Suppose that $X$ is a random variable, prove that $\operatorname{Var}(X) \geq 4.5$ Suppose that $X$ is a random variable for which $\mathbb E(X) = 10$, $\Pr(X \leq 7) = 0.2$, and $\Pr(X \geq 13) = 0.3$. Prove that $\operatorname{Var}(X) \geq 4.5$.
I'm not really sure how to start solving this; thanks for any help!

The way to start solving this is to figure out that $|10-7|=|13-10|=3$ so, plugging into the definition of variance,
$$
\text{Var}(X) = E(X-E(X))^2\geq E\left[1_{|X-E(X)|\geq 3}\right] \cdot 3^2 = 9(P(X\leq 7)+P(X\geq 13))=\frac{9}{2},
$$
where $1_{|X-E(X)|\geq 3}$ is the indicator function of the event $|X-E(X)|\geq 3$, i.e. $X \leq 7$ or $X \geq 13$, then $E\left[1_{|X-E(X)|\geq 3}\right] = P(X \leq 7$ or $X \geq 13)$.
