[SOLVED] Weibull distribution problem

**downthesun01** · April 11th 2010, 11:48 PM

Find the probability that X>2

$\alpha = 0.5, \beta = 2$

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

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

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

**mr fantastic** · April 12th 2010, 01:51 AM

Originally Posted by downthesun01

Find the probability that X>2

$\alpha = 0.5, \beta = 2$

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

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

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.

**downthesun01** · April 12th 2010, 02:39 AM

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.

**downthesun01** · April 14th 2010, 01:50 AM

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.

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