# Thread: [SOLVED] Weibull distribution problem

1. ## [SOLVED] Weibull distribution problem

Find the probability that X>2

$\alpha = 0.5, \beta = 2$

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

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

Using integration by substitution:
$=1 -(-e^{-0.5x^2}|_{0}^{2})$

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

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

2. Originally Posted by downthesun01
Find the probability that X>2

$\alpha = 0.5, \beta = 2$

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

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

Using integration by substitution:
$=1 -(-e^{-0.5x^2}|_{0}^{2})$

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

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
I get that answer using my CAS, so you're probably making a data entry error when using Mathematica.

3. Super weird. My calculations and my textbook have the Weibull distribution's cumulative mass function as:
$F(x)=1-e^{-\alpha x^\beta}$

But Mathematica has it listed as:

$F(x)=1-e^{(\frac{-x}{\beta})^\alpha}$

Are those two equations equal or something? If they are, I don't see it. I did a problem in the book and got the right answer so I'm going to go on assuming that I'm doing everything right.

4. Well the solutions to my assignment were posted and my solution to this problem was correct. I swear I entered everything correctly in Mathematica, but whatever, I got the problem right. Thank you to Mr. Fantastic for his quick response as always. Much appreciated.