10.10 Chebyshev s Inequality

Let X be an rv (continuous or discrete) with mean and variance ? ². Chebyshev s inequality states that

That is, the probability of X differing from its mean by more than some positive number ?, is less than or equal to the ratio of the variance ? ² to the square of the number ?. This inequality is often expressed as

where ? = k ? and k > 1.

If, in (10.109), we let k = 2, we obtain

and this implies that the probability of X differing from its mean by more than two standard deviations, is less than or equal to 0.25. That is, the probability that X will lie within two standard deviations of its mean is greater or equal to 0.75 since

Likewise, if k = 3,

or

that is, at least 8/9 or 89% of the data will fall between ? 3 ? and + 3 ?.