The height of 21-year old women is normally distributed and the population standard deviation is known to be $\sigma=$ 2 in. A random sample of 64 such women is taken and measured. It is found that their mean height is 63 inches. Find the $98 \%$ Confidence Interval for the population mean.

Step 1 ：The problem is asking for a confidence interval for the population mean. The confidence interval can be calculated using the formula: \[\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\] where: \(\bar{x}\) is the sample mean, \(Z\) is the Z-score corresponding to the desired confidence level, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step 2 ：In this case, we know that: \(\bar{x} = 63\) inches, \(\sigma = 2\) inches, \(n = 64\), and the Z-score for a 98% confidence level is approximately 2.33 (from Z-table or using a function to calculate it).

Step 3 ：We can substitute these values into the formula to find the confidence interval.

Step 4 ：Calculate the margin of error: \[Z \frac{\sigma}{\sqrt{n}} = 2.33 \frac{2}{\sqrt{64}} = 0.58\]

Step 5 ：Calculate the lower bound of the confidence interval: \[\bar{x} - \text{margin of error} = 63 - 0.58 = 62.42\]

Step 6 ：Calculate the upper bound of the confidence interval: \[\bar{x} + \text{margin of error} = 63 + 0.58 = 63.58\]

Step 7 ：Final Answer: The $98 \%$ Confidence Interval for the population mean is approximately \(\boxed{(62.42, 63.58)}\).