If $E(X)=15$, $P(X\le11)=0.2$, and $P(X\ge19)=0.3$, what can be $V(X)$?

Question

If $E(X)=15$, $P(X\le11)=0.2$, and $P(X\ge19)=0.3$, which of the following is impossible ?

1. $V(X)\le7$
2. $V(X)\le8$
3. $V(X)>8$
4. $V(X)>7$

I know $V(X)=E(X^2)-E(X)^2$.

Please, any other hints??

Answer 1

Notice that you have $P(|X-E(X)|\geqslant 4)$ and try to use Chebyshev’s inequality: $$\forall\alpha>0,\ P(|X-E(X)|\geqslant \alpha)\leqslant\dfrac{V(X)}{\alpha^2}$$