# How to find sampling distribution of sample mean

So suppose $$Y$$ takes values $$0$$ and $$1$$ with probabilities

$$Pr(Y=1)=p=0.78$$ and $$Pr(Y=0)=1-p=0.22$$

I calculated the mean of Y, which is $$0.78$$ and the variance of Y, which is $$0.1716$$.

I also know that the sampling distribution of the sample mean depends on n. For example, let $$n=2$$, and I want to calculate the sampling distribution of the sample mean .

$$\overline Y$$=sample mean

On the answers sheet, it states that the sampling distribution of the sample mean is :

$$Pr(\overline Y=0)=(1-p)^2=0.22^2=0.0484$$

$$Pr(\overline Y=0.5)=2·p(1-p)=2·0.22·0.78=0.3432$$

$$Pr(\overline Y=1)=p^2=0.78^2=0.6084$$

Now I have two questions:

1) where is the $$\overline Y=0, \overline Y=0.5,\overline Y=1$$ coming from ? Meaning, where are the $$0,0.5,1$$ coming from?

2) Once I know the $$0,0.5,1$$, how to i calculate the sampling distributions? (meaning, where are the $$(1-p)^2,2·p(1-p),p^2$$ coming from ?)

Thanks for the help, I would greatly appreciate simple and clear answers!

You have two (Bernoulli) independent random variables $$Y_1$$ and $$Y_2$$ (with probability of success $$p=0.78$$) and you want to compute the distribution of the sample mean $$\overline{Y} := \frac{Y_1+Y_2}{2}$$.
1) Since the variables $$Y_1$$ and $$Y_2$$ can take only the values $$0$$ and $$1$$, it is clear that the sample mean can take only the values $$0$$ (when $$Y_1 = Y_2 = 0$$), $$1$$ (when $$Y_1 = Y_2 = 1$$) or $$1/2$$ (when $$Y_1=0$$, $$Y_2 = 1$$ or viceversa).
2) $$P(\overline{Y} = 0) = P(Y_1 =0 \ \text{and}\ Y_2=0) = P(Y_1=0)\cdot P(Y_2=0) = (1-p)^2$$ (since $$Y_1$$ and $$Y_2$$ are independent). Similarly $$P(\overline{Y} = 1) = p^2$$. Finally, $$P(\overline{Y} = 1/2) = P(Y_1 =0 \ \text{and}\ Y_2=1) + P(Y_1 =1 \ \text{and}\ Y_2=0) = 2p(1-p).$$