I am studying Probability theory.
I met Chebyshev's inequality chapter.
I understood that Chebyshev's inequality is
$$ P\left({|X-\mu| \ge \epsilon}\right) \le \frac{Var(X)}{\epsilon^2} $$
And I tried to solve the question.
$\mbox{Given that }E(X)=100\mbox{ and } V(X)=625,\\ \mbox{Find the upper bound of }P(X\ge125)\mbox{ using Chebyshev's inequality.}$
My trial: $$ P(X\ge125) =P(X-100 \ge 25) \le P(|X-100| \ge 25) \le\frac{Var(X)}{25^2} = 1\\ \therefore P(X\ge125) \le 1 $$
But my teacher told me that the answer is $\displaystyle\frac12$.
Her trial: $$ P(X\ge125) =P(X-100 \ge 25)=\frac12P(|X-100| \ge 25) \le \frac{Var(X)}{25^2} = 1\\ \therefore \therefore P(X\ge125) \le \frac12 $$
Is is possible to change like the following without random variable $X$ is symmetry?
$$P(X-100 \ge 25) = \frac12P(|X-100| \ge 25)$$
I'm casting doubt on the following X.
I think $P(X-100 \ge 25) = \frac12P(|X-100| \ge 25)$ is impossible for random variable following the above picture.
One cannot in general turn the Chebyshev Inequality into a correct one-sided inequality by simply dividing the tail probability by $2$.
However, there are one-sided inequalities, for example the Cantelli Inequality. This says that $$\Pr(X-\mu\ge \epsilon)\le \frac{\sigma^2}{\sigma^2+\epsilon^2}.$$ In your case, the Cantelli Inequality yields a probability bound of $\frac{625}{625+625}$, exactly the bound mentioned by your teacher.