Say we have $ X_1,\ldots,X_n$ be a random sample from a population with mean $\mu$ and variance $\sigma ^2$, How can I got about proving that
$$ E\left(\sqrt n \, \frac{\overline{X^{(n)}}-\mu}{\sigma}\right)=0$$ and $$Var\left(\sqrt n \, \frac{\overline{X^{(n)}}-\mu}{\sigma}\right)=1$$
What I mean by $\overline{X^{(n)}}$ is the sample mean of $X_1, \ldots, X_n$ .
I guess my impression of finding expected value and variance has always been that there needs be some type of $f(x,y)$ function multiplied by this and summed over, but there isn't one. How can I go about proceeding with this proof?
\begin{align} & \operatorname{E} \overline{X}^{(n)} = \operatorname{E} \frac{X_1+\cdots+X_n} n = \frac 1 n \operatorname{E}(X_1+\cdots+X_n) = \frac 1 n (\operatorname{E}X_1+\cdots+\operatorname{E}X_n) \\[10pt] = {} & \frac 1 n (\mu+\cdots+\mu) = \frac 1 n \cdot n\mu = \mu. \end{align}
Then we can say: \begin{align} & \operatorname{E} \left(\sqrt n \, \frac{\overline{X^{(n)}}-\mu}{\sigma}\right) = \frac{\sqrt n} \sigma \left( \operatorname{E} \bar X^{(n)} - \operatorname{E}(\mu) \right) = \frac{\sqrt n} \sigma (\mu - \mu) = \cdots \end{align}
Next, the variance:
\begin{align} & \operatorname{var} (\bar X^{(n)}) = \operatorname{var}\frac{X_1 + \cdots+X_n} n = \frac 1 {n^2} \operatorname{var}(X_1+\cdots+X_n) \\[10pt] = {} & \frac 1 {n^2} \left( \operatorname{var}(X_1) + \cdots + \operatorname{var}(X_n) \right) \qquad \text{provided }X_1,\ldots,X_n \text{ are uncorrelated} \\[10pt] = {} & \frac 1 {n^2} (\sigma^2 + \cdots+\sigma^2) = \frac 1 {n^2} n\sigma^2 = \frac {\sigma^2} n. \end{align}
And finally:
\begin{align} & \operatorname{var} \left(\sqrt n \, \frac{\overline{X^{(n)}}-\mu}{\sigma}\right) = \left(\frac{\sqrt n} \sigma\right)^2 \operatorname{var} \left( \bar X^{(n)} - \mu \right) = \frac n {\sigma^2} \operatorname{var}(\bar X^{(n)}) = \cdots \end{align}