Fortran Creating a distribution with specific mean and variance in FORTRAN 90

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To create a normal distribution with a mean of 0.5 and a variance of 0.05 in FORTRAN 90, using the RANDOM_NUMBER function directly is not suitable, as it generates uniform distributions. Instead, methods like the Box-Muller transform can be employed to convert uniform random numbers into normally distributed ones. A suggested resource includes C source code for generating normally distributed random numbers, which can be adapted for FORTRAN. This approach is based on established algorithms found in "Numerical Recipes," which also offers a FORTRAN version. The discussion emphasizes the need for a proper transformation technique to achieve the desired normal distribution.
sue132
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Hi,

I'm trying to create a normal distribution with mean 0.5 and variance 0.05. I tried it initially with MATLAB, for which I used
Code: 
newdist=0.5+(randn(1,1000)*sqrt(0.05));

Could you please help me in doing this in FORTRAN 90? Would generating a sequence using RANDOM_NUMBER and using the above equation give me similar results?

Thank you.

(P.S. : The LINUX OS on my system needs to be replaced, and I'm writing some more code before I can run them on another system. It would be great to have your help in the meanwhile, so I could check everything together. Thanks again)
 
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You can't do it directly with random_number, which has a uniform distribution. There are different ways to produce a normal distribution from a uniform distribution, such as the Box-Muller transform.
 
Hi sue132! :smile:

http://www.ohio.edu/people/just/Escalate/source/random.ccsome C source code for nrandom that generates normally distributed random numbers (based on the Box-muller transform).

The algorithm is easy to port to any computer language.
It comes from Numerical Recipes in C, for which there is also a FORTRAN version.
 


Thank you for the replies
 
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