[Numpy-discussion] weibull distribution has only one parameter?

D.Hendriks (Dennis) D.Hendriks@tue...
Mon Nov 12 10:44:36 CST 2007


Alan G Isaac wrote:

>On Mon, 12 Nov 2007, "D.Hendriks (Dennis)" apparently wrote: 
>  
>
>>All of this makes me doubt the correctness of the formula 
>>you proposed. 
>>    
>>
>It is always a good idea to hesitate before doubting Robert.
><URL:http://en.wikipedia.org/wiki/Weibull_distribution#Generating_Weibull-distributed_random_variates>
>
>hth,
>Alan Isaac
>  
>
So, you are saying that it was indeed correct? That still leaves the 
question why I can't seem to confirm that in the figure I mentioned (red 
and green lines). Also, if you refer to X = lambda*(-ln(U))^(1/k) as 
'proof' for the validity of the formula, I have to ask if 
Weibull(a,Size) does actually correspond to (-ln(U))^(1/a)?
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