We use a chi-squared distribution which gives us a confidence interval that is not symmetric about the sample standard deviation.
We assume that the population of pipes is normally distributed, then
[tex]\sqrt[]{\frac{(n-1)s^2}{\chi^{2_{}}_{\text{right}}^{}}}\le\sigma\le\sqrt[]{\frac{(n-1)s^2}{\chi^{2_{}}_{\text{left}}}}[/tex]
where s is the sample standard deviation, and sigma is the population standard deviation and n represents the sample size.
We need to find the L & R values that make
[tex]P(L\le\chi^2\le R)=0.95[/tex]
true.
Using a table or calculator, we have
[tex]\begin{gathered} \chi^2_{\text{Right}}=20.483 \\ \chi^2_{\text{Left}}=3.247 \end{gathered}[/tex]
The other variables are given. Plugging all of those values in the formula, we have
[tex]\begin{gathered} \sqrt[]{\frac{(11-1)(14.6)^2}{20.483}}\le\sigma\le\sqrt[]{\frac{(11-1)(14.6)^2}{3.247}} \\ 10.2\le\sigma\le25.6 \end{gathered}[/tex]
The 95% confidence interval for the standard deviation is approximately (10.2, 25.6).
