Standard Deviation

For a discrete random variable: first determine the expected value μ = Σx_i·p_i. Then for each value form the deviation x_i − μ, square it and weight it with p_i; the sum is the variance Var(X) = Σ(x_i − μ)²·p_i. Finally take the square root: σ = √Var(X). Die example: $\mu{} = 3.5$, $E(X^{2}) =\frac{ 91}{6} \approx{} 15.17$, so Var = 15.17 − 3.5² = 2.92 and $\sigma{} \approx{} 1.71$. Often faster is the shortcut formula Var(X) = E(X²) − μ², where you only average the squared values. For binomially distributed variables you skip everything: $\sigma{} = \sqrt{(n\cdot{}p\cdot{}(1-p))}$. Important: take the root only at the very end, never of the individual deviations.

σ is the theoretical standard deviation of a distribution, computed from the probability model. s is the empirical standard deviation of a sample, computed from data, and there you divide by n − 1 instead of n: s = √(Σ(x_i − x̄)²/(n − 1)). The reason for the Bessel correction: the sample mean x̄ automatically lies "in the middle of" its own data, so deviations from x̄ come out systematically smaller than those from the true μ. Dividing by n − 1 compensates this underestimation and makes $s^{2}$ an unbiased estimator of the variance. For large n the difference is tiny. Calculators and spreadsheets offer both variants (σ_n and s_n−1, or STDEV.P and STDEV.S); choose by context: whole population or sample.

For a Bernoulli chain with n trials and success probability p, $\sigma{} = \sqrt{(n\cdot{}p\cdot{}(1-p))}$; q means 1 − p. Example: $n = 400$ applications with $p = 0.25$ gives $\mu{} = 100$ and $\sigma{} = \sqrt{75} \approx{} 8.66$. The normal-approximation interval μ ± 2σ runs roughly from 83 to 117. For those integer counts, the exact binomial sum is about 95.69%; the continuous 2σ rule itself gives 95.45%. The Laplace condition $\sigma{} > 3$ is met but does not guarantee identical coverage. The formula applies only to binomial distributions. Under the root is $n\cdot{}p\cdot{}q$, not its square. σ is maximal at $p = 0.5$, where Bernoulli trials vary most.

It measures how far outcomes typically lie from the expected value: small σ means tightly clustered, large σ widely spread values. Two distributions can have the same mean and completely different spread; only μ and σ together characterize the situation. σ becomes tangible via the sigma rules (for approximately bell-shaped distributions): about 68% of outcomes lie within μ ± σ, 95% within μ ± 2σ, 99.7% within μ ± 3σ. An outcome beyond 2σ is thus already remarkable, beyond 3σ very unusual; hypothesis tests and industrial quality control build exactly on this. Important for interpretation: σ carries the same unit as the measured quantity itself, unlike the variance with its squared unit.

Because simply averaging the deviations always gives zero: positive and negative distances from the expected value cancel exactly, Σ(x_i − μ)·p_i = 0 is a defining property of μ. So a trick is needed to remove the signs. Squaring does this and is mathematically well-behaved: it is differentiable, penalizes large outliers overproportionally and leads to elegant rules like the additivity of variance for independent variables, from which the √(npq) formula follows. The alternative of averaging absolute values gives the mean absolute deviation; it is intuitive but computationally unwieldy and plays no role in school stochastics. To restore the unit, you take the square root after squaring: that is exactly σ.