Answer: [77.658; 92.342]

The mean is 85, the standard deviation is 18, and the sample size is 40. The 99% z-value is 2.58. Confidence interval = $\bar{x} \pm z^* \left(\frac{\sigma}{\sqrt{n}}\right)$ = $85 \pm 2.58 \left(\frac{18}{\sqrt{40}}\right) = [77.658; 92.342]$ **Calculation breakdown:** - Standard error = $\frac{18}{\sqrt{40}} = \frac{18}{6.3246} = 2.846$ - Margin of error = $2.58 \times 2.846 = 7.342$ - Lower bound = $85 - 7.342 = 77.658$ - Upper bound = $85 + 7.342 = 92.342$ Therefore, the 99% confidence interval is [77.658; 92.342].

Author: Tanishq Prabhu