1. **State the problem:** We have a sample of size $n=20$ from a normal distribution with sample standard deviation $s=5.8$ mg. We want to find a 90% confidence interval for the population standard deviation $\sigma$. 2. **Formula and explanation:** For a normal population, the confidence interval for the variance $\sigma^2$ is based on the chi-square distribution: $$\left(\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}\right)$$ where $\alpha=1-0.90=0.10$, and $\chi^2_{p, n-1}$ is the chi-square critical value with $n-1$ degrees of freedom at probability $p$. 3. **Calculate degrees of freedom:** $$df = n-1 = 20-1 = 19$$ 4. **Find chi-square critical values:** $$\chi^2_{0.05, 19} = 10.117$$ $$\chi^2_{0.95, 19} = 30.144$$ 5. **Calculate confidence interval for variance:** $$\left(\frac{19 \times 5.8^2}{30.144}, \frac{19 \times 5.8^2}{10.117}\right) = \left(\frac{19 \times 33.64}{30.144}, \frac{19 \times 33.64}{10.117}\right)$$ $$= \left(\frac{639.16}{30.144}, \frac{639.16}{10.117}\right) = (21.21, 63.20)$$ 6. **Convert variance interval to standard deviation interval by taking square roots:** $$\left(\sqrt{21.21}, \sqrt{63.20}\right) = (4.60, 7.95)$$ **Final answer:** The 90% confidence interval for the population standard deviation $\sigma$ is approximately **(4.60 mg, 7.95 mg)**.