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Math Help - Standard Deviation

  1. #1
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    Standard Deviation

    How would I work out the Standard Deviation and the RMSD of

    X 3 4 5 6 7 8 9
    f 2 5 8 14 9 4 3

    Please show me the steps
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  2. #2
    Senior Member TriKri's Avatar
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    Have you checked this out as a start?

    Standard derivation at Wikipedia
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    Yeah I did but I still dont understand. Can someone show me how to work out the Standard deviation with those numbers please?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    Yeah I did but I still dont understand. Can someone show me how to work out the Standard deviation with those numbers please?
    \{ x_i \} is your data set and N is the number of data points.
    First find the mean (average) value, \bar{x}.
    Then calculate x_i - \bar{x} for each data point.
    Then calculate (x_i - \bar{x})^2 for each data point.
    Now add all the (x_i - \bar{x})^2 together to get \sum_{i=1}^N (x_i - \bar{x})^2.
    Now divide this by N to get \sigma ^2 = \frac{1}{N} \sum_{i=1}^N (x_i - \bar{x})^2. ( \sigma ^2 is called the "variance.")
    Now take the square root. \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \bar{x})^2}. This is your standard deviation.

    -Dan
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  5. #5
    Senior Member TriKri's Avatar
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    I think there is something that says you can divide by N-1 as well, probably because if you have only one value, the standard derivation would become 0 if you divided by 1, while it rather should be undefined (as in division by 0 undefined ).

    There is also an alternative way of calculation the standard derivation where you calculate the average value "at the same time" (though it really does the same):

    \{ x_i \} is your data set and n is the number of data points.
    Calculate the sum s of the values: s\ =\ \sum_{i\ =\ 1}^n x_i
    The average value a is obtained by the fraction s/n
    Calculate the sum S of the squares of the values: S\ =\ \sum_{i\ =\ 1}^n x_i^2
    Calculate what the sum of the squares would have been ( S_1) if every value x_i had been the same as the average value: S_1/n\ =\ a^2\ \Rightarrow\ S_1\ =\ (s/n)^2 \cdot n\ =\ s^2/n
    Now let's se how much is missing to obtain the actual sum of the squares: S - S_1
    Now if we divide the difference by n we'll get the variance: \sigma^2\ =\ \frac{S - S_1}{n}\ =\ \frac{S - s^2/n}{n}\ =\ \frac{S}{n} - \frac{s^2}{n^2}\ =\ \frac{\displaystyle{\sum_{i\ =\ 1}^n x_i^2}}{n} - \left(\frac{\displaystyle{\sum_{i\ =\ 1}^n x_i}}{n}\right)^2
    So, the standard derivation \sigma\ =\ \sqrt{\frac{\displaystyle{\sum_{i\ =\ 1}^n x_i^2}}{n} - \left(\frac{\displaystyle{\sum_{i\ =\ 1}^n x_i}}{n}\right)^2}
    And let's say you only want to divide by n-1: \sigma\ =\ \sqrt{\frac{\displaystyle{\sum_{i\ =\ 1}^n x_i^2}}{n-1} - \frac{\left(\displaystyle{\sum_{i\ =\ 1}^n x_i}\right)^2}{n \cdot (n-1)}}

    This method is useful if you have a set of numbers which you already have calculated the standard derivation for with this method, and want to ad a new number to the set and at the same time fast calculate the new standard derivation. Just be sure to store s and S so you are able to quickly update them when a new number is added.

    /Kristofer

    By the way, what does "-Dan" mean?
    Last edited by TriKri; November 17th 2006 at 10:58 AM.
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  6. #6
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    Quote Originally Posted by TriKri View Post
    I think there is something that says you can divide by N-1 as well, probably because if you have only one value, the standard derivation would become 0 if you divided by 1, while it rather should be undefined (as in division by 0 undefined ).

    There is also an alternative way of calculation the standard derivation where you calculate the average value "at the same time" (though it really does the same):

    \{ x_i \} is your data set and n is the number of data points.
    Calculate the sum s of the values: s\ =\ \sum_{i\ =\ 1}^n x_i
    The average value a is obtained by the fraction s/n
    Calculate the sum S of the squares of the values: S\ =\ \sum_{i\ =\ 1}^n x_i^2
    Calculate what the sum of the squares would have been ( S_1) if every value x_i had been the same as the average value: S_1/n\ =\ a^2\ \Rightarrow\ S_1\ =\ (s/n)^2 \cdot n\ =\ s^2/n
    Now let's se how much is missing to obtain the actual sum of the squares: S - S_1
    Now if we divide the difference by n we'll get the variance squared: \sigma^2\ =\ \frac{S - S_1}{n}\ =\ \frac{S - s^2/n}{n}\ =\ \frac{S}{n} - \frac{s^2}{n^2}\ =\ \frac{\displaystyle{\sum_{i\ =\ 1}^n x_i^2}}{n} - \left(\frac{\displaystyle{\sum_{i\ =\ 1}^n x_i}}{n}\right)^2
    So, the variance \sigma\ =\ \sqrt{\frac{\displaystyle{\sum_{i\ =\ 1}^n x_i^2}}{n} - \left(\frac{\displaystyle{\sum_{i\ =\ 1}^n x_i}}{n}\right)^2}
    \sigma^2 is the variance, and \sigma is the standard deviation.

    RonL
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by TriKri View Post
    I think there is something that says you can divide by N-1 as well, probably because if you have only one value, the standard derivation would become 0 if you divided by 1, while it rather should be undefined (as in division by 0 undefined ).
    The "N-1" method gives you an unbiased estimate of the population variance
    from a sample. If you have the distribution then its the "N" form that should
    be used.

    RonL
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    By the way, what does "-Dan" mean?
    Either I'm subtracting the variable D \cdot a \cdot n from somthing, or that would be my name.

    -Dan
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    Either I'm subtracting the variable D \cdot a \cdot n from somthing, or that would be my name.

    -Dan
    I know how I'm betting on this one

    RonL
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  10. #10
    Senior Member TriKri's Avatar
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    Quote Originally Posted by topsquark View Post
    Either I'm subtracting the variable D \cdot a \cdot n from somthing, or that would be my name.

    -Dan
    Okey! For a while I was a little insecure, but now I know.

    -Dan
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  11. #11
    Senior Member TriKri's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    \sigma^2 is the variance, and \sigma is the standard deviation.

    RonL
    Thanks! Fixed.
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