Answer: (124.5, 135.5)

## Explanation For a sample from a normally distributed population with known standard deviation, the confidence interval for the population mean is calculated as: **Formula:** \(\bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}\) Where: - \(\bar{x} = 130\) (sample mean) - \(\sigma = 20\) (known population standard deviation) - \(n = 50\) (sample size) - \(Z_{\alpha/2} = 1.96\) (for 95% confidence level, two-tailed) **Calculation:** 1. Standard error = \(\frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{50}} = \frac{20}{7.071} \approx 2.828\) 2. Margin of error = \(1.96 \times 2.828 \approx 5.5437\) 3. Confidence interval = \(130 \pm 5.5437\) 4. Lower bound = \(130 - 5.5437 = 124.4563 \approx 124.5\) 5. Upper bound = \(130 + 5.5437 = 135.5437 \approx 135.5\) Therefore, the 95% confidence interval is approximately **(124.5, 135.5)**. **Why other options are incorrect:** - **A (125, 135)**: Too narrow; doesn't account for the full margin of error - **B (1.96 ± 5.5)**: This is not a proper confidence interval format for the mean IQ - **C (130, 135.5)**: Only provides upper bound, not a proper two-sided interval

Author: Nikitesh Somanthe