Use the following sample to estimate a population mean $\mu$. \[ \begin{array}{|r|} \hline 58.8 \\ \hline 71.6 \\ \hline 82.4 \\ \hline 86.4 \\ \hline 94.1 \\ \hline 95.4 \\ \hline 101.9 \\ \hline 79.9 \\ \hline \end{array} \] \begin{tabular}{|r|} \hline 58.8 \\ \hline 71.6 \\ \hline 82.4 \\ \hline 86.4 \\ \hline 94.1 \\ \hline 95.4 \\ \hline 101.9 \\ \hline 79.9 \\ \hline \end{tabular} Assuming the population is normally distributed, find the $99.9 \%$ confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places. \[ 99.9 \% \text { C. } 1 .= \]

Step 1 ：Given a sample of data: 58.8, 71.6, 82.4, 86.4, 94.1, 95.4, 101.9, 79.9. We are asked to estimate a population mean $\mu$ and find the $99.9\%$ confidence interval about the population mean.

Step 2 ：First, we calculate the sample mean $\bar{x}$, which is the sum of all the values divided by the number of values. In this case, $\bar{x} = \frac{58.8 + 71.6 + 82.4 + 86.4 + 94.1 + 95.4 + 101.9 + 79.9}{8} = 83.8125$.

Step 3 ：Next, we calculate the sample standard deviation $s$, which is the square root of the variance. The variance is the average of the squared differences from the mean. In this case, $s = \sqrt{\frac{(58.8-83.8125)^2 + (71.6-83.8125)^2 + (82.4-83.8125)^2 + (86.4-83.8125)^2 + (94.1-83.8125)^2 + (95.4-83.8125)^2 + (101.9-83.8125)^2 + (79.9-83.8125)^2}{8-1}} = 13.969501218215143$.

Step 4 ：We then find the z-score corresponding to the desired confidence level. For a $99.9\%$ confidence level, the z-score is approximately 3.29.

Step 5 ：Finally, we calculate the confidence interval using the formula $\bar{x} \pm z \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score, $s$ is the sample standard deviation, and $n$ is the sample size. In this case, the confidence interval is $83.8125 \pm 3.29 \frac{13.969501218215143}{\sqrt{8}}$, which simplifies to approximately $(67.56, 100.06)$.

Step 6 ：\(\boxed{\text{Final Answer: The } 99.9\% \text{ confidence interval about the population mean is approximately } (67.56, 100.06). \text{ So, we are } 99.9\% \text{ confident that the true population mean falls within this interval.}}\)