1. **Problem statement:** We want to construct a 95% two-sided confidence interval for the population standard deviation $\sigma$ of titanium percentage in an alloy, based on a sample of size $n=51$ with sample standard deviation $s=0.37$. The population is assumed normal. 2. **Formula and explanation:** For a normal population, the confidence interval for the variance $\sigma^2$ is given by $$\left(\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}\right)$$ where $\chi^2_{\alpha/2, n-1}$ and $\chi^2_{1-\alpha/2, n-1}$ are the chi-square critical values with $n-1$ degrees of freedom. 3. Since we want the confidence interval for $\sigma$, take square roots: $$\left(\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}}\right)$$ 4. **Calculate parameters:** - $n=51$ - $s=0.37$ - Degrees of freedom $df = n-1 = 50$ - Confidence level $95\% \Rightarrow \alpha=0.05$ - Critical values from chi-square distribution: - $\chi^2_{0.025, 50} \approx 73.361$ - $\chi^2_{0.975, 50} \approx 31.555$ 5. **Calculate interval bounds:** - Lower bound: $$\sqrt{\frac{50 \times 0.37^2}{73.361}} = \sqrt{\frac{50 \times 0.1369}{73.361}} = \sqrt{\frac{6.845}{73.361}} = \sqrt{0.0933} \approx 0.3055$$ - Upper bound: $$\sqrt{\frac{50 \times 0.37^2}{31.555}} = \sqrt{\frac{6.845}{31.555}} = \sqrt{0.2169} \approx 0.4657$$ 6. **Interpretation:** We are 95% confident that the true population standard deviation $\sigma$ lies between approximately 0.3055 and 0.4657. **Final answer:** $$\boxed{(0.31, 0.47)}$$ (rounded to two decimal places)