Variance and standard deviation | Year 12

01 / LEARN

Understand the idea.

Spread about the mean

Variance measures spread by averaging squared deviations from the mean: $Var (X) = E ({(X - μ)}^{2})$ . Squaring stops positive and negative deviations cancelling. Standard deviation is the nonnegative square root of variance.

Check units when reporting spread. For values measured in dollars, variance has squared-dollar units, while standard deviation has dollar units. A smaller standard deviation means less spread about the mean, but does not determine the distribution's shape.

A practical calculation formula

For calculation, use $Var (X) = E (X^{2}) - {[E (X)]}^{2}$ . Compute the weighted average of the squared values separately from the weighted average of the values. Then square the mean and subtract it from the second moment.

Keep exact fractions or full calculator precision until the end. Variance cannot be negative; a negative result signals incorrect arithmetic, an invalid distribution or rounding loss. Do not replace the square of the mean by the mean itself.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

1
Compute the mean.
$E (X) = 0 (0.5) + 1 (0.25) + 3 (0.25) = 1$
This weighted sum locates the centre used in the variance.
2
Compute the second moment.
$E (X^{2}) = 0^{2} (0.5) + 1^{2} (0.25) + 3^{2} (0.25) = 2.5$
Square each possible value before weighting it.
3
Subtract the squared mean.
$Var (X) = 2.5 - 1^{2} = 1.5$
The second moment includes both location and spread; subtracting the squared mean isolates spread.
4
Take the nonnegative square root.
$σ = \sqrt{1.5} \approx 1.2247$
Standard deviation is reported on the original measurement scale.

Answer

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

EXAMPLE 02

A success indicator

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

Worked solution

1
Find the first and second moments.
$\begin{aligned} E (B) & = 0 (0.7) + 1 (0.3) = 0.3 \\ E (B^{2}) & = 0^{2} (0.7) + 1^{2} (0.3) = 0.3 \end{aligned}$
The indicator values satisfy $B^{2} = B$ , so these moments are equal.
2
Calculate the variance.
$Var (B) = 0.3 - {(0.3)}^{2} = 0.21$
Subtract the square of the expectation, rather than the expectation.
3
Give the associated standard deviation.
$SD (B) = \sqrt{0.21} \approx 0.4583$
The square root converts squared spread into standard deviation.

Answer

Variance 0.21; standard deviation about 0.4583.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

Use $Var (X) = 19 - 4^{2} = 19 - 16 = 3$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet 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 $E (X^{2}) = 14$ . 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 $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{1}{4}$ . Find $E (X^{2})$ 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
$E (X) = 1$ ; $E (X^{2}) = 2$ .
$Var (X) = 2 - 1^{2} = 1$ .
Mean 1; variance 1.
$E (X) = 3$ and $E (X^{2}) = 14$ . 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
$E (X) = \frac{- 1 + 1}{2} = 0$ .
$X^{2}$ 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 $= \sqrt{9} = 3$ .
Its units are dollars, unlike the squared-dollar variance.
3 dollars.
$X$ has values 1, 2 and 3 with probabilities $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{1}{4}$ . Find $E (X^{2})$ and $Var (X)$ .
3 marks
$E (X) = \frac{1}{4} + 1 + \frac{3}{4} = 2$ .
$E (X^{2}) = \frac{1}{4} + 2 + \frac{9}{4} = \frac{9}{2}$ .
$Var (X) = \frac{9}{2} - 4 = \frac{1}{2}$ .
$E (X^{2}) = \frac{9}{2}; Var (X) = \frac{1}{2}$ .
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.

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Spread about the mean

A practical calculation formula

See the method in action.

A three-point distribution

Compute the mean.

Compute the second moment.

Subtract the squared mean.

Take the nonnegative square root.

A success indicator

Find the first and second moments.

Calculate the variance.

Give the associated standard deviation.

Try it yourself.

Check your understanding

Keep practising