If my random variables $X_1,...,X_n$ are i.i.d. $N(\mu,\sigma^2)$, why isn't $\bar{X}\sim N(\mu,0)$? If my random variables $X_1,...,X_n$ are i.i.d. $N(\mu,\sigma^2)$, why isn't $\bar{X}\sim N(\mu,0)$?
In other words, if, as I understand it, $X_1,...,X_n$ all have the same mean, $\mu$, how can there be any variance at all in $\bar{X}$?

Let's look at a simple example in which $n=3.$ And instead of a normal distribution we'll say
$$
X_i = \begin{cases} 1 \\ {} & \text{each with probability }1/2. \\ 0 \end{cases}
$$
Then $\mu= \operatorname E(X_i) = 1/2$ and $\sigma= \sqrt{\operatorname{var}(X_i)} = 1/2.$
As in the question, we let $\overline X =\dfrac{X_1+X_2+X_3} 3.$
What is $\operatorname{var}(\,\overline X\,)$?
We have
$$
(X_1,X_2,X_3) = \begin{cases}
(0,0,0) & \text{so that } \overline X = 0 \\[6pt]
(0,0,1) & \text{so that } \overline X = 1/3 \\ (0,1,0) & \text{so that } \overline X = 1/3 & \\ (1,0,0) & \text{so that } \overline X = 1/3 \\[6pt] (0,1,1) & \text{so that } \overline X = 2/3 \\ (1,0,1) & \text{so that } \overline X = 2/3 \\ (1,1,0) & \text{so that } \overline X = 2/3 \\[6pt] (1,1,1) & \text{so that } \overline X = 1 
\end{cases} \text{ each with probability }1/8.
$$
What then is the variance of $\overline X$?
