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Variance and standard deviation of a discrete random variable: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

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

8 min read

Quick answer

The variance Var(X)=E((X−μ)2)=E(X2)−[E(X)]2\mathrm{Var}(X)=E((X-\mu)^{2})=E(X^{2})-[E(X)]^{2} measures spread around the mean; the standard deviation is its square root. Under a linear change, Var(aX+b)=a2Var(X)\mathrm{Var}(aX+b)=a^{2}\mathrm{Var}(X).

Jump to a section
  1. What this dot point is asking
  2. Definition of variance
  3. Why the standard deviation
  4. Linear transformations
  5. Why the shortcut works
  6. Interpreting the standard deviation
  7. Exam-style questions

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(X2)−μ2E(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.

Variance and standard deviation

Var(X)=σ2=∑x(x−μ)2P(X=x)=E(X2)−[E(X)]2\mathrm{Var}(X)=\sigma^{2}=\sum_x (x-\mu)^{2}P(X=x)=E(X^{2})-[E(X)]^{2}

SD(X)=σ=Var(X)\mathrm{SD}(X)=\sigma=\sqrt{\mathrm{Var}(X)}

The equivalent form E(X2)−μ2E(X^{2})-\mu^{2} is almost always faster by hand: compute E(X2)=∑x2P(X=x)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)E(X)

A variable XX has P(X=0)=0.5P(X=0)=0.5, P(X=2)=0.3P(X=2)=0.3, P(X=4)=0.2P(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.E(X) = 0(0.5)+2(0.3)+4(0.2) = 0+0.6+0.8 = 1.4.

Step 2: Compute E(X2)E(X^{2})

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

E(X2)=02(0.5)+22(0.3)+42(0.2)=0+1.2+3.2=4.4.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 Var(X)=E(X2)−[E(X)]2\mathrm{Var}(X)=E(X^{2})-[E(X)]^{2} and find σ\sigma

Subtracting the square of the mean from E(X2)E(X^{2}) gives the variance, and taking the square root gives the standard deviation:

Var(X)=E(X2)−[E(X)]2=4.4−1.42=4.4−1.96=2.44.\mathrm{Var}(X) = E(X^{2})-[E(X)]^{2} = 4.4 - 1.4^{2} = 4.4 - 1.96 = 2.44.

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

Final answer: Var(X)=2.44\mathrm{Var}(X)=2.44, σ≈1.562\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 XX, 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.

Variance under a linear change

Var(aX+b)=a2 Var(X),SD(aX+b)=∣a∣ SD(X).\mathrm{Var}(aX+b)=a^{2}\,\mathrm{Var}(X), \qquad \mathrm{SD}(aX+b)=|a|\,\mathrm{SD}(X).

For example, if Var(X)=2.44\mathrm{Var}(X)=2.44, then Var(3X+5)=9×2.44=21.96\mathrm{Var}(3X+5)=9\times 2.44=21.96. The added 55 has no effect on spread; only the multiplier 33 matters, squared.

Why the shortcut works

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

E((X−μ)2)=E(X2−2μX+μ2)=E(X2)−2μE(X)+μ2.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)=μE(X)=\mu, the middle term is −2μ2-2\mu^{2} and the last is +μ2+\mu^{2}, leaving
E(X2)−2μ2+μ2=E(X2)−μ2.E(X^{2})-2\mu^{2}+\mu^{2}=E(X^{2})-\mu^{2}.

This is why the computational form E(X2)−[E(X)]2E(X^{2})-[E(X)]^{2} always matches the definition, and why it is faster: you compute one extra weighted sum E(X2)E(X^{2}) rather than re-centring every value before squaring.

Interpreting the standard deviation

A small standard deviation means the values of XX cluster tightly around the mean; a large one means they are widely spread. Because the standard deviation shares the units of XX, 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 XX has E(X)=3E(X)=3 and Var(X)=1.5\mathrm{Var}(X)=1.5. A new variable is defined by Y=2X−1Y=2X-1. Find E(Y)E(Y), Var(Y)\mathrm{Var}(Y) and SD(Y)\mathrm{SD}(Y).

The mean transforms linearly:

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

The variance uses only the multiplier, squared, and ignores the constant:
Var(Y)=22 Var(X)=4×1.5=6.\mathrm{Var}(Y)=2^{2}\,\mathrm{Var}(X)=4\times 1.5=6.

The standard deviation scales by the absolute value of the multiplier:
SD(Y)=∣2∣ 1.5=2×1.225≈2.449.\mathrm{SD}(Y)=|2|\,\sqrt{1.5}=2\times 1.225\approx 2.449.

Markers reward applying E(aX+b)=aE(X)+bE(aX+b)=aE(X)+b for the mean, squaring the multiplier for the variance, and using ∣a∣|a| for the standard deviation while ignoring the constant −1-1 in both spread calculations.

Common errors
Forgetting to square the mean
The shortcut is E(X2)−[E(X)]2E(X^{2})-[E(X)]^{2}; subtract the square of the mean, not the mean itself.
Reporting the variance as the standard deviation
The standard deviation is Var(X)\sqrt{\mathrm{Var}(X)}; do not stop at the variance when a standard deviation is asked.
Using aa instead of a2a^{2} for variance
Under aX+baX+b the variance scales by a2a^{2}, while the standard deviation scales by ∣a∣|a|.
Letting the constant bb affect spread
Adding bb shifts the distribution but leaves variance and standard deviation unchanged.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks
Calculator-assumed. A discrete random variable XX has P(X=1)=0.2P(X=1)=0.2, P(X=2)=0.5P(X=2)=0.5, P(X=3)=0.3P(X=3)=0.3. (a) Find E(X)E(X). (b) Find Var(X)\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=2.1E(X)=1(0.2)+2(0.5)+3(0.3)=0.2+1.0+0.9=2.1.

(b) E(X2)=1(0.2)+4(0.5)+9(0.3)=0.2+2.0+2.7=4.9E(X^{2})=1(0.2)+4(0.5)+9(0.3)=0.2+2.0+2.7=4.9. Then Var(X)=E(X2)−[E(X)]2=4.9−2.12=4.9−4.41=0.49\mathrm{Var}(X)=E(X^{2})-[E(X)]^{2}=4.9-2.1^{2}=4.9-4.41=0.49.

(c) σ=0.49=0.70\sigma=\sqrt{0.49}=0.70.

Markers reward the mean, E(X2)E(X^{2}), the subtraction of μ2\mu^{2}, and the square root for the standard deviation.

WACE 2024-style3 marks
Calculator-free. A discrete random variable XX has Var(X)=4\mathrm{Var}(X)=4. Find (a) Var(3X)\mathrm{Var}(3X), (b) Var(X+7)\mathrm{Var}(X+7) and (c) the standard deviation of 3X3X.
Show worked answer →

A test of the linear-transformation rule for variance.

(a) Var(3X)=32Var(X)=9×4=36\mathrm{Var}(3X)=3^{2}\mathrm{Var}(X)=9\times 4=36.

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

(c) SD(3X)=∣3∣ SD(X)=34=3×2=6\mathrm{SD}(3X)=|3|\,\mathrm{SD}(X)=3\sqrt{4}=3\times 2=6.

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

Practise this

Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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