Standard Deviation

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February 22nd 2010, 11:25 PM
Punch

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.
February 22nd 2010, 11:32 PM
Punch

$4(4k-1)^2=k^2-4k+4+k^2+k^2+25k^2-10k+4-4(16k^2-8k+1)$
$4(4k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
$16k^2-32k+4=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$

this is my wrong attempt.

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