For a distribution taking on values x_{1}, ..., x_{N}, x_{i} occurring with
probability p_{i}, we know the expected value,
or average, is |

x_{1}p_{1} + x_{2}p_{2} + ... +
x_{N}p_{N} |

Then for a continuous distribution p(x) the expected value, or mean, mu is |

In general, the n^{th} moment of p(x) is |

So mu = M_{1}(p). |

The variance is the average value of
^{2} |

Expanding (x - mu)^{2} and distributing the integrals, we see |

Return to the standard deviation.