1. The problem asks for the 99% confidence interval for the population mean when a simple random sample of size $n$ is drawn from a normally distributed population with sample mean $\bar{x}$ and sample standard deviation $s$. 2. The formula for a confidence interval for the population mean when the population standard deviation is unknown and the sample size is large or the population is normal is: $$\bar{x} \pm z^* \cdot \frac{s}{\sqrt{n}}$$ where $z^*$ is the critical z-score corresponding to the desired confidence level. 3. From the table given, the $z^*$-score for a 99% confidence level is 2.58. 4. Therefore, the 99% confidence interval is: $$\bar{x} \pm 2.58 \cdot \frac{s}{\sqrt{n}}$$ 5. This means we take the sample mean $\bar{x}$ and add and subtract the margin of error $2.58 \cdot \frac{s}{\sqrt{n}}$ to get the interval estimate for the population mean with 99% confidence. 6. Among the options given, the correct one is: $$\bar{x} \pm 2.58 \cdot \frac{s}{\sqrt{n}}$$ which matches the last option. Final answer: $\boxed{\bar{x} \pm 2.58 \cdot \frac{s}{\sqrt{n}}}$