Hi Mattrnfnr!

Using functions that are available on graphical calculators, the formula would be:

$\displaystyle p = \text{normcdf}(z) - \text{normcdf}(-z)$

The inverse is:

$\displaystyle z = \text{norminv}(\frac 1 2 p + \frac 1 2)$

So if you want a 95% confidence interval, you need:

$\displaystyle z = \text{norminv}(\frac 1 2 \cdot 0.95 + \frac 1 2) = \text{norminv}(0.975)$

Btw, the normcdf(z) function is defined as:

$\displaystyle \text{normcdf}(z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^z e^{- \dfrac 1 2 x^2} dx$

Another name for the same function is $\displaystyle \Phi(z)$.

This is the formula Mark referred to (with a few slight modifications... )