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Prove that Pearson's second measure of Skewness that is

Skewness = 3(Mean - Median)/Standard deviation

lies between -3 and +3

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1 Answer 1

It would be simpler to state this question as proving that the difference between the mean and the median is less than or equal to one standard deviation.

There are various proofs of this. I think the easiest is to start from the one-sided version of Chebyschev's inequality

$$\Pr(X-\mu \gt k\sigma)\lt \frac{1}{1+k^2}$$

and let $k=1$ so

$$\Pr(X \gt \mu + \sigma)\lt \frac{1}{2}$$

so the median is less than or equal to $\mu + \sigma$ (and, by reversing the sign of $X$ and $\mu$, is also greater than or equal to $\mu - \sigma$). QED

Multiply by 3 for your statement.

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okay thank u so much :) –  linear Jun 22 '11 at 18:56

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