Step 1 :Given a random sample from a normal distribution: 3, 8, 10, 4, 6, 5.
Step 2 :We are asked to construct a 99% confidence interval for the population mean \(\mu\).
Step 3 :The formula for the confidence interval is \(\bar{x} \pm Z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z_{\alpha/2}\) is the Z-score corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 4 :First, calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\). For the given sample, \(\bar{x} = 6.0\) and \(s = 2.61\) (rounded to two decimal places).
Step 5 :The sample size \(n\) is 6, and the Z-score corresponding to the 99% confidence level \(Z_{\alpha/2}\) is approximately 2.576.
Step 6 :Substitute these values into the formula to get the confidence interval: \((3.26, 8.74)\).
Step 7 :Final Answer: The 99% confidence interval for the population mean \(\mu\) is approximately \(\boxed{(3.26, 8.74)}\).