1. **Stating the problem:** We are given a sample size $n = 100$, a population variance $\sigma^2 = 239$, and a confidence level of 95%. We want to find the confidence interval for the population mean $\mu$ when the sample mean is 48. 2. **Formula used:** The confidence interval for the population mean when the population variance is known is given by: $$\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean, - $z_{\alpha/2}$ is the z-score corresponding to the desired confidence level, - $\sigma$ is the population standard deviation, - $n$ is the sample size. 3. **Important rules:** - For a 95% confidence level, $z_{\alpha/2} = 1.96$ (from standard normal distribution tables). - The population standard deviation $\sigma$ is the square root of the variance: $\sigma = \sqrt{239}$. 4. **Intermediate work:** Calculate $\sigma$: $$\sigma = \sqrt{239} \approx 15.4596$$ Calculate the standard error: $$SE = \frac{\sigma}{\sqrt{n}} = \frac{15.4596}{\sqrt{100}} = \frac{15.4596}{10} = 1.54596$$ Calculate the margin of error: $$ME = z_{\alpha/2} \times SE = 1.96 \times 1.54596 \approx 3.0313$$ 5. **Calculate the confidence interval:** $$\text{Lower bound} = 48 - 3.0313 = 44.9687$$ $$\text{Upper bound} = 48 + 3.0313 = 51.0313$$ 6. **Final answer:** The 95% confidence interval for the population mean $\mu$ is: $$\boxed{(44.97, 51.03)}$$