The defect length of a corrosion defect in a pressurizedsteel pipe is normally distributed with mean value 30 mmand standard deviation 7.8 mm [suggested in the article“Reliability Evaluation of Corroding PipelinesConsidering Multiple Failure Modes and TimeDependent Internal Pressure” (J. of InfrastructureSystems, 2011: 216–224)].a. What is the probability that defect length is at most20 mm? Less than 20 mm?b. What is the 75th percentile of the defect length dis-tribution—that is, the value that separates the small-est 75% of all lengths from the largest 25%?c. What is the 15th percentile of the defect lengthdistribution?d. What values separate the middle 80% of the defectlength distribution from the smallest 10% and thelargest 10%?
Question
The defect length of a corrosion defect in a pressurizedsteel pipe is normally distributed with mean value 30 mmand standard deviation 7.8 mm [suggested in the article“Reliability Evaluation of Corroding PipelinesConsidering Multiple Failure Modes and TimeDependent Internal Pressure” (J. of InfrastructureSystems, 2011: 216–224)].a. What is the probability that defect length is at most20 mm? Less than 20 mm?b. What is the 75th percentile of the defect length dis-tribution—that is, the value that separates the small-est 75% of all lengths from the largest 25%?c. What is the 15th percentile of the defect lengthdistribution?d. What values separate the middle 80% of the defectlength distribution from the smallest 10% and thelargest 10%?
Solution
a. To find the probability that the defect length is at most 20 mm, we need to calculate the cumulative probability up to 20 mm using the given mean and standard deviation.
b. The 75th percentile of the defect length distribution represents the value that separates the smallest 75% of all lengths from the largest 25%. To find this value, we can use the cumulative distribution function (CDF) of the normal distribution with the given mean and standard deviation.
c. The 15th percentile of the defect length distribution represents the value that separates the smallest 15% of all lengths from the largest 85%. Similar to the previous question, we can use the CDF of the normal distribution to find this value.
d. To find the values that separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%, we need to calculate the percentiles that correspond to these boundaries. The lower boundary would be the 10th percentile, and the upper boundary would be the 90th percentile. Again, we can use the CDF of the normal distribution to find these values.
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