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Math Help - verification on formula

  1. #1
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    verification on formula

    I don't know how to write sigma notation here, so I am going to word it instead

    sum of square of the difference of the x and mean = sum of the square of the x minus square of the sum of x divided by n

    I have been using this formula, and I know it works.
    But now that I think about it...n has not been squared. I know x bar is equal to sum of x divided by n

    I don't know whether you can understand what I am trying to say

    I thank you all for your precious time,

    Judi
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  2. #2
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    Quote Originally Posted by Judi View Post
    I don't know how to write sigma notation here, so I am going to word it instead

    sum of square of the difference of the x and mean = sum of the square of the x minus square of the sum of x divided by n

    I have been using this formula, and I know it works.
    But now that I think about it...n has not been squared. I know x bar is equal to sum of x divided by n

    I don't know whether you can understand what I am trying to say

    I thank you all for your precious time,

    Judi
    \sum_{i=1}^{n} (x_i - \mu)^2 = \sum_{i=1}^{n} (x_i)^2 - 2 \sum_{i=1}^{n} x_i \mu + \sum_{i=1}^{n} \mu^2


    \sum_{i=1}^{n} (x_i)^2 - 2 \mu \sum_{i=1}^{n} x_i + n \mu^2


    \sum_{i=1}^{n} (x_i)^2 - 2 \mu (n \mu) + n \mu^2


    \sum_{i=1}^{n} (x_i)^2 - n \mu^2


    \sum_{i=1}^{n} (x_i)^2 - n \left( \frac{\sum_{i=1}^n x_i}{n}\right)^2


    \sum_{i=1}^{n} (x_i)^2 - \frac{\left(\sum_{i=1}^n x_i\right)^2}{n}
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