If $X_n ∼ N(0, 1/n)$, then why is $\sqrt{n}X_n ∼ N(0, 1)$? I saw the following in a textbook and I have difficulties understanding the concept. 
I understand that $X_n$ is normally distributed with E($X_n$) = 0 and Var($X_n$) = $\frac{1}{n}$.
However,   I do not understand why multiplying   $X_n$ by $\sqrt n$  would make  it standard normal.
 A: Because the variance is a second-order moment, related to squaring, so a factor gets squared. 
More precisely, since the expectation is a  linear operation, in general, if you have a centered $d$-order moment (yours is a  $2$-order moment):
$$\mu_{d}(X)=\operatorname {E} \left[X^{d}\right]=\int _{-\infty }^{\infty }x^{d}\,dF(x)\,$$
multiplying $X$ by a constant results in getting that constant  outside the integral, affected with power $d$ (as long as the integral is properly defined):  
$$\mu_{d}(aX) = a^d \mu_{d}(X)\,.$$
So if you multiply $X$ by $\sqrt{n}$, the variance (power of two moment) of $X$ is multiplied by $(\sqrt{n})^2$.
The mean is a first-order moment, so you would multiply it by $\sqrt{n}$, and as it is $0$ for $X$, the resulting variable still has zero mean.
A: Suppose $X\sim N(\mu, \sigma^{2})$. Then,
\begin{eqnarray*}
X-\mu &\sim& N(0, \sigma^{2})\\
\frac{X-\mu}{\sigma}&\sim&N(0,1).
\end{eqnarray*}
Similarly, if $X_{1},X_{2},\cdots X_{n}$ is a random sample from $N(\mu,\sigma^{2})$, then the sample mean, 
\begin{eqnarray*}
\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}&\sim & N\left(\mu,\frac{\sigma^{2}}{n}\right)\\
\bar{X}-\mu &\sim& N\left(0,\frac{\sigma^{2}}{n}\right)\\
\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}&\sim&N(0,1) 
\end{eqnarray*}
