Answer
$\sigma$ is between 5 and 8.5.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{57+...+47}{38}=54.76.$
Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(57-54.76)^2+...+(46-54.76)^2}{37}}=6.5613.$
$\alpha=1-0.98=0.02.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=38-1=37$.
$X_{L}^2= X_{0.95}^2=63.691$
$ X_{R}^2= X_{0.05}^2=12.592$
Hence the confidence interval:$\sigma$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(37)\cdot 6.5613^2}{63.691}}=5$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(37)\cdot 6.5613^2}{12.592}}=8.5.$
