Computation that underlies Norm.S.Inv

In writing this, I'm going to assume you have a couple of years of calculus behind you (I'm not sure how I would explain this with only algebra). From the help file for norm.s.inv http://office.microsoft.com/en-us/ma...892.aspx?CTT=1

Originally Posted by Excel help

Given a value for probability, NORM.S.INV seeks that value z such that NORM.S.DIST(z,TRUE) = probability. Thus, precision of NORM.S.INV depends on precision of NORM.S.DIST. This function uses an iterative search technique.

As you've found in the documentation, the probability density function used for norm.s.dist is f(z)=1/sqrt(2pi)*exp(-0.5z^2). If I remember correctly, the cumulative version is a "simple" integration of that function. I'm not enough of a statistician to be certain I'm doing this derivation correctly, but I believe it is close.

1) The cumulative integral is y=integral from -infinity to x of (1/sqrt(2pi)*exp(-0.5z^2)) dz (http://en.wikipedia.org/wiki/Cumulat...ution_function). The help file for the NORM.S.DIST function doesn't say whether they use an exact antiderivative to perform the integration or a numerical "riemann sum" (http://en.wikipedia.org/wiki/Riemann_sum) type technique for doing this integration.
2) The NORM.S.INV function is going to be the solution to that equation (y=integral from -infinity to x of (1/sqrt(2pi)*exp(-0.5z^2)) where we are given y and z and try to find x.
3) Again, not knowing the exact anti-derivative for that integrand, I don't see any way to find x other than to use numerical methods (above and beyond any numerical integration used to perform the integration). The help file only says that they are using an "iterative search technique" without further clarification. If I had to guess, I would guess that they are using something similar to the Newton Raphson method (http://en.wikipedia.org/wiki/Root-finding_algorithm), but I can't say that for sure.

I don't know if that is as specific as you want, but that would seem consistent with what little documentation we have about this function.