How to calculate cumulative normal distribution?

Here's an example:

>>> from scipy.stats import norm
>>> norm.cdf(1.96)
0.9750021048517795
>>> norm.cdf(-1.96)
0.024997895148220435

In other words, approximately 95% of the standard normal interval lies within two standard deviations, centered on a standard mean of zero.

If you need the inverse CDF:

>>> norm.ppf(norm.cdf(1.96))
array(1.9599999999999991)

Starting Python 3.8, the standard library provides the NormalDist object as part of the statistics module.

It can be used to get the cumulative distribution function (cdf - probability that a random sample X will be less than or equal to x) for a given mean (mu) and standard deviation (sigma):

from statistics import NormalDist

NormalDist(mu=0, sigma=1).cdf(1.96)
# 0.9750021048517796

Which can be simplified for the standard normal distribution (mu = 0 and sigma = 1):

NormalDist().cdf(1.96)
# 0.9750021048517796

NormalDist().cdf(-1.96)
# 0.024997895148220428

Adapted from here http://mail.python.org/pipermail/python-list/2000-June/039873.html

from math import *
def erfcc(x):
    """Complementary error function."""
    z = abs(x)
    t = 1. / (1. + 0.5*z)
    r = t * exp(-z*z-1.26551223+t*(1.00002368+t*(.37409196+
        t*(.09678418+t*(-.18628806+t*(.27886807+
        t*(-1.13520398+t*(1.48851587+t*(-.82215223+
        t*.17087277)))))))))
    if (x >= 0.):
        return r
    else:
        return 2. - r

def ncdf(x):
    return 1. - 0.5*erfcc(x/(2**0.5))

Taken from above:

from scipy.stats import norm
>>> norm.cdf(1.96)
0.9750021048517795
>>> norm.cdf(-1.96)
0.024997895148220435

For a two-tailed test:

Import numpy as np
z = 1.96
p_value = 2 * norm.cdf(-np.abs(z))
0.04999579029644087

Simple like this:

import math
def my_cdf(x):
    return 0.5*(1+math.erf(x/math.sqrt(2)))

I found the formula in this page https://www.danielsoper.com/statcalc/formulas.aspx?id=55