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From: Damien on 14 Jul 2010 09:54
I think the best way to explain my problem is by using an example:

I’ve tested 2 sets of parts at 2 different pressure levels, for example:
At P1 = 2800 bars, 40 % of the parts failed
At P2 = 3000 bars, 70 % of the parts failed

I know that my parts follow a Lognormal distribution.
I have 2 points: (2800, 40%) and (3000, 70%)
=> I want the 2 parameters (mu, sigma) of that distribution

I managed to get the parameters for a normal distribution, using the following lines:
b= glmfit(Pressure,FailureRate,'normal','link','probit');
Mean = -b(1)/b(2);
Sigma = 1/b(2);

But I can’t find how to do the same for a Lognormal

Thanks for your help !!!
From: Tom Lane on 14 Jul 2010 13:12
>I think the best way to explain my problem is by using an example:
>
> I’ve tested 2 sets of parts at 2 different pressure levels, for
> example:
> At P1 = 2800 bars, 40 % of the parts failed
> At P2 = 3000 bars, 70 % of the parts failed
>
> I know that my parts follow a Lognormal distribution.
> I have 2 points: (2800, 40%) and (3000, 70%)
> => I want the 2 parameters (mu, sigma) of that distribution
>
> I managed to get the parameters for a normal distribution, using the
> following lines:
> b= glmfit(Pressure,FailureRate,'normal','link','probit');
> Mean = -b(1)/b(2);
> Sigma = 1/b(2);
>
> But I can’t find how to do the same for a Lognormal

It is possible to write your own link function to use in place of probit.

But it's not clear to me what it is that follows a normal/lognormal
distribution. Is it that there is some measurement following this
distribution, and you judge pass/fail depending on whether it exceeds some
cutoff, and you determine this pass/fail for two different cutoffs?

If my guess is correct, here is some code that solves for the parameters:

>> f = @(ms) norm([normcdf(2800,ms(1),ms(2))-.4,
>> normcdf(3000,ms(1),ms(2))-.7]);
>> ms = fminsearch(f,[3000 100])
ms =
2865.14892166213 257.152854744613
>> normcdf((2800-ms(1))/ms(2))
ans =
0.400000013747603
>> normcdf((3000-ms(1))/ms(2))
ans =
0.699999985756403

You could easily use logncdf in place of normcdf.

-- Tom


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