1. **Problem statement:** We have a random sample $X_1, \ldots, X_n$ from a normal distribution with mean $\mu$ and variance $\sigma^2$. The sample variance is denoted by $S^2$. We want to find the smallest sample size $n$ such that: a. $\Pr\left(\frac{2S^2}{\sigma^2} \leq 1.5\right) \geq 0.95$ b. $\Pr\left(|5S^2 - \sigma^2| \leq \frac{1}{2}\sigma^2\right) \geq 0.8$ 2. **Key formula and facts:** For a normal distribution, the statistic $$\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}$$ where $\chi^2_{n-1}$ is a chi-square distribution with $n-1$ degrees of freedom. 3. **Part (a):** We want $$\Pr\left(\frac{2S^2}{\sigma^2} \leq 1.5\right) = \Pr\left(S^2 \leq \frac{1.5}{2} \sigma^2\right) = \Pr\left(S^2 \leq 0.75 \sigma^2\right) \geq 0.95$$ Using the chi-square variable, $$\Pr\left(\frac{(n-1)S^2}{\sigma^2} \leq (n-1) \times 0.75\right) \geq 0.95$$ Let $X = \chi^2_{n-1}$. Then $$\Pr(X \leq 0.75(n-1)) \geq 0.95$$ We want the 95th percentile of $X$ to be at least $0.75(n-1)$: $$0.75(n-1) \geq \chi^2_{n-1,0.95}$$ or equivalently $$n-1 \geq \frac{\chi^2_{n-1,0.95}}{0.75}$$ We find the smallest integer $n$ satisfying this by trial or using chi-square tables. 4. **Part (b):** We want $$\Pr\left(|5S^2 - \sigma^2| \leq \frac{1}{2} \sigma^2\right) = \Pr\left(\sigma^2 - \frac{1}{2} \sigma^2 \leq 5S^2 \leq \sigma^2 + \frac{1}{2} \sigma^2\right)$$ Simplify bounds: $$\Pr\left(0.5 \sigma^2 \leq 5S^2 \leq 1.5 \sigma^2\right) = \Pr\left(\frac{0.5}{5} \sigma^2 \leq S^2 \leq \frac{1.5}{5} \sigma^2\right) = \Pr\left(0.1 \sigma^2 \leq S^2 \leq 0.3 \sigma^2\right) \geq 0.8$$ Using chi-square variable $X = \frac{(n-1)S^2}{\sigma^2}$, $$\Pr\left(0.1(n-1) \leq X \leq 0.3(n-1)\right) \geq 0.8$$ We want $$F_{\chi^2}(0.3(n-1)) - F_{\chi^2}(0.1(n-1)) \geq 0.8$$ where $F_{\chi^2}$ is the CDF of $\chi^2_{n-1}$. We find the smallest $n$ satisfying this by trial or numerical methods. 5. **Summary:** - For (a), find smallest $n$ such that $0.75(n-1) \geq \chi^2_{n-1,0.95}$. - For (b), find smallest $n$ such that $F_{\chi^2}(0.3(n-1)) - F_{\chi^2}(0.1(n-1)) \geq 0.8$. These conditions can be solved numerically using chi-square distribution tables or software.