# Mathematica function to calculate equivalent NormalDistribution from a WeibullDistribution

I have a WeibullDistribution with mean 12.0 and standard deviation 3.2. Is there any built-in Mathematica function that calculates an equivalent NormalDistribution from the WeibullDistribution? Additionally, how do you use Mathematica to find a WeibullDistribution having a given mean and standard deviation?

• Whoever voted the question down: This is a fine question and should be answered. The true question is how to find the "equivalent" parameters for a Weibull and a Normal distribution, given that the parameterization of a Weibull is not in the form of mean and variance (or standard deviation). – David G. Stork Oct 1 '15 at 15:42
• I didn't vote, but it really is not clear what this means. If you know the mean/sd the equivalent would be just the normal distribution with the same mean/sd. LeoRon7, can you edit the question to be more clear? Perhaps you know the weibull parameters and want the mean/sd, which would be essentially the inverse of the answer you got. – george2079 Oct 1 '15 at 15:53
• @DavidG.Stork, well if that's the question, it belongs on math.stackexchange imo... – ciao Oct 1 '15 at 16:53
• There is no equivalent Normal distribution to match your Weibull distribution. For starters, the Weibull has a strictly positive domain of support and is skewed, whereas the Normal is defined on the real line and is symmetrical. – wolfies Oct 1 '15 at 16:57
• @DavidG.Stork the downvote was mine and it was because it is not clear what "equivalent" means in the sense of two different distributions. What should one do about the higher moments? I upvoted your answer, but objectively one cannot be sure that it answers the question. OP also asks for a built-in function, for which I do not understand the motivation at all. It did not seem to me that the question met the threshold of closure, so I thought a downvote was the best way to indicate that it seems poorly posed. I will gladly retract it if the question is improved. – Oleksandr R. Oct 1 '15 at 17:01

If by "equivalent" you mean having the same mean and standard deviation:

NormalDistribution[12, 3.2]


What is the relationship between the mean and variance of a Weibull distribution and its parameters $\alpha$ and $\beta$?

Mean[WeibullDistribution[α, β]]


$\beta \Gamma \left(1+\frac{1}{\alpha}\right)$

Variance[WeibullDistribution[α, β]]


$\beta ^2 \left(\Gamma \left(1+\frac{2}{\alpha }\right)-\Gamma \left(1+\frac{1}{\alpha}\right)^2\right)$

So, if you are given the mean and variance of a Weibull distribution, you must solve (numerically) for $\alpha$ and $\beta$:

FindRoot[{β Gamma[1 + 1/α] == 12,
β^2 (-Gamma[1 + 1/α]^2 + Gamma[1 + 2/α]) == 3.2*3.2},
{{α, 5}, {β, 10}}]


$\{\alpha \to 4.23053,\beta \to 13.1967\}$

(Note that the desired variance is the square of the standard deviation.)

So here is a NormalDistribution and a WeibullDistribution having the same means and standard deviations:

Plot[{PDF[WeibullDistribution[4.23053, 13.1967], x],
PDF[NormalDistribution[12, 3.2], x]},
{x, 0, 20}] • The diagram illustrates neatly that the two distributions are not equivalent ... and thus too that the OP's first question should be dropped or modified, because there is NO equivalent Normal distribution. David's answer above addresses the OP's second question very nicely. However, perhaps the OP should ponder why he is only making use of the first 2 (sample?) moments as the basis to fit his Weibull pdf ... and ignoring higher moments that may readily be available. – wolfies Oct 2 '15 at 17:39