Random values from a Triangular Distribution

[leebean337]

I am trying to create a function that will extract random values from a triangular distribution (with min=.92, mode=.95, and max=.96). Any help most appreciated. Thanks!

[Jerry W. Lewis]

Explicit formulas for the pdf and cdf are given at
http://mathworld.wolfram.com/Triangu...tribution.html
Specifically, with your values,
CDF = (x-92)^2/12 for 92<=x<=95
= 1-(96-x)^2/4 for 95<=x<=96

The inverse is then
INV = 92+2*sqrt(3*p) for 0<=p<=0.75
= 96-2*sqrt(1-p) for 0.75<=p<=1

You can generate random variates by INV(Rnd()). Implementation is
straightforward in VBA. As cell formulas without a VBA UDF, it would require
a helper column to avoid multiple calls to RAND() for each number.

Jerry

"leebean337" wrote:

>
> I am trying to create a function that will extract random values from a
> triangular distribution (with min=.92, mode=.95, and max=.96). Any
> help most appreciated. Thanks!

[leebean337]

Thanks Jerry - very helpful!