Evaluate $P(0.65S < \sigma < 2.37S)$

Question

Consider a sample size n=5, X₁, X₂,...,X₅ ~ $N(\mu, \sigma^2)$. Let's define: $$\bar{X}=\frac{1}{n}(X_1+X_2+...+X_n)$$ $$S^2=\frac1{n-1}\sum_{i=1}^n\left(X_i-\overline{X}\right)^2$$

Evaluate $P(0.65S < \sigma < 2.37S)$.

I have in mind to use $(n-1){\frac{S^2}{\sigma^2}} $~$ \chi^2$_n-1 but I have a hard time applying it to this problem. Can anyone help me? Thanks in advance.

Answer 1

$$0.65 S < \sigma <2.37S\Leftrightarrow \frac{1}{0.65 S} >\frac{1}{\sigma}>\frac{1}{2.37S}\Leftrightarrow \frac{1}{0.65 } >\frac{S}{\sigma}>\frac{1}{2.37}\Leftrightarrow \frac{1}{0.65^2 } >\frac{S^2}{\sigma^2}>\frac{1}{2.37^2}\Leftrightarrow\frac{4}{0.65^2 } >\frac{4S^2}{\sigma^2}>\frac{4}{2.37^2}\Leftrightarrow 0.71 <\frac{4S^2}{\sigma^2}< 9.47$$

Hence

$$P(0.65 S < \sigma <2.37S )=P\left(0.71 <\frac{4S^2}{\sigma^2}< 9.47 \right) =F_{\chi_4 }(9.47 ) -F_{\chi_4 }(0.71 )$$