Finding standard devition of unknown but non-normal distribution I have a sample of ~3,000 star systems, and I've found the distribution of a particular parameter.  I'd like to find the standard deviation of this distribution, and so I've used the standard deviation equation I'm familiar with:
$\sigma = \sqrt{\frac{1}{N - 1}\sum_{n = 0}^{N}{(x_n^2 - \bar{x}^2)}}\tag{1}$
However, the standard deviation this gives is much too large.  The two graphs below show the distribution and a zoomed-in view of the distribution:

Equation 1 gives $\sigma = 0.0083$, but the graphs above show that $\sigma \approx 0.001$.  The distribution is clearly non-Gaussian.  I'm not sure if Equation 1 holds for non-Gaussian distributions, but I'm unsure about how I would go about calculating a standard deviation.  I'm also not sure if things like the central limit theorem apply here, because that seems limited to random selections from a population, while this is simply the entire population.  What's the correct methodology to find $\sigma$ for this case?

By saying that you bound too much too much of your data, you seem to be saying that you want to bound the middle $68\%$ of the data the way that $(\mu-\sigma, \mu+\sigma)$ does for Gaussian distributions.
To do that, you calculate the empirical quantiles. For the lower endpoint, you want the $0.16$ quantile ($16$th percentile), as $50-(68/2)=16$. For the upper endpoint, you want the $0.84$ quantile, as $50+(68/2)=84$.
If you want a one-number summary, you could say that the middle $68\%$ of the data are contained within $(q_{0.84}-q_{0.16})/2$, where $q_k$ is the $k$th quantile.
(It could be argued that you should take the quantile corresponding to the mean, rather than taking the median, as there are no guarantees that the mean and median are equal. Seeing your graph, I am comfortable assuming them to be equal, however.)
Your calculation of the standard deviation is correct. As you’ve found out, however, the value does not tell you the same information as it would in a Gaussian setting. Unless we make some distribution assumptions, the best we can do it the Chebyshev inequality.
