in reply to Curved Random Distribution

If all you care about is that your probability distribution qualitatively looks like the classic bell-curve, you can use this quick-and-dirty approximation: 
sub almost_normal {
    my ($mean, $variance) = @_;
    return sqrt($variance) * atan2( rand(2) - 1, 1 ) + $mean;
}

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The reason it works is because the normal distribution is shaped like exp(-x2). A reasonable first-order approximation of exp(x) is (1+x), so we can try using 1/(1+x2) to approximate the normal curve. This function's integral (accumulating probability distribution) is none other than arctan.

This gives you a bell-shaped curve in the shape of 1/(1+x2), which is qualitatively "bell-shaped", but has a lot more probability in its "tails" than the normal distribution.

Update: Augh, this is completely backwards, please ignore. In any case, it should be taking tangents of a random angle (or something?), but I can't get the scaling factors to work out right now. So much for a quick and easy alternative. ;)

Can someone who actually knows this math say (correctly) what I was trying to say? I want the accumulating probability distribution to be shaped like arctan.

Update 2: (trying to salvage this node) After thinking about it for a while, this seems to work (stealing code from BrowserUk's reply): 

#!/usr/bin/perl -slw

my $PI = 4*atan2(1,1);
sub almost_normal {
    my ($mean, $variance) = @_;
    my $x = rand(2*$PI);
    return sqrt($variance) * sin($x)/cos($x) + $mean;
}

our $PRECISION ||= 1;
our $ITERS = 1000 * $PRECISION;

my %plot;

$plot{ int( $PRECISION * almost_normal( 100, 5 ) ) / $PRECISION }++
    for 1 .. $ITERS;

print "Rand 100 +-5";
printf "%7.2f : %-3d : %s\n", $_, 
    $plot{ $_ }, '#' x ( $plot{ $_ } / $ITERS * 100 * $PRECISION )
    for sort{ $a <=> $b } keys %plot;

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The variance is pretty wide, though ... you might be well-served to scale it down a bit. Although it's nice that the tails do extend really far.

blokhead

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