Calculating following probabilities when $X_i\sim N(\mu,\sigma^2)$

Question

Consider $X_1, ...,X_5 \sim N(\mu, \sigma^2)$ with $\bar{X} = \frac{1}{5} \sum_{i=1}^{5} X_i$ and the variance given as $S^2 = \frac{1}{4} \sum_{i=1}^{5} (X_i - \bar{X})^2$. How can I then calculate the following three probabilities:

(1) $\mathbb{P}[\bar{X} - 2.059 S < \mu < \bar{X} + 2.059 S ]$

(2) $\mathbb{P}[0.65 S < \sigma < 2.37 S ]$

(3) $\mathbb{P}[\{ \bar{X} - 2.059 S < \mu < \bar{X} + 2.059 S \} \cap \{ 0.65 S < \sigma < 2.37 S \}]$

My ansatz for (1) is

(1) $\mathbb{P}[ - 2.059 < (\mu - \bar{X})/S < 2.059 ] = \Phi(2.059)- \Phi(-2.059) \approx 0.9605$,

since $(\mu - \bar{X})/S \sim N(0,1)$.

Answer 1

Preassuming that the $X_i$ are independent here:

• $\frac{4S^2}{\sigma^2}$ has chi-squared distribution with $4$ degrees of freedom.
• $T=\frac{\overline{X}-\mu}{S}$ has a Student's t-distribution with $n-1=5-1=4$ degrees of freedom.
• $T^{2}$ has an F-distribution with parameters $1$ and $4$.

Moreover $\overline X$ and $S^2$ are independent so that the outcome in $(3)$ is actually the product of the outcomes in $(1)$ and $(2)$

Hint on $(1)$:

You can rewrite $P\left(\overline{X}-2.059S\leq\mu\leq\overline{X}-2.059S\right)=P\left(T^{2}\leq2.059^{2}\right)$