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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.

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  • $\begingroup$ "the (length of the) pipes follow a normal law"? $\endgroup$
    – user9464
    Commented Jun 24, 2017 at 20:23
  • $\begingroup$ Yes, sorry for the misunderstanding $\endgroup$
    – malloc
    Commented Jun 24, 2017 at 20:26
  • $\begingroup$ It looks right to me. My standard deviation with more decimal places is $\sigma=4.7478$ $\endgroup$ Commented Jun 24, 2017 at 20:44
  • $\begingroup$ 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"? $\endgroup$
    – Cardinal
    Commented Jun 24, 2017 at 20:46

1 Answer 1

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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) $

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