probability of sample variance lying between values

Question

Let $X_1,\ldots,X_n$ be a random sample of size $n = 10$ from a population which is normally distributed with mean $48$ and variance $36$.

What is the probability that the sample variance of such a sample lies between $25$ and $60$?

Answer 1

For a normal population, the quantity $${{(n-1)s^2} \over{\sigma^2} }$$ is known to have a chi-squared distribution with $n-1$ degrees of freedom.

So then $$ P[25 < s^2 < 60]=P\left[{{9(25)} \over {36} } < {{9s^2} \over 36} < {{9(60)} \over {36} } \right]=P \left[ {25 \over 4}<\chi^2_{(9)}< 15 \right].$$

Using spreadsheet software, this evaluates to 0.6237