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In proof testing of circuit boards, the probability that

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 88E 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 88E

In proof testing of circuit boards, the probability that any particular diode will fail is .01. Suppose a circuit board contains 200 diodes. a. ?How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? b. ?What is the (approximate) probability that at least four diodes will fail on a randomly selected board? c. ?If five boards are shipped to a particular customer, how likely is it that at least four of them will work properly? (A board works properly only if all its diodes work.)

Step-by-Step Solution:

Answer : Step 1 of 3 : Given, In proof testing of circuit boards, the probability that any particular diode will fail is .01. Suppose a circuit board contains 200 diodes. a) The claim is to find the number of diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail We have, n = 200 and p = 0.01 Let X~ B(200, 0.01) Mean of the binomial distribution is np = 200×0.01 = 2 Standard deviation = np(1 p) = 200 × 0.01(1 0.01) = 1.4071 We expect 2 diodes expect to fail and standard deviation is 1.4071. Step 2 of 3 : b) The claim is to find the the probability that at least four diodes will fail on a randomly selected board Therefore, P(4) = 1 - P(x3) Let X~ B(n,p,q) n P(X = x) = ( )(p) (x p) nx x=0 Therefore, P(4) = 1 - P(x3) = 1 - [ P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) ] = 1 - [(( 200)(0.01) (1 0.01) 2000 ) + ( ( 200)(0.01) (1 0.01) 2001 ) + 0 1 200 2 2002 200 3 2003 (( 2 )(0.01) (1 0.01) ) + (( 3 )(0.01) (1 0.01) ) ] = 1 - [ 0.134 + 0.2706 + 0.2720 + 0.1814] = 1 - 0.858 = 0.142 Therefore, P(4) = 0.142

Step 3 of 3

Chapter 3, Problem 88E 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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