Results 1 to 4 of 4

Math Help - [SOLVED] Weibull distribution problem

  1. #1
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    [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<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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by downthesun01 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. CDF of the Weibull Distribution
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 17th 2013, 03:06 AM
  2. Weibull distribution
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 17th 2013, 03:05 AM
  3. Weibull & Exponential distribution
    Posted in the Statistics Forum
    Replies: 3
    Last Post: June 19th 2010, 05:14 AM
  4. Failure analysis in weibull distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 23rd 2009, 02:50 AM
  5. one-parameter weibull distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 4th 2008, 09:16 PM

Search Tags


/mathhelpforum @mathhelpforum