Answer: (71.96, 78.04)

## Explanation To construct a 99% confidence interval for the population mean, we use the formula: $$\mu = \bar{x} \pm Z_{\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}}$$ Where: - $\bar{x} = 75$ (sample mean) - $\sigma = 10$ (population standard deviation) - $n = 72$ (sample size) - $\alpha = 0.01$ (for 99% confidence level) From the standard normal distribution table, $Z_{0.005} = 2.58$ Now calculate the margin of error: $$\text{Margin of Error} = 2.58 \times \frac{10}{\sqrt{72}} = 2.58 \times \frac{10}{8.4853} = 2.58 \times 1.1785 = 3.04$$ Therefore, the confidence interval is: $$75 \pm 3.04 = (71.96, 78.04)$$ This means we are 99% confident that the true population mean score lies between 71.96 and 78.04.

Author: Tanishq Prabhu