FindFit, Weibull

• To: mathgroup at smc.vnet.net
• Subject: [mg92868] FindFit, Weibull
• From: P_ter <petervansummeren at gmail.com>
• Date: Thu, 16 Oct 2008 05:04:48 -0400 (EDT)

```I found some data which seemed to fit a WeibullDistribution. The data are:
pr= {16, 34, 53, 75, 93, 120};
After adding the median rank to the data:
praxis={{16, 0.109101}, {34, 0.26445}, {53, 0.421407}, {75, 0.578593}, {93, 0.73555}, {120, 0.890899}};
I checked with LeastSquares the parameters:
\[Alpha]=1.4301;\[Beta] = 76.318.
So far ok.
But the following did not work:
fit2 = {a, b} /. FindFit[praxis, CDF[WeibullDistribution[a, b], x], {a ,b}, {x}]

So, I used the parameters in the Weibull CDF and the times from pr to get:
praxis1= {{16, 0.106157}, {34, 0.27559}, {53, 0.451308}, {75, 0.623149}, {93, 0.73256}, {120, 0.848082}}
(something went wrong with editing, so maybe I sent a by accident an unfinished message)
I checked with FindFit and it worked. But then I gave extra some variations:
praxis2 = {#, RandomReal[{0.95, 1.05}] CDF[WeibullDistribution[1.4, 76.318], #]} & /@ pr;
fit4 = {a,b} /. FindFit[praxis2, CDF[WeibullDistribution[a,b], x], {a,b}, {x}]
After a few times trying the two above lines FindFit did not work.
It could be ok. But it seems to me that FindFit with a such a small relative variation in the data from 0.95 to 1.05 should still work.
How can I use FindFit in the case of praxis in a safe way?