About the width of the distribution of the kinetic energy in a gas In this lecture about statistical mechanics, page $10$, the author said that the kinetic energy $E$ of a gas can be viewed as a random variable (because it is a sum of squared velocities, which themselves are random variables), and that its probability distribution is very sharp around its mean $U=\langle E \rangle$.

He said about the above distribution,

The distribution of the system's total energy $E$ is very sharply peaked around its mean $U=\langle E\rangle$: the width of this peak is $∼ΔE_{rms}/U$,

where $\Delta E _{rms}=\langle (E-\textrm {U})^{2} \rangle ^{1/2}=\sigma$.
What justifies the author's claim that the width of the distribution of $E$ can be approximated by $ΔE_{rms}/U$?
 A: The claim follow from central limit theorem. If you consider your energy per particle $E/N$ to be defined as $$\frac{E}{N} = \frac{1}{N} \sum_i \frac{mv_i^2}{2} $$
then using the central limit theorem (https://en.wikipedia.org/wiki/Central_limit_theorem) you will find that the width of the peak of the distribution of $P(E/N )$ distribution is then $\approx 1/\sqrt{N}$.
Note the supplemental factor $1/N$ used in my formula. This would be the correct way of expressing that the distribution of energy is very peaked.
Now in your lectures, then instead compared the root-mean-square of the distribution of $E$ to its average $U$, in this case the width of your distribution should be $\Delta E _\text{rms}$. As the ratio of $\Delta E _\text{rms} / U \simeq 1/\sqrt{N}$ is very small when you effectively draw your distribution, you will get a very peaked distribution. So the claim in your lecture notes should be that the aspect ratio of your distribution is indeed $\Delta E _\text{rms} / U $, not that the width of the distribution is this ratio.
