1. **Problem statement:** We have a sample from a normal population with unknown variance. The sample mean is $\bar{x} = 40$. The 97% lower confidence bound for the population mean $\mu$ is given as $\mu \geq 38.5$. We need to find the 97% upper confidence bound for $\mu$. 2. **Understanding confidence bounds:** - A 97% lower confidence bound means that with 97% confidence, the true mean $\mu$ is greater than or equal to 38.5. - The confidence interval is one-sided here, so the lower bound is 38.5. 3. **Formula for one-sided confidence bound when variance is unknown:** $$\bar{x} - t_{\alpha, n-1} \frac{s}{\sqrt{n}} \leq \mu$$ where $t_{\alpha, n-1}$ is the t-critical value for confidence level $1-\alpha$ with $n-1$ degrees of freedom, $s$ is the sample standard deviation, and $n$ is the sample size. 4. Since the lower bound is 38.5 and the sample mean is 40, the margin of error is: $$40 - 38.5 = 1.5$$ 5. The 97% lower bound corresponds to $\alpha = 0.03$ (3% in the lower tail). The upper bound at 97% confidence corresponds to the same margin of error but on the other side: $$\mu \leq \bar{x} + t_{\alpha, n-1} \frac{s}{\sqrt{n}}$$ 6. Because the confidence level and sample are the same, the margin of error is symmetric. Therefore, the 97% upper confidence bound is: $$40 + 1.5 = 41.5$$ **Final answer:** The 97% upper confidence bound for $\mu$ is $\boxed{41.5}$.