Mathematical Methods · sample proportions · manipulative

Does (p̂ − p)/√(p̂(1−p̂)/n) look standard normal?

Rule of thumb: need np and n(1−p) both decently large. Extreme p or tiny n → z stays skewed, and p̂ = 0/1 trials are skipped (SE = 0).

population this sample z of p̂ N(0,1) guide

Population — Bernoulli(p)

Standardised proportions z = (p̂ − p)/√(p̂(1−p̂)/n)

Draw samples — each contributes one z below

0.50
10

Changing n recomputes every stored sample’s z from its raw 0/1 draws.

Working

samples
0
mean of z
–
skipped (p̂=0/1)
0
np · n(1−p)
–