1. **Problem Statement:** We want to estimate the true average completion time of a training module based on a sample average of 58 minutes from 49 users, with a known population standard deviation of 10 minutes and a 95% confidence level (z = 1.96). 2. **Formula for Standard Error (SE):** The standard error of the sample mean is given by: $$SE = \frac{\sigma}{\sqrt{n}}$$ where $\sigma$ is the population standard deviation and $n$ is the sample size. 3. **Calculate Standard Error:** $$SE = \frac{10}{\sqrt{49}} = \frac{10}{7} = 1.4286$$ 4. **Formula for Confidence Interval (CI):** The 95% confidence interval for the population mean is: $$\bar{x} \pm z \times SE$$ where $\bar{x}$ is the sample mean, $z$ is the z-score for 95% confidence (1.96), and $SE$ is the standard error. 5. **Calculate Confidence Interval:** $$58 \pm 1.96 \times 1.4286 = 58 \pm 2.8$$ So, $$\text{Lower bound} = 58 - 2.8 = 55.2$$ $$\text{Upper bound} = 58 + 2.8 = 60.8$$ 6. **Interpretation:** The 95% confidence interval for the true average completion time is from 55.2 to 60.8 minutes. Since 62 minutes is outside this interval, it is not a reasonable estimate of the true average completion time based on this data. **Final answers:** - Standard Error: $1.4286$ minutes - Confidence Interval: $[55.2, 60.8]$ minutes - Interpretation: 62 minutes is not within the 95% confidence interval, so it is unlikely to be the true average.