Answer
df=39, $X_{L}^2=24.433$, $ X_{R}^2=59.342$, $\mu$ is between 52.86 and 82.37.
Work Step by Step
$\alpha=1-0.95=0.05.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=40-1=39$.
$X_{L}^2= X_{0.975}^2=24.433$
$ X_{R}^2= X_{0.025}^2=59.342$
Hence the confidence interval:$\mu$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(39)\cdot 65.2^2}{59.342}}=52.86$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(39)\cdot 65.2^2}{24.433}}=82.37.$
