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Math Help - Standard Deviation

  1. #1
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    Standard Deviation

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

    Attempt

    Mean= 4k-1

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

    Since mean=standard deviation, 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; February 23rd 2010 at 12:44 AM.
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  2. #2
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    4(4k-1)^2=k^2-4k+4+k^2+k^2+25k^2-10k+4-4(16k^2-8k+1)
    4(4k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4
    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; February 23rd 2010 at 12:44 AM.
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