# 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). – Glen_b -Reinstate Monica Apr 30 '14 at 22:28

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