Random numbers

Random numbers
You are encouraged to solve this task according to the task description, using any language you may know.

The goal of this task is to generate a 1000-element array (vector, list, whatever it's called in your language) filled with normally distributed random numbers with a mean of 1.0 and a standard deviation of 0.5

Ada

with Ada.Numerics.Float_Random; use Ada.Numerics.Float_Random;
with Ada.Numerics.Generic_Elementary_Functions;

procedure Normal_Random is
   Seed : Generator;
   function Normal_Distribution(Seed : Generator) return Float is
      package Elementary_Flt is new 
         Ada.Numerics.Generic_Elementary_Functions(Float);
      use Elementary_Flt;
      use Ada.Numerics;
      R1 : Float;
      R2 : Float;
      Mu : constant Float := 1.0;
      Sigma : constant Float := 0.5;
   begin
      R1 := Random(Seed);
      R2 := Random(Seed);
      return Mu + (Sigma * Sqrt(-2.0 * Log(X => R1, Base => 10.0)) * 
         Cos(2.0 * Pi * R2));
   end Normal_Distribution;
      
   type Normal_Array is array(1..1000) of Float;
   Distribution : Normal_Array;
begin
   Reset(Seed);
   for I in Distribution'range loop
      Distribution(I) := Normal_Distribution(Seed);
   end loop;
end Normal_Random;

C plus plus

#include <cstdlib>   // for rand
#include <cmath>     // for atan, sqrt, log, cos
#include <algorithm> // for generate_n

double const pi = 4*std::atan(1.0);

// simple functor for normal distribution
class normal_distribution
{
public:
  normal_distribution(double m, double s): mu(m), sigma(s) {}
  double operator() // returns a single normally distributed number
  {
    double r1 = (std::rand() + 1.0)/(RAND_MAX + 1.0); // gives equal distribution in (0, 1]
    double r2 = (std::rand() + 1.0)/(RAND_MAX + 1.0);
    return mu + sigma * std::sqrt(-2*std::log(r1))*std::cos(2*pi*r2);
  }
private:
  double mu, sigma;
};

int main()
{
  double array[1000];
  std::generate_n(array, 1000, normal_distribution(1.0, 0.5));
}

E

accum [] for _ in 1..1000 { _.with(entropy.nextGaussian()) }

IDL

result = 1.0 + 0.5*randomn(seed,1000)

Java

double[] list = new double[1000];
Random rng = new Random();
for(int i = 0;i<list.length;i++) {
  list[i] = 1.0 + 0.5 * rng.nextGaussian()
}

MAXScript

randList = #()
for i in 1 to 1000 do append randList (random 0.0 2.0)

Perl

map {.5 + rand} 1..1000

Pop11

;;; Choose radians as arguments to trigonometic functions
true -> popradians;

;;; procedure generating standard normal distribution
define random_normal() -> result;
lvars r1 = random0(1.0), r2 = random0(1.0);
     cos(2*pi*r1)*sqrt(-2*log(r2)) -> result
enddefine;

lvars array, i;

;;; Put numbers on the stack
for i from 1 to 1000 do 1.0+0.5*random_normal() endfor;
;;; collect them into array
consvector(1000) -> array;

Python Interpreter: Python 2.5

import random
randList = [random.gauss(1, .5) for i in range(1000)]

Tcl

proc nrand {} {return [expr sqrt(-2*log(rand()))*cos(4*acos(0)*rand())]}
for {set i 0} {$i < 1000} {incr i} {lappend result [expr 1+.5*nrand()]}