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**downthesun01** Find the probability that X>2

$\displaystyle \alpha = 0.5, \beta = 2 $

$\displaystyle p(X>2)=1-p(X<2)=\int_{\alpha}^{\beta} \alpha \beta x^{\beta-1}e^{-\alpha x^\beta}dx

$

$\displaystyle =1 - \int_{0.5}^{2} x e^{-0.5 x^2}dx $

Using integration by substitution:

$\displaystyle =1 -(-e^{-0.5x^2}|_{0}^{2}) $

$\displaystyle 1-(-e^-2 - (-1)) = 1-(-e^-2 +1) -= e^{-2} \approx 0.1353$

But that's not the answer that I'm getting when I use Mathematica.

I believe my answer is right but I'm not too sure. Can someone double check it for me? Thanks