The z- is a N(0, 1) distribution, given by the equation:
\[ f(z) = \frac{1}{2 \pi} e^{\frac{-x^2}{2}} \]
The area within an interval (a,b) = normalcdf(a,b) = \[ \int_a^b e^{\frac{-z^2}{2}} \, dz \] (It is not integrable algebraically.)
The Taylor expansion of the above assists in speeding up the calculation:\[ normalcdf(-\infty,z) = \frac{1}{2} + \frac{1}{\sqrt{2 \pi}} \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)2^k k!} \]
Java Normal Probability Calculator
(required JavaScript)
To find the area P under the normal probability curve N(mean,
standard_deviation) within the interval (left, right), type in the 4 parameters
and press "Calculate". The standard normal curve N(0,1) has a mean=0 and
s.d.=1. Use -inf and +inf for infinite limits.
