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?