To compute a confidence interval for the population mean $\mu$ when the population standard deviation $\sigma$ is known, we use the formula:

$CI = \bar{x} \pm Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$

Where:

• $\bar{x}$ is the sample mean,
• $Z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the confidence level (99% confidence level implies $Z_{\alpha/2} = 2.576$),
• $\sigma$ is the population standard deviation,
• $n$ is the sample size.

Part 2: Given values

• $n = 200$,
• $\bar{x} = 117$,
• $\sigma^2 = 23$ (variance, so $\sigma = \sqrt{23} \approx 4.796$).

1. Calculate the standard error: $SE = \frac{\sigma}{\sqrt{n}} = \frac{4.796}{\sqrt{200}} \approx \frac{4.796}{14.142} \approx 0.339.$

2. Calculate the margin of error: $ME = Z_{\alpha/2} \cdot SE = 2.576 \cdot 0.339 \approx 0.873.$

3. Calculate the confidence interval: $CI = \bar{x} \pm ME = 117 \pm 0.873.$

   $CI = (117 - 0.873, 117 + 0.873) = (116.13, 117.87).$

Final Answer for Part 2:

$(116.13, 117.87)$

If you need further details or have any questions, let me know!

Here are five related questions to explore:

1. What does the 99% confidence level mean in the context of this problem?
2. How would the confidence interval change if the confidence level were reduced to 95%?
3. What happens to the confidence interval if the sample size increases while keeping everything else constant?
4. How does using the sample standard deviation (instead of the population standard deviation) affect the calculation?
5. Why is $Z_{\alpha/2}$ different for different confidence levels?

Tip: Always check whether you’re using population standard deviation ($\sigma$) or sample standard deviation ($s$) when calculating confidence intervals!