# Solved-The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 30 mm and st

#### ByDr. Raju Chaudhari

Oct 6, 2020

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" (J. 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%?

### Solution

Given that $\mu = 30$ and $\sigma = 7.8$.

(a) The probability that defect length is at most 20 mm

 \begin{aligned} P(X\leq 20) &= P\bigg(\frac{X-\mu}{\sigma} \leq \frac{20-30}{7.8}\bigg)\\ &= P\bigg(Z \leq -1.28\bigg)\\ &=0.0999 \end{aligned}

(b) Let the 75th percentile of the defect length distribution be $a$.

 \begin{aligned} &P(X\leq a) =0.75\\ &\Rightarrow P(X\leq a) =0.75\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}< \frac{a-30}{7.8}\big)=0.75\\ &\Rightarrow P(Z< \frac{a-30}{7.8}\big)=0.75\\ &\Rightarrow \frac{a-30}{7.8}= 0.674\\ &\Rightarrow a = 30 + 0.674* 7.8\\ &\Rightarrow a = 35.2572 \end{aligned}

(c) Let the 15th percentile of the defect length distribution be $b$.

 \begin{aligned} &P(X\leq b) =0.15\\ &\Rightarrow P(X\leq b) =0.15\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}< \frac{b-30}{7.8}\big)=0.15\\ &\Rightarrow P(Z< \frac{b-30}{7.8}\big)=0.15\\ &\Rightarrow \frac{b-30}{7.8}= -1.036\\ &\Rightarrow b = 30 + -1.036* 7.8\\ &\Rightarrow b = 21.9192 \end{aligned}

(d) Let the values that separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10% be $-a$ and $a$ respectively.

Thus we have $P(X\leq -a) = 0.10$ and $P(X\leq a) = 0.90$.

 \begin{aligned} &P(X\leq a) =0.9\\ &\Rightarrow P(X\leq a) =0.9\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}< \frac{a-30}{7.8}\big)=0.9\\ &\Rightarrow P(Z< \frac{a-30}{7.8}\big)=0.9\\ &\Rightarrow \frac{a-30}{7.8}= 1.282\\ &\Rightarrow a = 30 + 1.282* 7.8\\ &\Rightarrow a = 39.9996 \end{aligned}