Let $X$ be continuous with pdf $f(x)=e^{-x}$ if $0<x<\infty$, and $0$ otherwise. How can I use Chebychev's Inequality to find a lower bound? I have very few examples work from, and I do understand that I could find the exact probability with the information given but the idea is to use Chebychev.
According to my calculations $\mu=1$ and $\sigma^2=2$. With this in mind I set up the inequality to read:
$$P(|x-1|<2.5)\geq 1-\dfrac{1}{(\dfrac{2.5}{\sqrt2})^2}$$
Which would yield:
$$P(|x-1|<2.5) \geq0.68$$
I have absolutely no examples to work from dealing with Chebychev specifically, and I believe the actual probability is about $0.97$. Is there anyway to provide guidance as to whether my process was correct or not?