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Thread: Standard Deviation

  1. #1
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    Standard Deviation

    Find the value of $\displaystyle k$ for which the mean is equals to the standard deviation for the following quantities. $\displaystyle k-2$, $\displaystyle k$, $\displaystyle k$ and $\displaystyle 5k-2$ where $\displaystyle k$ is not negative.

    Attempt

    Mean= $\displaystyle 4k-1$

    Standard deviation= $\displaystyle \sqrt{\frac{(k-2)^2+(k)^2+(k)^2+(5k-2)^2}{4}-(4k-1)^2}$

    Since mean=standard deviation, $\displaystyle 4k-1=\sqrt{\frac{(k-2)^2+(k)^2+(k)^2+(5k-2)^2}{4}-(4k-1)^2}$

    Facing problems with simplifying this... Need some help with simplifying it and solving for k.
    Last edited by Punch; Feb 22nd 2010 at 11:44 PM.
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  2. #2
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    $\displaystyle 4(4k-1)^2=k^2-4k+4+k^2+k^2+25k^2-10k+4-4(16k^2-8k+1)$
    $\displaystyle 4(4k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
    $\displaystyle 16k^2-32k+4=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$

    this is my wrong attempt.
    Last edited by Punch; Feb 22nd 2010 at 11:44 PM.
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