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Random values from a Triangular Distribution
I am trying to create a function that will extract random values from triangular distribution (with min=.92, mode=.95, and max=.96). An help most appreciated. Thanks -- leebean33 ----------------------------------------------------------------------- leebean337's Profile: http://www.excelforum.com/member.php...fo&userid=3109 View this thread: http://www.excelforum.com/showthread.php?threadid=50806 |
#2
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Random values from a Triangular Distribution
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! |
#3
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Random values from a Triangular Distribution
Thanks Jerry - very helpful! -- leebean337 ------------------------------------------------------------------------ leebean337's Profile: http://www.excelforum.com/member.php...o&userid=31099 View this thread: http://www.excelforum.com/showthread...hreadid=508069 |
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