Question about calculating the rms noise from sampled DC measurement
The standard equation for computing \$\sigma\$ is:

$$\sigma^2=\frac1N\sum \left(x_i-\overline{x}\right)^2$$

The above is the population variance.

For sample variance there is a slightly different form:

$$\sigma^2=\frac1{N-1}\sum \left(x_i-\overline{x}\right)^2$$

Taking the population variance equation above, for example, a more computationally efficient algorithm is:

$$\sigma^2=\frac1N\left(\sum x_i^2-\frac1N\left(\sum x_i\right)^2\right)$$

Here, you compute two sums. One sum of \$x_i^2\$ and another of \$x_i\$. The problem here is that you are taking the difference of two "large" values. So long as these values of \$x_i\$ are integers and so long as your summation accumulator is sufficiently large enough to avoid overflow, there's no difficulty with the efficient algorithm. 

Also, in most cases, the values of \$x_i\$ are close to each other. So even if the values of \$x_i\$ are in floating point notation, there may be no difficulties.

However, if your values of \$x_i\$ are in floating point notation and if their dynamic range is wide (spans from small values to large values as might be found over the large range of stellar distances), then you will need to pre-sort the values (smallest to largest) prior to performing the summations, so that the smaller values have a change to accumulate sufficiently to have an impact when the larger values begin to be summed into the result.

The more robust algorithm will pre-sort, smallest to largest, all the \$x_i\$ values before performing the summations. It will also be aware that the final result is the difference of two relatively large values. So the remaining precision of the result is limited by the residual significant bits of the result. Just a note to the wise.