A random sample of 87 observations produced a mean $\bar{x}=25.8$ and a standard deviation $s=2.5$. a. Find a $95 \%$ confidence interval for $\mu$. b. Find a $90 \%$ confidence interval for $\mu$. c. Find a $99 \%$ confidence interval for $\mu$. a. The $95 \%$ confidence interval is (Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.)

Step 1 ：We are given a random sample of 87 observations which produced a mean of 25.8 and a standard deviation of 2.5. We are asked to find the 95% confidence interval for the population mean, denoted as \(\mu\).

Step 2 ：The formula for a confidence interval is \(\bar{x} \pm Z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z\) is the Z-score (which depends on the confidence level), \(s\) is the standard deviation, and \(n\) is the sample size.

Step 3 ：For a 95% confidence level, the Z-score is approximately 1.96.

Step 4 ：Substituting the given values into the formula, we get \(25.8 \pm 1.96 \frac{2.5}{\sqrt{87}}\).

Step 5 ：Calculating the margin of error, we get approximately 0.53.

Step 6 ：Subtracting this margin of error from the sample mean, we get the lower bound of the confidence interval, which is approximately 25.27.

Step 7 ：Adding the margin of error to the sample mean, we get the upper bound of the confidence interval, which is approximately 26.33.

Step 8 ：Thus, the 95% confidence interval for \(\mu\) is approximately \(\boxed{(25.27, 26.33)}\).