Step 1 :The problem provides the sample mean \(\bar{x} = 18.7\), the sample standard deviation \(s = 5.1\), and the sample size \(n = 39\). It asks for a 95% confidence interval for the population mean \(\mu\).
Step 2 :The formula for a confidence interval is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(z\) is the z-score corresponding to the desired level of confidence.
Step 3 :For a 95% confidence interval, the z-score is approximately 1.96.
Step 4 :Substitute the given values into the formula to calculate the margin of error, E: \(E = z \frac{s}{\sqrt{n}} = 1.96 \frac{5.1}{\sqrt{39}}\).
Step 5 :Calculate the lower and upper bounds of the confidence interval: Lower bound = \(\bar{x} - E = 18.7 - E\), Upper bound = \(\bar{x} + E = 18.7 + E\).
Step 6 :The 95% confidence interval for the population mean, when the sample size is 39, is approximately \(\boxed{[17.10, 20.30]}\).