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Let $x_1,x_2,\ldots,x_n$ be a random sample from a normal distribution with mean $\mu$ and and variance $\sigma^2$. show that $E[\bar x]=\mu$ and $V[\bar x]=\sigma^2 /n$, where $\bar x = \frac1n\sum_i x_i$ is the arithmetic mean of the $x_i$.

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Also what is the distribution of xbar –  david Apr 4 '13 at 16:54
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mew = $\$$\mu$\$$ –  alex.jordan Apr 4 '13 at 16:57
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To show that $E[\bar{x}]=\mu$ and $\mathrm{Var}(\bar{x})=\frac{\sigma^2}{n}$ you need to use that: For any random variables $X$ and $Y$ and constants $a,b$ the following is true $$ {\rm E}[aX+bY]=a{\rm E}[X]+b{\rm E}[Y], $$ and if furthermore $X$ is independent of $Y$ then $$ \mathrm{Var}(aX+bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y). $$ This naturally generalizes to linear combinations of $n$ independent random variables.

To find the distribution of $\bar{x}$, you need to use that if $X\sim\mathcal{N}(\mu_1,\sigma_1^2)$ and $Y\sim\mathcal{N}(\mu_2,\sigma_2^2)$ are two independent normal distributions, then $$ aX+bY\sim \mathcal{N}(a\mu_1+b\mu_2,a^2\sigma_1^2+b^2\sigma_2^2) $$ for any constants $a,b\in\mathbb{R}$. Again, this naturally generalizes to $n$ independent normal distributed variables.

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