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

Answer:

14.69% probability that defect length is at most 20 mm

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 28}{7.6}$

$Z = -1.05$

$Z = -1.05$ has a pvalue of 0.1469

14.69% probability that defect length is at most 20 mm