1. **State the problem:** We have a sample of 900 days, with 100 foggy days. We want to find the 99% confidence interval for the percentage of foggy days in the district. 2. **Formula used:** For a proportion $p$, the confidence interval is given by $$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z_{\alpha/2}$ is the z-value for the desired confidence level. 3. **Calculate sample proportion:** $$\hat{p} = \frac{100}{900} = \frac{1}{9} \approx 0.1111$$ 4. **Find z-value for 99% confidence:** For 99% confidence, $\alpha = 0.01$, so $z_{0.005} \approx 2.576$. 5. **Calculate standard error:** $$SE = \sqrt{\frac{0.1111 \times (1 - 0.1111)}{900}} = \sqrt{\frac{0.1111 \times 0.8889}{900}} = \sqrt{\frac{0.09877}{900}} = \sqrt{0.0001097} \approx 0.01047$$ 6. **Calculate margin of error:** $$ME = 2.576 \times 0.01047 \approx 0.02697$$ 7. **Calculate confidence interval:** $$0.1111 \pm 0.02697 = (0.0841, 0.1381)$$ 8. **Convert to percentage:** $$8.41\% \text{ to } 13.81\%$$ **Final answer:** The 99% confidence interval for the percentage of foggy days is approximately **8.41% to 13.81%**.