Question

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 28 mm and standard deviation 7.6 mm.(a)What is the probability that defect length is at most 20 mm

Answer 1

14.69% probability that defect length is at most 20 mm

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\mu = 28, \sigma = 7.6`

What is the probability that defect length is at most 20 mm

This is the pvalue of Z when X = 20. So

`Z = (X - \mu)/(\sigma)`

`Z = (20 - 28)/(7.6)`

`Z = -1.05`

`Z = -1.05` has a pvalue of 0.1469

14.69% probability that defect length is at most 20 mm