Chevyshev inequality for difference of random variables

Question

Let $X$, $Y$ be two random samples with replacement of a population such that $E(X)$= 9 and $\sigma(X)$=2. Find a number $c$ such that $P(|X-Y|>c)\ge 0.99$

Using triangle inequality: $$P(|X-Y|>c)=1-P(|X-Y|\le c)=1-P(|X-9+9-Y|\le c)=1-P(|X-9|+|Y-9|\le c)$$

I want to use Chevyshev inequality but I don´t know how to proceed from here. I would really appreciate and hint or suggestion.

Answer 1

Following the hint from @jameselmore

$$P(|X-Y| \gt c) = P(X-Y\lt -c)+P(X-Y\gt c) = P(Y-X \gt c) + P(X-Y \gt c) $$

Due to symmetry, $P(X-Y \gt c) = P(Y-X \gt c)$

$$ = 2P(X-Y \gt c) \gt .99$$

$$ =P(X-Y \gt c) \gt 0.495$$

$$P(\frac{Z}{\sqrt{8}} \gt \frac{c}{\sqrt{8}}) > 0.495$$

$$P(z \gt \frac{c}{\sqrt{8}}) >0.495$$

$$ P(z \lt \frac{c}{\sqrt{8}}) \lt .505$$

$$\frac{c}{\sqrt{8}} = \Phi^{-1}(0.505) = 0.01253347$$

$$ c = \sqrt{8}\times 0.012533$$

$$ c = 0.03545$$