# How to compute Chebyshev bounds on probabilities: one- or two-sided inequality?

Suppose the distribution of scores of a test has mean 100 and standard deviation 16. Calculate an upper bound for the probability $P\{X>148\text{ or }X<52\}$.

Here is my progress:

By the addivity axiom, $P\{X>148\text{ or }X<52\}=P\{X>148\}+P\{X<52\}$.
Can I use Chebyshev's Inequality on both probabilities or do I use the one-sided Chebyshev inequality? Or do I use the corollary from the one-sided Chebyshev inequality (stated below)?

$P\{X\ge \mu+a\} \le \frac{\sigma^2}{\sigma^2+a^2}$ (1)

$P\{X\le \mu-a\} \le \frac{\sigma^2}{\sigma^2+a^2}$ (2)

Being that the problem states $\mu$, $\sigma$ and $a$, I believe I should use (1) for $P\{X>148\}$ and (2) for $P\{X<52\}$ to get an upper bound.

• Given that all these inequalities are applicable, why not do the question both ways and see which result is better (that is, smaller)? There's a certain ambiguity to this question in any event: what does this "probability" mean? Although you seem to interpret it as a proportion of actual test scores (and there's nothing wrong with that), in similar circumstances it could be interpreted as an estimate of a property of some underlying distribution assumed to describe the test outcomes.
– whuber
Apr 30 '14 at 21:19
• Since it's symmetric about the mean, the two sided calculation would be easiest. If you calculate it both ways (as whuber suggested, and which I strongly suggest you try), it should also be tighter, for fairly obvious reasons (even without looking at the formula, it's clear from general reasoning that it should be so). Apr 30 '14 at 22:28

Notice how $148-100=100-52=3\cdot16$...
and here's the inequality: $Pr(|X-\mu|\ge k\sigma)\le\frac{1}{k^2}$