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It depends ultimately on the underlying distribution, but for most we can assume an approximately normal distribution. For a normal distribution, the confidence intervals are approximately 68.27% for $\mu \pm 1\sigma$ and 95.45 for $\mu \pm 2\sigma$ . As the mean $\mu=50$ and standard deviation $\sigma=10$ , this gives the range 40 to 60 as $\mu \pm \sigma$ , so we should expect about $0.6827 \cdot 1000 = 683$ in this range. Similarly, 30 to 70 is $\mu \pm 2\sigma$ , so we expect about $0.9545 \cdot 1000 = 955$ within that range. These formulae are ultimately derived from numerical integration of the probability density function of the standard normal distribution (normal distribution with mean $\mu=0$ and variance $\sigma^2 =1$ ), which is
$\dfrac{1}{\sqrt{2\pi}}e^{-\tfrac{x^2}{2}}$ .
Thus the fraction of the population within n standard deviations of the mean is approximated by
$\int_{-n}^{n}\dfrac{1}{\sqrt{2\pi}}e^{-\tfrac{x^2}{2}}\, dx$

--Kevin C.