Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Re: [mg5309] Mathematica: Weibull distribution fnc.
Replies: 0

 BobHanlon@aol.com Posts: 906 Registered: 12/7/04
Re: [mg5309] Mathematica: Weibull distribution fnc.
Posted: Nov 26, 1996 3:36 AM

This shows how to work with Weibull distribution.

Bob Hanlon (Col, USAF (Retired))
______________________________________

In[1]:=
Needs["Statistics`ContinuousDistributions`"]
Needs["Statistics`DiscreteDistributions`"]
n /: Negative[n] = False;
n /: IntegerQ[n] = True;
Off[General::spell1, General::intinit]

Let the number of failures, X, be a Poisson process with mean (mu t) where mu

is given by

In[6]:=
mu = (1/beta) (t/beta)^(alpha - 1);

In[7]:=
PDF[PoissonDistribution[mu t], n] // Simplify

Out[7]=
t alpha n
((----) )
beta
-----------------
alpha
(t/beta)
E n!

Let T be the time until the first failure. Then

Pr{T <= t} = 1 - Pr{T > t} = 1 - PDF[PoissonDistribution[mu t], 0]

In[8]:=
cdf = 1 - PDF[PoissonDistribution[mu t], 0]

Out[8]=
-1 + alpha
-((t (t/beta) )/beta)
1 - E

The PDF for T is then

In[9]:=
pdf = D[cdf, t] // Simplify

Out[9]=
t alpha
alpha (----)
beta
-----------------
alpha
(t/beta)
E t

In[10]:=
PDF[WeibullDistribution[alpha, beta], t] - pdf == 0 //
PowerExpand

Out[10]=
True

Consequently, T has a Weibull distribution.

In[11]:=
Domain[WeibullDistribution[]]

Out[11]=
{0, Infinity}

The exponential and Rayleigh distributions are special cases of the Weibull
distribution with alpha equal to 1 and 2 respectively:

In[12]:=
PDF[WeibullDistribution[1, 1/lambda], x] ==
PDF[ExponentialDistribution[lambda], x]

Out[12]=
True

In[13]:=
PDF[WeibullDistribution[2, Sqrt[2] sigma], x] ==
PDF[RayleighDistribution[sigma], x]

Out[13]=
True

In[14]:=
Mean[WeibullDistribution[alpha, beta]]

Out[14]=
1
beta Gamma[1 + -----]
alpha

In[15]:=
Variance[WeibullDistribution[alpha, beta]]

Out[15]=
2 1 2 2
beta (-Gamma[1 + -----] + Gamma[1 + -----])
alpha alpha

Generalized moment:

In[16]:=
Integrate[x^t PDF[WeibullDistribution[alpha, beta], x],
{x, 0, Infinity}] // PowerExpand

Out[16]=
t t
beta Gamma[1 + -----]
alpha

FORWARDED MESSAGE:

Subj: [mg5309] Mathematica: Weibull distribution fnc.
Date: Sat, Nov 23, 1996 7:43 AM EDT
From: riglinbd@ml.wpafb.af.mil
X-From: riglinbd@ml.wpafb.af.mil (Brian)
To: mathgroup@smc.vnet.net

I am trying to employ a Weibull distribution in one of the
notebooks I am working on. However, Mathematica provides
scant if any information on how to use this function. Could
some one with more experience than I please enlighten me as
to how the Weibull distribution function found in the
Statistics'ContinuousDistribution package is used.