Reason for taking the mean of experimental observations Whenever we do some experiment in Physics, e.g. measure Planck's constant, we take the mean of several experimental observations. This seems to be the most obvious way to report a result.
Is there a reasoning based on MLE or other statistics concepts that can be used to justify that the mean is the best way to measure the value over several observations?
 A: If $n$ (independent) measurements are made of some quantity $\mu$, then, assuming
that systematic errors (e.g. bias in the measuring instrument such that the
measurement is always $0.3$ larger than the actual value) have been eliminated,
a suitable model for the measurements is $n$ independent identically distributed random variables $X_1$, $X_2, \ldots, X_n$, with mean $\mu$ and variance $\sigma^2$.
If we ignore $X_2, \ldots, X_n$ and report just the observed value of $X_1$ 
as the value of $\mu$, the variance of the measurement is $\sigma^2$.  If we report the sample mean $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ as the value of $\mu$, then the variance is
$\sigma^2/n$.  Why is the reduction in variance important?  Well, the Chebyshev
inequality says that for a random variable $X$ with mean $\mu$ and variance
$\sigma^2$,
$$
P\{\vert X - \mu \vert \geq \alpha \sigma \} \leq \frac{1}{\alpha^2}.
$$ 
Choosing $\alpha = 5$, we have that if $x_1$ is the observed value of $X_1$,
the $96\%$ confidence interval for $\mu$ is of length $10\sigma$, that is,
$(x_1 - 5\sigma, x_1 + 5\sigma)$ is the $96\%$ confidence interval.
On the other hand, the much shorter interval 
$(\bar{x} - 5\sigma/\sqrt{n}, \bar{x} + 5\sigma/\sqrt{n})$ can be used
if we report $\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$ as the value of $\mu$.
Note the reduction in length by a factor of $\sqrt{n}$.
There are theories going back to Gauss and perhaps even earlier that after things like systematic errors (e.g.  bias in instruments that always give higher readings than the actual value) have been eliminated, the residual error in a measurement can be modeled as a zero-mean normal random variable.  In this case,
we get $(x_1 - 1.97\sigma, x_1 + 1.97\sigma)$ as the $95\%$ confidence interval
for one measurement, and 
$(\bar{x} - 1.97\sigma/\sqrt{n}, \bar{x} + 1.97\sigma/\sqrt{n})$ as the
$95\%$ confidence interval.  Once again, we have the same reduction in
the length of the confidence interval by a factor of $\sqrt{n}$ when 
the average of $n$ measurements is used instead of just one measurement.
