Do the mean lifetimes of short-lived particles follow a Gaussian (or 'normal', or 'standard') distribution? When plotted? For instance, the average 'mean' lifetime of a muon is just over 2 microseconds....
If I plotted many, many muon lifetimes after careful experimentation, would the chart show a Gaussian distribution?
Or would it show some other type of statistical distribution?
 A: A particle's lifespan is has a memoryless and hence exponential distribution. If $\tau$ is the mean lifetime, the lifetime has CDF $1-e^{-t/\tau}$ (so the half-life is $\tau\ln2$) and PDF $\frac{1}{\tau}e^{-t/\tau}$ for $t\ge0$. Therefore, the proportion of surviving particles after a time $t$ is on average $e^{-t/\tau}$, and the mean lifetime of one particle is indeed$$\int_0^\infty\frac{1}{\tau}e^{-t/\tau}tdt\stackrel{x:=t/\tau}{=}\int_0^\infty\tau xe^{-x}dx=\tau.$$Muons have one value of $\tau$; other species will have another.
A: Yes, indeed, the average lifetime will tend to the normal distribution, as required by the Central limit theorem. A caveat: one has to distinguish between the distribution of lifetime and the distribution of the average lifetime.
Distribution of lifetimes can be taken as exponential:
$$
w(t)=\frac{1}{\tau}e^{-\frac{t}{\tau}},$$
where $\tau$ is the mean lifetime. This distribution is obviously not Gaussian/normal.
If we perform $n$ measurements, with results $\{t_1, t_2,...,t_n\}$, we can define the average lifetime
$$
\bar{t}=\frac{1}{n}\sum_{i=1}^nt_i,$$
where each of the $t_i$ is distributed according to the above exponential law. $\bar{t}$ is itself a random variable which varies from an experiment to an experiment, and we can calculate its distribution as:
$$
w_{ave}(\bar{t})=\left\langle \delta\left(\frac{1}{n}\sum_{i=1}^nt_i-\bar{t}\right)\right\rangle=\\
\int_0^{+\infty}dt_1...\int_0^{+\infty}dt_nw(t_1)...w(t_n)\delta\left(\frac{1}{n}\sum_{i=1}^nt_i-\bar{t}\right)=\\
\left(\frac{n}{\tau}\right)^n\frac{\bar{t}^{n-1}}{(n-1)!}e^{-\frac{n\bar{t}}{\tau}},$$
(See the full derivation in the Appendix.)
which is a Gamma distribution with mean $\tau$ and standard deviation $\frac{\tau}{\sqrt{n}}$. In the limit of large $n$ it approaches Gaussian/normal distribution with these parameters.
Remarks
To answer the comments

*

*The distribution approaches the normal distribution in the sense that it can be approximated by the Gaussian to any order of accuracy (provided that we choose $n$ sufficiently high). It however does not converge to the normal distribution in strictly mathematical sense.

*According to the central limit theorem, distribution of a sum of many i.i.d. quantities (independent and identically distributed) approaches normal distribution. A product of many such quantities approaches a log-normal distribution - simply because the log of this product is a sum of many i.i.d. quantities approaching the normal.

Appendix: Gaussian/normal limit
$$
w_{ave}(\bar{t})=
\left(\frac{n}{\tau}\right)^n\frac{\bar{t}^{n-1}}{(n-1)!}e^{-\frac{n\bar{t}}{\tau}}=
\left(\frac{n}{\tau}\right)^n\frac{1}{(n-1)!}e^{-\frac{n\bar{t}}{\tau}+(n-1)\log\bar{t}}
$$
The "phase" $\phi(\bar{t})=-\frac{n\bar{t}}{\tau}+(n-1)\log\bar{t}$
has a minimum at
$$\phi'(\bar{t})=-\frac{n}{\tau}+\frac{n-1}{\bar{t}}=0\Rightarrow \bar{t}^*=\frac{n-1}{n}\tau.\\
$$
Its second derivative at the minimum is
$$\phi''(\bar{t})=-\frac{n-1}{\bar{t}^2}=-\frac{n^2}{(n-1)\tau^2}=-\frac{1}{\sigma^2},
$$
and the Taylor expansion around the minimum is
$$\phi(\bar{t})\approx -(n-1)+(n-1)\log\left(\frac{n-1}{n}\tau\right)-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}\\
$$
We thus can write
$$
w_{ave}(\bar{t})\approx
\left(\frac{n}{\tau}\right)^n\frac{1}{(n-1)!}e^{-(n-1)+(n-1)\log\left(\frac{n-1}{n}\tau\right)-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}=\\
\left(\frac{n}{\tau}\right)^n\frac{1}{(n-1)!}\left(\frac{(n-1)\tau}{ne}\right)^{n-1}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}=
\\
\frac{n}{\tau}\frac{1}{(n-1)!}\left(\frac{n-1}{e}\right)^{n-1}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}.
$$
We can now make use of the Stirling approximation
$$
(n-1)!\approx \sqrt{2\pi(n-1)}\left(\frac{n-1}{e}\right)^{n-1},$$
which gives us
$$
w_{ave}(\bar{t})\approx
\frac{1}{\sqrt{2\pi(n-1)\tau^2/n^2}}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}=
\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}},
$$
which is the normal distribution.
