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The defect length of a corrosion defect in a pressurized

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 39E Chapter 4

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 39E

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 30 mm and standard deviation 7.8 mm [suggested in the article ?"Reliability Evaluation of Corroding Pipelines Considering Multiple Failure Modes and Time-Dependent Internal Pressure" („I. of Infrastructure Systems, 2011: 216-224)]. a. What is the probability that defect length is at most 20 mm? Less than 20 mm? b. What is the 75th percentile of the defect length distribution—that is, the value that separates the smallest 75% of all lengths from the largest 25%? c. What is the 15th percentile of the defect length distribution? d. What values separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%?

Step-by-Step Solution:

Answer : Step 1: From the given information the defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 30 mm and standard deviation 7.8 mm. Normal Distribution Mean = 30 Standard Deviation ( sd ) = 7.8 Normal Distribution ~ N(0,1) The formula is given by, x Z = a). Now we have to find the probability that defect length is at most 20 mm. Less than 20 mm. Here x=20. x P(X < x) = ( ) P(X < 20) = ( 2030) 7.8 P(X < 20) = ( 10) 7.8 P(X < 20) = 1.2821 P(X < 20) =P ( Z < -1.2821) Then from the Standard Normal Table is P(X < 20) = 0.0999 Therefore the probability of less than 20 mm is 0.0999. Step 2: b). Now we have to find the 75th percentile of the defect length distribution—that is, the value that separates the smallest 75% of all lengths from the largest 25%. Here 75% = 75/100 =0.75 Then P ( Z < x ) = 0.75 The value that separates the smallest 75% of all lengths from the largest 25%. So 0.75-0.25 =0.45 From normal table. The value of z to the cumulative probability of 0.75 is 0.67364 0.674 x x30 P( < 7.8 ) That is, x30 7.8 = 0.674 So x = 0.674 (7.8) + 30 x = (5.2572) + 30 x = 35.2572 Therefore 35.2572 is the 75th percentile of the defect length distribution—that is, the value that separates the smallest 75% of all lengths of the largest 25%.

Step 3 of 4

Chapter 4, Problem 39E 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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