Let $Y1,Y2,...,Y11$ be iid $N(μ,σ^2)$ so that n= 11. Find $P(0.325<S^2/σ^2<2.048)$
My approach was: $P(S^2/σ^2<2.048) - P(S^2/σ^2<0.325)$
The degrees of freedom would be 10. This is the table I'm attempting to use.
However, I'm not entirely sure where to go from here. I did use R software to get the solution of 0.004040321 - but I don't know if this is correct.
You have to convert your random variable to a $\chi^2_{n-1}$ random variable. Recall $$\dfrac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}$$ so you have to write $$\mathbb{P}\left(\dfrac{S^2}{\sigma^2} < 2.048 \right) = \mathbb{P}\left(\dfrac{(11-1)S^2}{\sigma^2} < 2.048 \cdot (11-1)\right) = \mathbb{P}\left(\chi^2_{10} < 20.48\right)$$ and similarly, $$\mathbb{P}\left(\dfrac{S^2}{\sigma^2} < 0.325 \right) = \mathbb{P}(\chi^2_{10} < 3.25)\text{.}$$ Can you finish this up?