To compute the 95% confidence interval for a population with a known standard deviation, we can use the formula for the confidence interval of the population mean when the standard deviation is known. This is typically done using a Z-distribution.

The formula for the confidence interval is:

$\text{CI} = \bar{X} \pm Z_{\alpha/2} \times \left(\frac{\sigma}{\sqrt{n}}\right)$

Where:

• $\bar{X}$ is the sample mean (which is given as 98).
• $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z_{\alpha/2} = 1.96$).
• $\sigma$ is the population standard deviation (given as 22).
• $n$ is the sample size.

To compute the confidence interval, we need to know the sample sizes for each part. Could you provide those sample sizes for parts (a) through (d), or would you like me to assume some sample sizes for demonstration?

Let me know if you have any questions or need further details.

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Five related questions:

1. How does the sample size affect the width of a confidence interval?
2. Why do we use the Z-distribution instead of the T-distribution in this case?
3. What happens to the confidence interval if the confidence level increases to 99%?
4. How does the confidence interval change if the population standard deviation is not known?
5. Why is a normal distribution assumption important for calculating confidence intervals?

Tip:

Remember that a larger sample size will result in a narrower confidence interval, making the estimate of the population mean more precise.