Statistics

The variance engine

$\sigma^2 = \frac{1}{n}\sum(x_i - \bar{x})^2, \qquad \sigma = \sqrt{\sigma^2}$ For small sets: find the mean, subtract, square, average, root. Keep squares exact — fractions beat decimals. Variance and SD are asked interchangeably; read the question twice before squaring or rooting.

The shift-scale law (the CGL favourite)

If every value becomes $ax + b$: $\text{new mean} = a\bar{x} + b, \qquad \text{new SD} = |a|\,\sigma$ Adding $b$ slides the whole distribution — spread unchanged. Only the multiplier touches the SD. Median scales identically; the range too. A question saying "5 is added to every number" is testing exactly this: the SD does not move.

SD of standard sets

• First $n$ naturals: mean $\frac{n+1}{2}$, SD $\sqrt{\frac{n^2-1}{12}}$ (n = 10: SD $\sqrt{8.25}$).
• An AP with common difference $d$ has SD $= d\sqrt{\frac{n^2-1}{12}}$ — consecutive integers 1..n are just $d = 1$.
• Two-value set $\{a, b\}$: SD $= \frac{|a-b|}{2}$ (deviations $\pm\frac{a-b}{2}$). These closed forms answer "the SD of 1, 2, ..., 50" style questions in one line.

Range and its transform

Range $= \max - \min$. Under $ax + b$: new range $= |a| \times$ old range ($b$ cancels). For combined sets, the new range needs the actual extreme values, not a formula — but the exam usually asks the transform version.

Coefficient of variation

$CV = \frac{\sigma}{\bar{x}} \times 100\%$ — spread relative to level. Comparing two data sets: the one with the lower CV is more consistent. One substitution each way; watch which mean goes with which SD.

Worked example: SD from scratch

Data 2, 4, 4, 4, 5, 5, 7, 9: mean $= \frac{40}{8} = 5$. Squared deviations 9, 1, 1, 1, 0, 0, 4, 16 sum to 32, so variance $= \frac{32}{8} = 4$ and SD $= 2$. Shortcut form: $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 = \frac{232}{8} - 25 = 4$ — faster when the mean is not a whole number.

Worked example: comparing consistency

Batsman A: mean 50, SD 10 → CV 20%. Batsman B: mean 40, SD 6 → CV 15%. B is more consistent even though A has the higher average. Questions ask one or the other — read which before answering. Lower CV means less relative spread, so it is the "more consistent" or "more stable" set.

Quick revision

• $\sigma^2 =$ mean of squared deviations; SD = root.
• SD$(ax + b) = |a|\,\sigma$; shifts never change spread.
• Naturals 1..n: mean $\frac{n+1}{2}$, SD $\sqrt{\frac{n^2-1}{12}}$; AP: multiply by $d$.
• Two values $\{a, b\}$: SD $= \frac{|a-b|}{2}$.
• Range$(ax + b) = |a| \times$ range; CV $= \frac{\sigma}{\bar{x}} \times 100$.