Binomial Distribution Variance

Each trial contributes p(1 − p) of variance, and independent variances add, so n trials give np(1 − p). The factor peaks at p = 0.5, where a coin is at its most unpredictable, and vanishes at p = 0 or 1, where the outcome is certain. One hundred fair coin tosses have σ² = 100 × 0.5 × 0.5 = 25, so σ = 5: seeing 60 heads is two standard deviations out, unusual but hardly impossible. Enter p as a decimal from 0 to 1, or use the % unit.

This is where poll margins of error come from: a sample of 1000 with p ≈ 0.5 has σ ≈ 15.8 respondents, about 1.6 percentage points, and the familiar "±3 points" is roughly two of those. Notice that the standard deviation grows like $\sqrt{n}$ while the mean grows like n, which is why doubling a sample only cuts relative uncertainty by about 30%. Solving for p has no unique answer, a variance of 24 in 100 trials fits both p = 0.4 and p = 0.6, so the solver returns the one at or below one half and names the other; use the mean μ = np to pin p down.