Random Gaussian Variables

Random Gaussian Variables

Jarrett's suggestion of using a Box-Muller transform is good for a quick-and-dirty solution. A simple implementation:

Random rand = new Random(); //reuse this if you are generating many
double u1 = 1.0-rand.NextDouble(); //uniform(0,1] random doubles
double u2 = 1.0-rand.NextDouble();
double randStdNormal = Math.Sqrt(-2.0 * Math.Log(u1)) *
             Math.Sin(2.0 * Math.PI * u2); //random normal(0,1)
double randNormal =
             mean + stdDev * randStdNormal; //random normal(mean,stdDev^2)

Generate random numbers following a normal distribution in C/C++

There are many methods to generate Gaussian-distributed numbers from a regular RNG.

The Box-Muller transform is commonly used. It correctly produces values with a normal distribution. The math is easy. You generate two (uniform) random numbers, and by applying an formula to them, you get two normally distributed random numbers. Return one, and save the other for the next request for a random number.

How to generate a random normal distribution of integers

One other way to get a discrete distribution that looks like the normal distribution is to draw from a multinomial distribution where the probabilities are calculated from a normal distribution.

import scipy.stats as ss
import numpy as np
import matplotlib.pyplot as plt

x = np.arange(-10, 11)
xU, xL = x + 0.5, x - 0.5 
prob = ss.norm.cdf(xU, scale = 3) - ss.norm.cdf(xL, scale = 3)
prob = prob / prob.sum() # normalize the probabilities so their sum is 1
nums = np.random.choice(x, size = 10000, p = prob)
plt.hist(nums, bins = len(x))

Here, np.random.choice picks an integer from [-10, 10]. The probability for selecting an element, say 0, is calculated by p(-0.5 < x < 0.5) where x is a normal random variable with mean zero and standard deviation 3. I chose a std. dev. of 3 because this way p(-10 < x < 10) is almost 1.

The result looks like this:

Generating random numbers with normal distribution in Excel

Use the NORMINV function together with RAND():

=NORMINV(RAND(),10,7)

To keep your set of random values from changing, select all the values, copy them, and then paste (special) the values back into the same range.

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Sample output (column A), 500 numbers generated with this formula:

JavaScript Math.random Normal distribution (Gaussian bell curve)?

I want to know if the JavaScript function Math.random is normal distribution or not

Javascript Math.random is not a Normal Distribution(Gaussian bell curve). From ES 2015, 20.2.2.27 "Returns a Number value with positive sign, greater than or equal to 0 but less than 1, chosen randomly or pseudo randomly with approximately uniform distribution over that range, using an implementation-dependent algorithm or strategy. This function takes no arguments." So the provided collection when n is high enough we will get approximately uniform distribution. All values in the interval will have equal probability of appearance(straight line parallel to the x axis, denoting number between 0.0 and 1.0).

how can I get numbers which are normal distribution

There are several ways of getting collection of numbers with a normal distribution. As answered by Maxwell Collard the Box-Muller transform
does transform uniform distribution to normal distribution(the code can be found in Maxwell Collard answer).

An answer to another stackoverflow answer to a question has a reply with other uniform distribution to normal distribution algorithms. Such as:
Ziggurat,
Ratio-of-uniforms,
Inverting the CDF
Besides one of the answers says that: says:

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

And finally

I want to rebuilt a Schmidt-machine (German physicist), the machine produces random numbers of 0 or 1 and they have to be normal distributed so I can draw them in Gaussian bell curve.

When we have only two values (0 or 1) Gaussian curve looks the same as uniform distribution with 2 possible values. That is why a simple

function randomZero_One(){
    return Math.round(Math.random());
}

would suffice. It would return pseudo-randomly with approximately equal probability values 0 and 1.

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