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Re: weibull distribution

  • To: mathgroup at
  • Subject: [mg31399] Re: [mg31375] weibull distribution
  • From: Tomas Garza <tgarza01 at>
  • Date: Wed, 31 Oct 2001 19:59:30 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

Of course there is! You'll find a lot of material in the on-line Help
Browser, especifically on the subject of nonlinear fitting. Fitting problems
are also described extensively in David Withoff's tutorial on Statistics
Examples, in Wolfram's Selected Tutorial Notes on Mathematica. It is
worthwhile getting hold of a copy (very inexpensive, too).  Let me just
reproduce one of Withoff's examples as applied to your question.

This is the Weibull PDF with shape and scale parameters (5,1):

orig = Plot[5*x^4*Exp[-x^5], {x, 0, 1.5}];

These are data from said PDF plus error:

data = Table[{x, 5*x^4*Exp[-x^5] + Random[Real,
      {-0.2, 0.2}]}, {x, 0.3, 1.4, 0.1}];
grdat = ListPlot[data]

This is combined graph of PDF and data:

Show[orig, grdat]

This is Withoff's algorithm for obtaining sum of squares of residuals:

SumOfSquares[data_, model_, vars_] :=
  Module[{func, prediction, response, residuals},
   func = Evaluate /@ Function[vars, model];
    prediction = Apply[func, data, 1];
    response = Last /@ data; residuals =
     response - prediction; residuals . residuals]

This is application of Withoff's function to our data and model with scale
parameter 1:

SS[a_] = SumOfSquares[data, a*x^(a - 1)*Exp[-x^a], x]

This is graph of residuals as a function of shape parameter

Plot[SS[a], {a, -1, 8}]

This finds estimate of shape parameter:

FindMinimum[SS[a], {a, 4}]
{0.13621, {a -> 4.9398}}

Estimate is 4.9398 vs true value of 5, with only 12 observations. Not bad.

Tomas Garza
Mexico City

----- Original Message -----
From: "Sarat Pothuri" <chandrika at>
To: mathgroup at
Subject: [mg31399] [mg31375] weibull distribution

> Hi!
> I am trying to fit data to weibul distribution. Is there a way to do
> using mathematica?
> -
> sarat

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