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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!

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1 Answer 1

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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, $$ P(\overline{Y} = 1/2) = P(Y_1 =0 \ \text{and}\ Y_2=1) + P(Y_1 =1 \ \text{and}\ Y_2=0) = 2p(1-p). $$

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  • $\begingroup$ Thankyou!!!!! You are awesome. Greatly appreciated- $\endgroup$
    – BM97
    Dec 29, 2018 at 17:51

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