Given the population 2, 3, 6, 8, 11, N=5, n=2, sampling with replacement, calculate the population mean, population standard deviation, and the mean of the sampling distribution of means.

$$\mu = \frac{\sum x}{N} = \frac{30}{5} = 6$$

The mean of the population is $\boxed{6}$.

$$(x - \mu)^2$$ $$(2 - 6)^2 = (-4)^2 = 16$$ $$(3 - 6)^2 = (-3)^2 = 9$$ $$(6 - 6)^2 = (0)^2 = 0$$ $$(8 - 6)^2 = (2)^2 = 4$$ $$(11 - 6)^2 = (5)^2 = 25$$

$$\sigma = \sqrt{\frac{\sum (x - \mu)^2}{N}} = \sqrt{\frac{54}{5}} = \sqrt{10.8}$$ $$\sigma \approx 3.286$$

The standard deviation of the population is $\boxed{3.29}$.

$$\mu_{\bar{x}} = \mu$$ $$\mu_{\bar{x}} = 6$$

The mean of the sampling distribution of means is $\boxed{6}$.

$$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$

Where $\sigma = \sqrt{10.8}$ (from part ii) and $n = 2$.

$$\sigma_{\bar{x}} = \frac{\sqrt{10.8}}{\sqrt{2}} = \sqrt{\frac{10.8}{2}} = \sqrt{5.4}$$ $$\sigma_{\bar{x}} \approx 2.324$$

The standard deviation of the sampling distribution of means is $\boxed{2.32}$.