hey i was trying to integrate the standard normal distribution to find where t wil equal one half of the distribution
$\displaystyle \int_{-\infty}^{t} \frac{1}{\sqrt{2\pi}}e^{(-x^{2})/2} dx = \frac{1}{2}$
don't you have one of these http://sweb.cz/business.statistics/normal.jpg ?