# A word problem on probability

I'm doing some probability exercise and I got stuck. This is the problem:

I have the average length of a pipe in a set of pipes: 14.5cm. Then I know that the pipes follow a normal law. I must calculate the standard deviation knowing that the probability to find pipes of length between 11.3cm and 14.5cm is 25%. Also I need to calculate the probability that a casual pipe taken from 70 pipes taken from the original set of pipes, has the average length less then 13.2cm.

I did the following things to find the standard deviation:

$$P(11.3<x<14.5)=0.25=P(x<14.5)-P(x<11.3)$$ $$P(x<14.5)=0.5$$ $$P(x<11.3)=0.25=P(z<\frac{11.3-14.5}{\sigma})=P(z<\frac{-3.2}{\sigma})$$ $$P(z<\frac{-3.2}{\sigma})=1-P(z>\frac{3.2}{\sigma})$$ $$P(z>\frac{3.2}{\sigma})=0.75$$ So $z\simeq0.68$ looking on the table; then $\sigma=4.7$.

Is what I did logically correct? Looking at the draw that I did it should be right.

What I need to do to proceed? I tried to calculate $P(x<13.2)$ but this don't give me what the exercise ask.

• "the (length of the) pipes follow a normal law"?
– user9464
Commented Jun 24, 2017 at 20:23
• Yes, sorry for the misunderstanding Commented Jun 24, 2017 at 20:26
• It looks right to me. My standard deviation with more decimal places is $\sigma=4.7478$ Commented Jun 24, 2017 at 20:44
• I have a hard time imagining "taken from 70 pipes taken from the original set of pipes", would you reworded it to plain English? What do you mean by "taken from 70 pipes"? Commented Jun 24, 2017 at 20:46

Your calculation in $a)$ looks right to me. My standard deviation with more decimal places is $σ=4.7478$.
As I understand it. In $b)$ you take a sample with a size of $70$. The sizes of the pipes are independent. Thus you have 70 independent and identically distributed random variables.
The average of the sample is distributed as $\overline X\sim\mathcal N\left(\mu, \frac{\sigma^2}{n}\right)=\mathcal N\left(14.5, \frac{4.7478^2}{70}\right)$
Thus $P(\overline X <13.2)=\Phi\left( \frac{13.2-14.5}{\sqrt{\frac{4.7478^2}{70}}} \right)$