Having followed Mike’s lead (thanks Mike) I have obtained the following:
The NORMSINV function returns the inverse of the standard normal cumulative distribution function for a particular probability. In other words, given a probability, p, the NORMSINV function returns the value, x, of the normally distributed random variable, Z(0,1), which attains that probability. The Excel function adheres to the following mathematical approximation, x(p), of the inverse of the standard normal CDF:
x(p) = t – (c0+c1*t+c2*t^2)/(1+d1*t+d2*t^2+d3*t^3) + error(p) where: t = sqrt(ln(1/p^2))c0 = 2.515517c1 = 0.802853c2 = 0.10328d1 = 1.432788d2 = 0.1889269d3 = 0.001308
With these parameters, abs(error(p))<4.5*10^-4.
It does not however return the same value as NORMSINV. There is no obvious correlation between the result of this formula and the output of NORMSINV!!I am tantilisingly close now so if you know the reason why, please let me know!
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