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This is a question that I happen to think of when looking at the Chebyshev's Inequality. In the inequality, it has this: $$ P(\left| X-\mu \right| \ge k\sigma )\le \frac { 1 }{ { k }^{ 2 } } $$

Suppose I have an expectation of $67.5$ and standard deviation of $7.5$. I want to find what probability of $P(30\le X\le 105)$.

$ P(30\le X\le 105)\quad \le \quad P(\left| X-\mu \right| \le 37.5)\\ \qquad \qquad \qquad \qquad \le \quad P(\left| X-\mu \right| \le 5\sigma )\\ \qquad \qquad \qquad \qquad \le \quad 1-P(\left| X-\mu \right| >5\sigma ) $

At this stage, there is a problem. By using Chebyshev's inequality, I can say $P(\left| X-\mu \right| \ge 5\sigma )\le \frac { 1 }{ { 5 }^{ 2 } } $ but in the above steps that I have done, I reached $1-P(\left| X-\mu \right| >5\sigma )$, which is just "more than", not "more than or equals". If I still use $\frac { 1 }{ { 5 }^{ 2 } } $ as its result, the result will not be accurate because something extra is minus out.

What should I do here?

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It seems your problem is not the strictness of the inequality, but that you should be using Chebychev's to put a lower bound on the probability $P(30\leq X \leq 105)$, not an upper bound. – William DeMeo Feb 22 '12 at 9:24
up vote 1 down vote accepted

You have

$$P(30 \leq X \leq 105) = P(|X-\mu| \leq 5 \sigma) = 1 - P(|X-\mu|>5\sigma).$$

By Chebychev's, you have

$$ P(|X-\mu|>5\sigma) \leq P(|X-\mu|\geq5\sigma) \leq 1/5^2.$$

Therefore, $$P(30 \leq X \leq 105) = 1 - P(|X-\mu|>5\sigma) \geq 1 - 1/5^2$$

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If $P(\left| X-\mu \right| \ge k\sigma )\le \dfrac { 1 }{ { k }^{ 2 } }$ then clearly you have $P(\left| X-\mu \right| \gt k\sigma ) \le P(\left| X-\mu \right| \ge k\sigma )\le \dfrac { 1 }{ { k }^{ 2 } }$. You can go further than this and show that you cannot have equality simultaneously in both cases, leading to

$$P(\left| X-\mu \right| \gt k\sigma )\lt \frac { 1 }{ { k }^{ 2 } }$$

but for $k \ge 1$ you can get arbitrarily close to equality, so without more information you cannot do better than this.

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