Calculate and interpret the variance and standard deviation of a discrete random variable, including the effect of a linear transformation

WACE Year 12 Mathematics Methods Unit 3 variance and standard deviation of a discrete random variable: the definition, the E of X squared minus mean squared shortcut, the linear transformation rule, with worked SCSA-style examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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What this dot point is asking
Definition of variance
Why the standard deviation
Linear transformations

What this dot point is asking

SCSA Unit 3 measures the spread of a discrete random variable with its variance and standard deviation. This dot point asks you to compute both, use the efficient $E(X^{2})-\mu^{2}$ form, and apply the linear transformation rule. It is examined in both sections of the WACE written examination.

Definition of variance

The variance is the expected squared distance from the mean, so it captures how widely the values are spread.

The equivalent form $E(X^{2})-\mu^{2}$ is almost always faster by hand: compute $E(X^{2})=\sum x^{2}P(X=x)$ and subtract the square of the mean.

Variance using the shortcut

Find $\mathrm{Var}(X)$ and $\sigma$

A variable $X$ has $P(X=0)=0.5$ , $P(X=2)=0.3$ , $P(X=4)=0.2$ .

First the mean.

E(X) = 0(0.5)+2(0.3)+4(0.2) = 0+0.6+0.8 = 1.4.

Now $E(X^{2})$ .

E(X^{2}) = 0^{2}(0.5)+2^{2}(0.3)+4^{2}(0.2) = 0+1.2+3.2 = 4.4.

Apply the shortcut.

\mathrm{Var}(X) = E(X^{2})-[E(X)]^{2} = 4.4 - 1.4^{2} = 4.4 - 1.96 = 2.44.

So $\sigma=\sqrt{2.44}\approx 1.562$ . Markers reward $E(X^{2})$ , the subtraction of $\mu^{2}$ , and the square root for the standard deviation.

Why the standard deviation

Variance is in squared units, which are awkward to interpret. Taking the square root returns the standard deviation to the same units as $X$ , giving a directly interpretable measure of typical distance from the mean.

Linear transformations

Adding a constant shifts every value but does not change the spread, so it leaves the variance unchanged. Multiplying by a constant scales every distance from the mean, so the variance scales by the square of the constant.

For example, if $\mathrm{Var}(X)=2.44$ , then $\mathrm{Var}(3X+5)=9\times 2.44=21.96$ . The added $5$ has no effect on spread; only the multiplier $3$ matters, squared.