[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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