Koji was correct.
When you estimate the variance of the population, you have to use unbiased variance (not sample variance). So, the estimate to dx in the equations Koji wrote is
dx = sqrt(sum(xi-xavg)/(n-1))
It is interesting because when n=2, statistical error of the averaged value will be the same as the standard deviation.
dXavg = dx/sqrt(n) = stdev/sqrt(n-1)
In most cases, I think you don't need 10 % precision for statistical error estimation (you should better do correlation analysis if you want to go further). You can simply use dx = stdev if n is sufficiently large (n > 6 from plot below).
Makes sense. I mixed up n and n-1
Probability function: X = (x1 + x2 + ... + xn)/n, where xi = xavg +/- dx
Xavg = xavg*n/n = xavg
dXavg^2 = n*dx^2/n^2
=> dXavg = dx/sqrt(n)
Xavg +/- dXavg = xavg +/- dx/sqrt(n)