Normal CDF, Abramowitz–Stegun

easyMathsacceptance —~15 min

Implement norm_cdf(x) — the standard normal CDF Φ(x) with the Abramowitz–Stegun 26.2.17 approximation, no libraries: for x >= 0, t = 1 / (1 + 0.2316419·x) and Φ = 1 - φ(x)·(b1·t + b2·t² + b3·t³ + b4·t⁴ + b5·t⁵) with b = [0.319381530, -0.356563782, 1.781477937, -1.821255978, 1.330274429] and φ the standard normal density. For negative x use symmetry: Φ(x) = 1 - Φ(-x).

This one polynomial is why the workbench prices a whole options chain in under a hundred milliseconds — it is the CDF inside lib/pricing.ts. Return Φ rounded to 6 decimal places.

example 1

input

x=1.0

output

0.841345

example 2

input

x=-1.0

output

0.158655

note

Φ(−1) must equal 1 − Φ(1).

constraints

-8 ≤ x ≤ 8

teacher · 2 hints · the first is free

↗ read the theory

the chapter this problem is cut from

▶ open in the workbench

run the same math on a real tape

●solution.py· utf-8 · spaces:4

from math import exp, pi, sqrt


def norm_cdf(x: float) -> float:
    """Standard normal CDF by Abramowitz–Stegun 26.2.17, rounded to 6 dp."""
    # TODO: the polynomial is only valid for x >= 0 — reflect negative x
    t = 1 / (1 + 0.2316419 * x)
    poly = t * (0.319381530 + t * (-0.356563782 + t * (1.781477937 + t * (-1.821255978 + t * 1.330274429))))
    pdf = exp(-0.5 * x * x) / sqrt(2 * pi)
    return round(1 - pdf * poly, 6)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to norm_cdf()

x=0.0

expected

0.5

the judge calls norm_cdf() with exactly these inputs · outputs are compared to 6 decimal places

● standing bypython 3.13 in your browser · pyodide 0.28.3 · judged on the server to 6 dp⌘↵ run · ⌘⇧↵ submit

paper only · educational · not advice