http://www.quora.com/Why-is-pi-in-the-normal-distribution
but very simply: there’s a relationship between the way the normal distribution curve is mathematically defined and how circles can be defined so you can use descriptive maths of circles (which are of course derided from pi) to describe the shape of the normal curve.
(this bit: Since x2 + y2 is the squared distance from the origin to the point (x, y), this function is constant on each circle of radius r centered at the origin)
The other question of course, is why is the normal distribution curve shaped the way it is in the first place? And that’s described by the Central Limit Theorem, “In probability theory, the central limit theorem (CLT) states conditions under which the mean of a sufficiently large number of independent random variables, each with finite mean and variance, will be approximately normally distributed”
the important bit of which is : “The central limit theorem (in its common form) requires the random variables to be identically distributed.”
http://en.wikipedia.org/wiki/Central_limit_theorem
A simpler decription in that link is:
Forget about the exponential part of the distribution for a moment and focus on the: 1/sqrt2pi
This is equivalent to the radius of a circle whose area is 1/2. Since we know we need to create a distribution that is symmetric about the y-axis we know we need the area under the curve to be 1/2 on each side.
Therefore, and perhaps this is a little simplistic, we can think of exp( -x^2 / 2) as a transformation of the unit circle (circle around (0,0) with area 1 and radius 1/sqrt(pi) ).
In other words, we are stretching a circle infinitely across the x-axis while maintaining an area under the curve of 1, a maximum y value at x=0, and a exponential curve to fit it all together.
and more here
http://math.stackexchange.com/questions/28558/what-do-pi-and-e-stand-for-in-the-normal-distribution-formula