1. **Problem Statement:** Find the mean and variance of the sampling distribution of the mean. Given the population variance $\sigma^2 = 2$ and sample size $n = 169$, find the standard error of the mean. 2. **Formula and Explanation:** - The mean of the sampling distribution of the sample mean is equal to the population mean $\mu$. - The variance of the sampling distribution of the sample mean is given by: $$\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$$ - The standard error of the mean (SEM) is the standard deviation of the sampling distribution of the sample mean: $$\text{SEM} = \sqrt{\text{Var}(\bar{X})} = \frac{\sigma}{\sqrt{n}}$$ 3. **Calculations:** - Given $\sigma^2 = 2$, so $\sigma = \sqrt{2}$. - Sample size $n = 169$. - Variance of sampling distribution: $$\text{Var}(\bar{X}) = \frac{2}{169} = \frac{2}{169}$$ - Standard error of the mean: $$\text{SEM} = \frac{\sqrt{2}}{\sqrt{169}} = \frac{\sqrt{2}}{13} = \frac{1.4142}{13} \approx 0.1088$$ 4. **Interpretation:** - The mean of the sampling distribution is the same as the population mean. - The variance of the sampling distribution is smaller than the population variance by a factor of the sample size. - The standard error quantifies the variability of the sample mean around the population mean. **Final answer:** - Mean of sampling distribution = $\mu$ (population mean) - Variance of sampling distribution = $\frac{2}{169}$ - Standard error of mean = approximately $0.1088$