What distribution does the mean of a random sample from a Uniform distribution follow? For example, let $X_1,\cdots,X_n$ be a random sample from $f(x|\theta)=1,\theta-1/2 < x < \theta +1/2$.
Clearly, $X_i \sim U(\theta-1/2 , \theta +1/2)$. Some intuition would suggest that $\bar{X}\sim f(x|\theta)=1,\theta-1/2 < x < \theta +1/2$. However, I do not think that this is actually correct. What distribution would $\bar{X}$ follow?
 A: No, it's not uniform. Intuitively, you would expect that the uncertainty over $\bar X$ decreases as $n$ increases. Also the central limit theorem suggests, as $n$ increases, the distribution approaches normal distribution. Which means, you'll have a peak around $\theta$, and it's going to narrow down as $n\rightarrow\infty$.
For a simple counter-example, if $n=2$, $\bar X$ is going to have triangular distribution, with its center at $\theta$, with the same limits.
A: The Irwin-Hall distribution is the distribution of a sum of $n$ uniform random variables. Therefore, an analytic expression for the density of the mean of $n$ uniform random variables is
$$\frac{1}{n!} \sum_{k=0}^n (-1)^k \binom{n}{k} (x-k)_+^{n-1}$$
By shifting this expression, you get the density of yours.
A: This is one case where using Fourier transforms makes for simple solutions.  Your density function is $\mathrm{rect}(\theta)$ with its Fourier transform $\mathrm{sinc}(f)$ (where $\mathrm{sinc}(f)=\frac{\sin \pi f}{\pi f}$ with the obvious continuation $\mathrm{sinc}(0)=1$).  Adding $n$ variables with that distribution leads to convolving the distribution $n$ times with itself (and dividing by $n$), so the resulting distribution has the Fourier transform $\bigl(\mathrm{sinc}(f)\bigr)^n\over n$.  Doing the inverse transform then delivers
$$\int_{-\infty}^\infty \cos(2\pi f\theta){\bigl(\mathrm{sinc}(f)\bigr)^n\over n}\,\mathrm{d}f$$.  In contrast to the piece-wise defined function in the $\theta$ domain, this is a single expression and thus properties like the moments of the function can be derived from this representation through the Fourier domain.
A: First, you might want to look at Wikipedia on Irwin-Hall distribution.
Unless $n$ is very small $A = \bar X = 
\frac{1}{n}\sum_{i=1}^{n} X_i,$ where
$X_i$ are independently $\mathsf{Unif}(\theta-.5,\theta+.5)$ has $A \stackrel{aprx}{\sim}\mathsf{Norm}(\mu = \theta, \sigma = 1/\sqrt{12n}).$
[The approximation is quite good for $n \ge 10.$ In fact, in the early days of computation when it was expensive to do operations other than pain arithmetic, a common way to simulate a standard normal random variable was to evaluate $Z = \sum_{1=1}^{12} X_i - 6,$ where $X_i$ were generated as independently standard uniform.]
The following simulation in R uses a million samples of size $n = 12$ with $\theta = 5.$
set.seed(2020)  # for reproducibility
m = 10^6;  n = 12;  th = 5
a = replicate(m, mean(runif(n, th-.5,th+.5)))
mean(a);  sd(a); 1/sqrt(12*n)
[1] 5.000153      # aprx 5
[1] 0.08339642    # aprx 1/12
[1] 0.08333333    # 1/12

Thus the mean and standard deviations are consistent
with the results of the Central Limit Theorem.
In R, the Shapiro-Wilk normality test is limited to
5000 observations. We show results for the first 5000
simulated sample means. Those observations are consistent with a normal distribution.
shapiro.test(a[1:5000])

    Shapiro-Wilk normality test

data:  a[1:5000]
W = 0.99979, p-value = 0.9257

The histogram below compares the simulated distribution of $\bar X$ with the PDF of $\mathsf{Norm}(\mu=5, \sigma=1/12).$
hdr = "Simulated Dist'n of Means of Uniform Samples: n = 12"
hist(a, br=30, prob=T, col="skyblue2", main=hdr)
 curve(dnorm(x, 5, 1/sqrt(12*n)), add=T, lwd=2)
 abline(v=5+c(-1,1)*1.96/sqrt(12*n), col="red")


This suggests that $$P\left(-1.96 < \frac{\bar X - \theta}{1/\sqrt{12n}} < 1.96\right) = 0.95,$$ so that a very good approximate 95% confidence interval for $\theta$ is of the form $(\bar X \pm 1.96/\sqrt{12n}).$
