Given a percentile input, calculate the number of standard deviations from 0 using a normal distribution

I need to reverse engineer an old function.  Before I came into the picture my friend had a hard drive crash and lost the original documentation and source code, no backups, OH MY!  :

We only have a short description left:

Function: NormalCDF(Percentile), returns the number of standard deviations "Percentile" is from 0
using the standard normal distribution with mean 0 and standard deviation 1.

The name of the function seems misleading to me, At first I thought it referred to the cumulative distribution function (CDF) of a normal distribution.  But that doesn't seem to make sense from the description.

I'll use Mathematica to implement the new version of this.  It has an internal function to generate a Normal distribution with mean 0 and standard deviation 1:

    NormalDistribution[0, 1]

Maybe the cumulative distribution function has some equivalence to "the number of standard deviations "Percentile" is from 0..."

I can also plot a CDF of the normal distribution like this:

Plot[CDF[NormalDistribution[0, 1], x], {x, -4, 4}]

but I don't feel like this gets me on the right track.

Regardless of the name of the function I need to do this:

Given a percentile input, calculate the number of standard deviations from 0 using a normal distribution

Any ideas?
AlphaSquaredAsked:
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ozoCommented:
It sounds like this function, scaled from 0 to 1
http://mathworld.wolfram.com/InverseErf.html
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aburrCommented:
I may be misunderstanding the problem
Given a percentile input, calculate the number of standard deviations from 0 using a normal distribution
But it seems to me the difficulty is in finding the percentile P not in finding x in terms of standard deviation.
Percentile P is that value of x which divides p(x) into two parts with P% of the total number having values lower than x. If mean = 0 and sd = 1 then x is x standard deviations from 0.
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AlphaSquaredAuthor Commented:
Figured it out. thanks to both of you for moving my thoughts along.

The inverse of the cumulative normal distribution does the trick

Then      
     z=Q^{-1}(x/100)
where x = percentile (x = 75) then the z score corresponding to the 75 percentile =:
     z=Q^{-1}(0.75)
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