1. **State the problem:** We have a random sample of 15 standardized test scores with a sum of 1042. We want to find the mean and sample standard deviation, then use the normal distribution and empirical rule to estimate a 95% confidence interval for the true population mean. 2. **Calculate the sample mean:** The sample mean $\bar{x}$ is given by $$\bar{x} = \frac{\text{sum of samples}}{\text{number of samples}} = \frac{1042}{15} = 69.4667 \approx 69.5$$ 3. **Sample standard deviation:** Since the problem states to use a statistics calculator, assume the sample standard deviation $s$ is calculated from the sample data. (If not given, it cannot be computed exactly here.) 4. **Empirical rule and confidence interval:** For a 95% confidence interval using the normal distribution, the interval is approximately $$\bar{x} \pm 2 \times \frac{s}{\sqrt{n}}$$ where $n=15$ is the sample size. 5. **Summary:** - Sample mean $\bar{x} = 69.5$ - Sample standard deviation $s$ (from calculator) - Margin of error $= 2 \times \frac{s}{\sqrt{15}}$ 6. **Final 95% confidence interval:** $$\left(69.5 - 2 \times \frac{s}{\sqrt{15}},\ 69.5 + 2 \times \frac{s}{\sqrt{15}}\right)$$ Round the interval endpoints to the nearest tenth once $s$ is known. **Note:** Without the exact sample standard deviation value, the numeric interval cannot be completed here.