Variance and standard deviation | Year 12

01 / LEARN

The idea, step by step

Spread about the mean

Variance averages squared deviations from the mean: $Var (X) = E ({(X - μ)}^{2})$ . The squares prevent positive and negative deviations cancelling. Standard deviation is the non-negative square root of variance.

If X is measured in dollars, variance has squared-dollar units but standard deviation has dollar units. A small standard deviation means values tend to be relatively close to the mean; it does not alone describe the full shape.

A practical calculation formula

Expanding the square gives $Var (X) = E (X^{2}) - {[E (X)]}^{2}$ . First compute two weighted sums, one using x and one using x². Subtract the square of the first sum from the second.

Do not round the mean before squaring it. Exact fractions or full calculator precision avoid a small negative variance caused by rounding. A truly negative result means there is an error in the distribution or arithmetic.

A common trap

Variance is E(X²) − E(X).

Instead: Square the mean: E(X²) − [E(X)]². In general E(X²) and [E(X)]² are different quantities.

02 / WORKED EXAMPLES

See the thinking

Example 1

A three-point distribution

X is 0, 1 or 3 with probabilities 0.5, 0.25 and 0.25. Find its mean, variance and standard deviation.

E(X) = 0(0.5) + 1(0.25) + 3(0.25) = 1.
E(X²) = 0 + 1(0.25) + 9(0.25) = 2.5.
Var(X) = 2.5 − 1² = 1.5; take its positive square root.

E (X) = 1, Var (X) = 1.5, σ = \sqrt{1.5} \approx 1.2247.

Example 2

A success indicator

B is 1 with probability 0.3 and 0 otherwise. Find Var(B).

E(B) = 0.3. Since B² = B for values 0 and 1, E(B²) = 0.3.
Var(B) = 0.3 − 0.3² = 0.21.
The standard deviation is √0.21 ≈ 0.4583.

Variance 0.21; standard deviation about 0.4583.

Check your understanding

If E(X) = 4 and E(X²) = 19, find Var(X).

Show answer

Subtract 4² = 16 from 19.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

X is 0 or 2 with equal probabilities. Find its mean and variance.
2 marks
E(X) = 3 and E(X²) = 14. Find variance and standard deviation.
2 marks
X is −1 or 1 with equal probabilities. Find its mean and standard deviation.
2 marks
A payout in dollars has variance 9. State its standard deviation and units.
2 marks
X has values 1, 2 and 3 with probabilities 1/4, 1/2 and 1/4. Find E(X²) and Var(X).
3 marks
A student obtains Var(X) = −0.8 from a valid exact probability table. Explain why this cannot be accepted.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

X is 0 or 2 with equal probabilities. Find its mean and variance.
2 marks
E(X) = 1; E(X²) = 2.
Var(X) = 2 − 1² = 1.
Mean 1; variance 1.
E(X) = 3 and E(X²) = 14. Find variance and standard deviation.
2 marks
Var(X) = 14 − 9 = 5.
Standard deviation is the non-negative square root.
$5; \sqrt{5} \approx 2.2361$ .
X is −1 or 1 with equal probabilities. Find its mean and standard deviation.
2 marks
E(X) = (−1 + 1)/2 = 0.
X² is always 1, so variance is 1 and standard deviation is 1.
Mean 0; standard deviation 1.
A payout in dollars has variance 9. State its standard deviation and units.
2 marks
Standard deviation = √9 = 3.
Its units are dollars, unlike the squared-dollar variance.
3 dollars.
X has values 1, 2 and 3 with probabilities 1/4, 1/2 and 1/4. Find E(X²) and Var(X).
3 marks
$E (X) = 1 / 4 + 1 + 3 / 4 = 2$ .
$E (X^{2}) = 1 / 4 + 2 + 9 / 4 = 9 / 2$ .
$Var (X) = 9 / 2 - 4 = 1 / 2$ .
$E (X^{2}) = 9 / 2; Var (X) = 1 / 2$ .
A student obtains Var(X) = −0.8 from a valid exact probability table. Explain why this cannot be accepted.
2 marks
Every squared deviation is non-negative and probabilities are non-negative.
Their weighted sum cannot be below zero; recheck the calculation.
Variance cannot be negative.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

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