A random sample of $\mathrm{n}$ measurements was selected from a population with unknown mean $\mu$ and known standard deviation. Calculate a $95 \%$ confidence interval for $\mu$ for each of the situations given in parts a through e.
b. $n=200, \bar{x}=101, \sigma^{2}=21$
(Round to two decimal places as needed)

Step 1 ：Given that the sample size \(n = 200\), the sample mean \(\bar{x} = 101\), and the population variance \(\sigma^{2} = 21\). We are asked to calculate a 95% confidence interval for the population mean \(\mu\).

Step 2 ：First, we need to calculate the standard deviation \(\sigma\) from the given variance. The standard deviation is the square root of the variance, so \(\sigma = \sqrt{21} = 4.58257569495584\).

Step 3 ：The formula for the confidence interval for a population mean, given a sample, is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\), where \(Z\) is the Z-score, which corresponds to the desired confidence level. For a 95% confidence level, the Z-score is approximately 1.96.

Step 4 ：Substituting the given values into the formula, we get the lower and upper bounds of the confidence interval as \(101 - 1.96 \frac{4.58257569495584}{\sqrt{200}}\) and \(101 + 1.96 \frac{4.58257569495584}{\sqrt{200}}\) respectively.

Step 5 ：Calculating these expressions, we find that the lower bound is approximately 100.36 and the upper bound is approximately 101.64.

Step 6 ：So, the 95% confidence interval for \(\mu\) is \(\boxed{[100.36, 101.64]}\).