mean and std deviation of a population equal? Hypothetically, if we have a population of size $n$ whose mean and std deviation are equal, I think with some work we have a constraint that the ratio, (Sum of squared points)/(Sum of points$)^2$ $= \frac{(2n-1)}{n^2}$, which gets small quickly as $n$ gets large. Are there heuristic considerations that might render such a population plausible as an extension of, say, the binomial distribution (as with the Poisson distribution, although that distribution the mean is equal to the variance)? 
Does this property (mean = Sqrt[variance] ) suggest anything about the population generally, if that question is not too vague? I have not encountered a population with this property in any texts, but am fairly sure it has been considered...?

One situation in which the mean is equal to the standard deviation is with the exponential distribution whose probability density is
$$
f(x) = \begin{cases} \frac1\theta e^{-x/\theta} & \text{if }x>0, \\  \\  \\  0 & \text{if } x < 0. \end{cases}
$$
The mean and the standard deviation are both equal to $\theta$.
These are the only distributions that are "memoryless" in the sense that if (capital) $X$ is a random variable that is so distributed, then
$$
\Pr(X > x + y \mid X > y) = \Pr(X > x)
$$
for all positive numbers $x$ and $y$.
How you got $\dfrac{2n-1}{n^2}$ escapes me.
