Chebyshev's inequality: "It is obvious ..."

Having trouble internalizing the last sentence from my textbook's snippets:

Quote:

For a binomially distributed random variable $\displaystyle X$, it can be shown, that the following holds:

$\displaystyle Var(X)=np(1-p)=npq $

**(16)** [...]

Chebyshev: $\displaystyle t>0 \rightarrow P\big(E(X)-t < X < E(X)+t\big)\geq 1-\frac{D(X)}{t^2}$

[...]

From **(16)** it is obvious, that the probability of a complementary event, i.e. the random variable $\displaystyle X$ having value which differs from $\displaystyle E(X)$ for at least $\displaystyle t$, is at most $\displaystyle \frac{Var(X)}{t^2}$.

I would like to understand it visually (e.g.: probability distribution graph, $\displaystyle E(X)$ is at MAX., now look at the $\displaystyle E(X)\pm t$ points, ...).(Happy)

BTW, I've also found

a similar thread, but could not work out the proof for

*Theorem 4* from

.pdf **mr fantastic** provided:

$\displaystyle Pr(|R-Ex(R)| \geq x) = Pr\big((R-Ex(R))^2 \geq x^2) \leq \frac{Ex\big((R-Ex(R))^2}{x^2} = \frac{Var[R]}{x^2}$. Why does the $\displaystyle \leq$ hold? (Thinking)

PS: Please, feel free to correct any grammatical mistake you find. (Nod)