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!
 A: 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 $X \geq 13)$.
A: Apply Chebyshev's inequality:
$\mathbb{P}\left(|X-\mu| \geq k\sigma \right) \leq k^{-2}$
Here set $\mu=10$, $k\sigma = 3$ so 
$\mathbb{P}\left(|X-10| \geq 3\right) = 0.2+0.3 = 0.5\leq k^{-2}$
Rearranging $0.5 \leq \left(\dfrac{3}{\sigma}\right)^{-2}$ and
$Var(X) = \sigma^2 \geq \dfrac{9}{2}$.
A: The variance is the average of $(X-\mathbb E(X))^2=(X-10)^2$. Let's call that random variable $Y$. What you have is that 30% of the time, $Y>3^2$ (because $X>13$), and 20% of the time, $Y\geq3^2$ (because $X\leq7$). The rest of the time, $Y$ is between $0$ and $3^2$. In summary, $Y$ is greater than or equal to $9$ at least half of the time, and the rest of the time it's positive. What does that tell you about its average value?
