The general results about the distribution of means of random samples can be applied to the uniform distribution. If \(\overline{X}\) is the mean of a random sample of \(n\) values from a \(\UniformDistn(a, b)\) distribution,
\[\begin{align} E[\overline{X}] & \;=\; E[X] \;=\; \frac{a+b} 2 \\[0.3em] \Var(\overline{X}) & \;=\; \frac {\Var(X)} n \;=\; \frac{ (b - a + 1)^2 - 1} {12n} \end{align} \]Example: Rolling a six-sided die
If \(X \sim \UniformDistn(1, 6)\),
\[\begin{align} E[X] & = \frac{1+6} 2 = 3.5 \\ \Var(X) & = \frac{ (6 - 1 + 1)^2 - 1} {12} = \frac {35} {12} = 2.917 \end{align} \]so for the average of a sample of size \(n\),
\[ \begin{align} E[\overline{X}] &= E[X] = 3.5 \\[0.4em] \Var(\overline{X}) &= \frac {\Var(X)} n = \frac {2.917} n \end{align} \]The barcharts below show the distribution of \(\overline{X}\) for a few sample sizes.
This illustrates that:
The limiting normal distribution should be expected from the Central Limit Theorem.