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Sums, the central limit theorem and normal approximation

Sums of many independent variables are approximately normal; standardise, look up Φ, and apply a continuity correction if the summands are integers.

The formulas

Sum moments

$$E\left[\sum X_i\right]=n\mu ,\mathrm{Var}\left(\sum X_i\right)=n{\sigma }^2$$

Standardisation

$$Z=\frac{\sum X_i-n\mu }{\sigma \sqrt{n}}\stackrel{d}{\to }N(0,1)$$

Sample mean

$$\bar{X}\approx N\left(\mu ,\frac{{\sigma }^2}{n}\right)$$

Continuity correction

$$P(S\ge k)\approx 1-\Phi \left(\frac{k-0.5-n\mu }{\sigma \sqrt{n}}\right)$$

Where it comes from

1. Independence makes variances add, so the sum's standard deviation grows like $\sqrt{n}$ while its mean grows like $n$.
2. Standardising therefore stabilises the distribution, and the CLT says the limit is standard normal whatever the summands' shape (given finite variance).
3. The continuity correction accounts for approximating a lattice-valued sum by a continuous density.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• Variances add, standard deviations do NOT. Everything else follows from that one line.
• σ of the sum grows like √n; σ of the MEAN shrinks like √n.

Traps

• Using n·σ instead of √n·σ for the standard deviation of a sum.
• Omitting the continuity correction when approximating a binomial or Poisson count.

Related

• Normal distribution
• Covariance, correlation and sums
• Binomial distribution

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