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 \(\boxed{(62.42, 63.58)}\).