K A random sample of $n$ measurements was seleciea irom 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. a. $n=75, \bar{x}=35, \sigma^{2}=14$ $(34.16,35.84)$ (Round to two decimal places as needed) b. $n=200, \bar{x}=101, \sigma^{2}=21$ (Round to two decimal places as needed)

Step 1 ：The problem is asking for a 95% confidence interval for the population mean. The formula for a confidence interval is given by: \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\) where \(\bar{x}\) is the sample mean, \(z\) is the z-score (which for a 95% confidence interval is approximately 1.96), \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size.

Step 2 ：For part a, we are given: \(n = 75\), \(\bar{x} = 35\), and \(\sigma^{2} = 14\), so \(\sigma = \sqrt{14}\).

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

Step 4 ：Calculate the lower bound of the confidence interval: \(\bar{x} - z \frac{\sigma}{\sqrt{n}} = 35 - 1.96 \frac{\sqrt{14}}{\sqrt{75}} = 34.15318321540804\).

Step 5 ：Calculate the upper bound of the confidence interval: \(\bar{x} + z \frac{\sigma}{\sqrt{n}} = 35 + 1.96 \frac{\sqrt{14}}{\sqrt{75}} = 35.84681678459196\).

Step 6 ：Round the lower and upper bounds to two decimal places to get the final answer: \((34.16, 35.84)\).

Step 7 ：Final Answer: The 95% confidence interval for the population mean is \(\boxed{(34.16, 35.84)}\).