1. **Problem Statement:** Find the mean and variance of the sampling distribution of the mean given a population variance of 2 and a sample size of 169. Also, 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, denoted as $\mu_{\bar{x}} = \mu$. - The variance of the sampling distribution of the sample mean is given by: $$\sigma^2_{\bar{x}} = \frac{\sigma^2}{n}$$ where $\sigma^2$ is the population variance and $n$ is the sample size. - The standard error of the mean (SEM) is the standard deviation of the sampling distribution of the mean: $$SEM = \sigma_{\bar{x}} = \sqrt{\frac{\sigma^2}{n}}$$ 3. **Given Data:** - Population variance $\sigma^2 = 2$ - Sample size $n = 169$ 4. **Calculations:** - Variance of sampling distribution of mean: $$\sigma^2_{\bar{x}} = \frac{2}{169} = \frac{2}{169} \approx 0.01183$$ - Standard error of mean: $$SEM = \sqrt{0.01183} \approx 0.1087$$ 5. **Interpretation:** - The mean of the sampling distribution of the mean is the same as the population mean (not given explicitly). - The variance of the sampling distribution of the mean is approximately 0.01183. - The standard error of the mean is approximately 0.1087, which measures the variability of the sample mean around the population mean. **Final answers:** - Variance of sampling distribution of mean: $0.01183$ - Standard error of mean: $0.1087$