(X = x) = ex , where, x = 0, 1, 2, 3, ….. x! a) The claim is to find the probability that during a given 1-min period, the first operator receives no requests P(receive no request) = P(x = 0 ) = ex x! Where, x= 0 and = 2 = e220 0! = 0.1353 Therefore, the probability of first operator receives no requests is 0.1353 Step 2 of 3 : b) The claim is to find the probability that during a given 1-min period, exactly four of the five operators receive no requests P( exactly 4 receives no request) = ( ) p (1 p) 54 4 Where, p = 0.1353 Therefore, 5 4 1 P( exactly 4 receives no request) = ( ) (4.1353) (1 0.1353) = 5 (0.0003351) (0.8647) = 0.001449 The probability value is 0.001449.

Step 3 of 3

Chapter 3, Problem 109E 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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