Calculating percentile of normal distribution See this Wikipedia page: Binomial proportion confidence interval.
To get the Agresti-Coull Interval, one needs to calculate a percentile of the normal distribution, called $z$. How do I calculate the percentile? Is there a ready-made function that does this in Wolfram Mathematica and/or Python/NumPy/SciPy?
 A: Well, you didn't ask about R, but in R you do it using ?qnorm
(It's actually the quantile, not the percentile, or so I believe)
> qnorm(.5)
[1] 0
> qnorm(.95)
[1] 1.644854

A: In Python, you can use the stats module from the scipy package (look for cdf(), as in the following example).
(It seems the transcendantal package also includes usual cumulative distributions).
A: For Mathematica $VersionNumber > 5 you can use
Quantile[NormalDistribution[μ, σ], 100 q]

for the q-th percentile.
Otherwise, you have to load the appropriate Statistics package first.
A: John Cook's page, Distributions in Scipy, is a good reference for this type of stuff:
In [15]: import scipy.stats

In [16]: scipy.stats.norm.ppf(0.975)
Out[16]: 1.959963984540054

A: You can use the inverse erf function, which is available in MatLab and Mathematica, for instance.
For the normal CDF, starting from
$$y=\Phi\left(x\right)=\frac{1}{2}\left[1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$$
We get
$$x=\sqrt{2}\ \text{erf}^{-1}\left(2y-1\right)$$
For the log-normal CDF, starting from
$$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}\text{erfc}\left(\frac{-\log x-\mu}{\sigma\sqrt{2}}\right)$$
We get
$$-\log \left(x\right)=\mu+\sigma\sqrt{2}\ \text{erfc}^{-1}\left(2y\right)$$
