Converting a Uniform Distribution to a Normal Distribution

The Ziggurat algorithm is pretty efficient for this, although the Box-Muller transform is easier to implement from scratch (and not crazy slow).

Changing the distribution of any function to another involves using the inverse of the function you want.

In other words, if you aim for a specific probability function p(x) you get the distribution by integrating over it -> d(x) = integral(p(x)) and use its inverse: Inv(d(x)). Now use the random probability function (which have uniform distribution) and cast the result value through the function Inv(d(x)). You should get random values cast with distribution according to the function you chose.

This is the generic math approach - by using it you can now choose any probability or distribution function you have as long as it have inverse or good inverse approximation.

Hope this helped and thanks for the small remark about using the distribution and not the probability itself.

Here is a javascript implementation using the polar form of the Box-Muller transformation.

/*
 * Returns member of set with a given mean and standard deviation
 * mean: mean
 * standard deviation: std_dev 
 */
function createMemberInNormalDistribution(mean,std_dev){
    return mean + (gaussRandom()*std_dev);
}

/*
 * Returns random number in normal distribution centering on 0.
 * ~95% of numbers returned should fall between -2 and 2
 * ie within two standard deviations
 */
function gaussRandom() {
    var u = 2*Math.random()-1;
    var v = 2*Math.random()-1;
    var r = u*u + v*v;
    /*if outside interval [0,1] start over*/
    if(r == 0 || r >= 1) return gaussRandom();

    var c = Math.sqrt(-2*Math.log(r)/r);
    return u*c;

    /* todo: optimize this algorithm by caching (v*c) 
     * and returning next time gaussRandom() is called.
     * left out for simplicity */
}

Use the central limit theorem wikipedia entry mathworld entry to your advantage.

Generate n of the uniformly distributed numbers, sum them, subtract n*0.5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to (1/12) * (1/sqrt(N)) (see wikipedia on uniform distributions for that last one)

n=10 gives you something half decent fast. If you want something more than half decent go for tylers solution (as noted in the wikipedia entry on normal distributions)

I would use Box-Muller. Two things about this:

1. You end up with two values per iteration
   Typically, you cache one value and return the other. On the next call for a sample, you return the cached value.
2. Box-Muller gives a Z-score
   You have to then scale the Z-score by the standard deviation and add the mean to get the full value in the normal distribution.

The standard Python library module random has what you want:

normalvariate(mu, sigma)
Normal distribution. mu is the mean, and sigma is the standard deviation.

For the algorithm itself, take a look at the function in random.py in the Python library.

The manual entry is here

This is my JavaScript implementation of Algorithm P (Polar method for normal deviates) from Section 3.4.1 of Donald Knuth's book The Art of Computer Programming:

function normal_random(mean,stddev)
{
    var V1
    var V2
    var S
    do{
        var U1 = Math.random() // return uniform distributed in [0,1[
        var U2 = Math.random()
        V1 = 2*U1-1
        V2 = 2*U2-1
        S = V1*V1+V2*V2
    }while(S >= 1)
    if(S===0) return 0
    return mean+stddev*(V1*Math.sqrt(-2*Math.log(S)/S))
}

I have the following code which maybe could help:

set.seed(123)
n <- 1000
u <- runif(n) #creates U
x <- -log(u)
y <- runif(n, max=u*sqrt((2*exp(1))/pi)) #create Y
z <- ifelse (y < dnorm(x)/2, -x, NA)
z <- ifelse ((y > dnorm(x)/2) & (y < dnorm(x)), x, z)
z <- z[!is.na(z)]

It is also easier to use the implemented function rnorm() since it is faster than writing a random number generator for the normal distribution. See the following code as prove

n <- length(z)
t0 <- Sys.time()
z <- rnorm(n)
t1 <- Sys.time()
t1-t0

function distRandom(){
  do{
    x=random(DISTRIBUTION_DOMAIN);
  }while(random(DISTRIBUTION_RANGE)>=distributionFunction(x));
  return x;
}