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Math Help - Standard Deviation - Different way to get value?

  1. #1
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    Standard Deviation - Different way to get value?

    So I learned that standard deviation is as such that if there is a set:
    \{x_1,x_2,x_3, ...,  x_n\}

    with the average:
    a = \displaystyle{\sum_{i=1}^n \frac{x_i}n

    Then the standard deviation is:
    \sqrt{ \frac{(x_1 - a)^2 + (x_2 - a)^2 + ... + (x_n - a)^2}n }



    But a while ago, I was told that this works too (and I checked it, and it seems to work):
    If given the set:
    \{x_1,x_2,x_3, ...,  x_n\}

    with the average
    a = \displaystyle{\sum_{i=1}^n \frac{x_i}n

    and the average of the square of each \tiny x_n:
    s = \displaystyle{\sum_{i=1}^n \frac{(x_i)^2}n

    then the standard deviation is as follows:
    \sqrt{s - a^2}

    Why is it that the second way of getting standard deviation right? In other words, how do I derive the second way from the first way? The main reason I am having trouble figuring this out is that it seems to make the assumption that:
    \frac{(x_n - a)^2}n = \frac{(x_n)^2}n - a^2

    Which if I recall correctly, is not true because:
    \frac{(x_n - a)^2}n = \frac{(x_n)^2 - 2ax_n + a^2}n
    Last edited by ZeroVector; February 17th 2011 at 05:25 PM.
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