SCSA WACE 2024-style practice: Calculator-free. A discrete random variable X has Var(X)=4. Find (a) Var(3X), (b) Var(X+7) and (c) the standard deviation of 3X.

A test of the linear-transformation rule for variance. (a) Var(3X)=3^2Var(X)=9× 4=36. (b) Adding a constant does not change spread, so Var(X+7)=Var(X)=4. (c) SD(3X)=|3|\,SD(X)=3sqrt(4)=3× 2=6. Markers reward squaring the multiplier in (a), recognising the constant has no effect in (b), and the standard-deviation scaling by |a| in (c).

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do we measure the spread of a discrete random variable around its mean?

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 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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

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

Step 1: Find the mean $E(X)$

A variable $X$ has $P(X=0)=0.5$ , $P(X=2)=0.3$ , $P(X=4)=0.2$ . The mean is required before the variance shortcut can be applied:

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

Step 2: Compute $E(X^{2})$

Weighting the squared values by their probabilities gives the second moment, which is the other ingredient in the shortcut formula:

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

Step 3: Apply the shortcut $\mathrm{Var}(X)=E(X^{2})-[E(X)]^{2}$ and find $\sigma$

Subtracting the square of the mean from $E(X^{2})$ gives the variance, and taking the square root gives the standard deviation:

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

\sigma=\sqrt{2.44}\approx 1.562.

Final answer: $\mathrm{Var}(X)=2.44$ , $\sigma\approx 1.562$

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.

Why the shortcut works

The two forms of the variance are equal by expanding the squared bracket:

E\bigl((X-\mu)^{2}\bigr)=E(X^{2}-2\mu X+\mu^{2})=E(X^{2})-2\mu E(X)+\mu^{2}.

Since

E(X)=\mu

, the middle term is

-2\mu^{2}

and the last is

+\mu^{2}

, leaving

E(X^{2})-2\mu^{2}+\mu^{2}=E(X^{2})-\mu^{2}.

This is why the computational form

E(X^{2})-[E(X)]^{2}

always matches the definition, and why it is faster: you compute one extra weighted sum

E(X^{2})

rather than re-centring every value before squaring.

Interpreting the standard deviation

A small standard deviation means the values of $X$ cluster tightly around the mean; a large one means they are widely spread. Because the standard deviation shares the units of $X$ , it can be compared directly with the mean. Two distributions with the same mean but different standard deviations have the same centre but different reliability, which matters in any decision that weighs an average against its risk.

Variance of a transformed variable

Apply the linear-transformation rule

A discrete random variable $X$ has $E(X)=3$ and $\mathrm{Var}(X)=1.5$ . A new variable is defined by $Y=2X-1$ . Find $E(Y)$ , $\mathrm{Var}(Y)$ and $\mathrm{SD}(Y)$ .

The mean transforms linearly:

E(Y)=2E(X)-1=2(3)-1=5.

The variance uses only the multiplier, squared, and ignores the constant:

\mathrm{Var}(Y)=2^{2}\,\mathrm{Var}(X)=4\times 1.5=6.

The standard deviation scales by the absolute value of the multiplier:

\mathrm{SD}(Y)=|2|\,\sqrt{1.5}=2\times 1.225\approx 2.449.

Markers reward applying

E(aX+b)=aE(X)+b

for the mean, squaring the multiplier for the variance, and using

|a|

for the standard deviation while ignoring the constant

-1

in both spread calculations.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marksCalculator-assumed. A discrete random variable

X

has

P(X=1)=0.2

P(X=2)=0.5

P(X=3)=0.3

. (a) Find

E(X)

. (b) Find

\mathrm{Var}(X)

. (c) Find the standard deviation, to two decimal places.

Show worked answer →

A full mean-variance computation.

(a) $E(X)=1(0.2)+2(0.5)+3(0.3)=0.2+1.0+0.9=2.1$ .

(b) $E(X^{2})=1(0.2)+4(0.5)+9(0.3)=0.2+2.0+2.7=4.9$ . Then $\mathrm{Var}(X)=E(X^{2})-[E(X)]^{2}=4.9-2.1^{2}=4.9-4.41=0.49$ .

Markers reward the mean, $E(X^{2})$ , the subtraction of $\mu^{2}$ , and the square root for the standard deviation.

WACE 20243 marksCalculator-free. A discrete random variable

X

has

\mathrm{Var}(X)=4

. Find (a)

\mathrm{Var}(3X)

, (b)

\mathrm{Var}(X+7)

and (c) the standard deviation of

3X

Show worked answer →

A test of the linear-transformation rule for variance.

(a) $\mathrm{Var}(3X)=3^{2}\mathrm{Var}(X)=9\times 4=36$ .

(b) Adding a constant does not change spread, so $\mathrm{Var}(X+7)=\mathrm{Var}(X)=4$ .

Markers reward squaring the multiplier in (a), recognising the constant has no effect in (b), and the standard-deviation scaling by $|a|$ in (c).