Show that $P(0<X<6)>\dfrac{2}{3}$

Question

Show that $P(0<X<6)>\dfrac{2}{3}$, if $X$ has probability density function $f(x)=\dfrac{e^{-x}x^2}{2}$, if $x>0$ and $0$, otherwise.

Now, $E(X)=3$ and $Var(X)=3$, so I can not use Chebyshev's inequality. Moreover I used the inequality, $e^{-x}\geq \dfrac{x^2}{2}-x+1$, but also this not working here.

What inequality I should use here? Thanks for any help.

Answer 1

$\mathbb{P}(0<X<6) = 1 - \mathbb{P}(|X-\mathbb{E}(X)|\geq 3)$.

Now you can try with Chebyshev's inequality.

Answer 2

We have that $$P(0<X<6)=\int_{0}^{6}f(x)\mathrm{d}x$$ $$=\frac{1}{2}\int_{0}^{6}e^{-x}x^2\mathrm{d}x$$ $$=\frac{1}{2}\left.e^{-x}(-x^2-2x-2)\right|_{x=0}^{6}$$ $$=1-25e^{-6} \approx 0.93803>\frac{2}{3}$$