The variance measures spread about the mean. Its definition is Var(X) = E[(X - μ)^2], but the computational formula is almost always easier:

If E(X) = 5, find E(2X + 3). [1 mark]

If Var(X) = 4, find Var(3X - 7). [1 mark]

SEAB Original-style practice: A discrete random variable X has P(X = x) = kx for x = 1, 2, 3, 4. Find k, then E(X) and Var(X).

Probabilities sum to 1: sum kx = k(1 + 2 + 3 + 4) = 10k = 1, so k = 0.1. The distribution is P(X = x) = 0.1x: P(1) = 0.1, P(2) = 0.2, P(3) = 0.3, P(4) = 0.4. E(X) = sum x\,P(X = x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3. E(X^2) = sum x^2\,P(X = x) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10. Var(X) = E(X^2) - [E(X)]^2 = 10 - 9 = 1. Markers reward finding k = 0.1, the expectation 3, E(X^2) = 10, and the variance 1 via the computational formula.

SEAB Original-style practice: A random variable X has E(X) = 4 and Var(X) = 2. Find E(3X - 1) and Var(3X - 1).

For a linear function, expectation is linear: E(aX + b) = a\,E(X) + b. So $E(3X - 1) = 3\,E(X) - 1 = 3(4) - 1 = 11.$ Variance scales by the square of the multiplier and ignores the constant: Var(aX + b) = a^2Var(X). So $Var(3X - 1) = 3^2Var(X) = 9(2) = 18.$ Markers reward the linear expectation rule giving 11, and the variance rule a^2Var(X) (with b having no effect) giving 18.

SingaporeFurther MathsSyllabus dot point

How do we describe a discrete random variable and compute its expectation and variance?

Work with discrete random variables, their probability distributions, expectation, variance, and the expectation and variance of linear functions

A focused answer to the H2 Further Mathematics outcome on discrete random variables. Probability distributions, expectation and variance, the computational formula for variance, and the rules for the expectation and variance of a linear function aX + b.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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The answer
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Try this

What this dot point is asking

SEAB wants you to describe a discrete random variable through its probability distribution, to compute its expectation (mean) and variance, to use the computational formula for variance, and to apply the rules for the expectation and variance of a linear function $aX + b$ . These are the foundation of all the distribution work that follows.

The answer

The probability distribution

A discrete random variable $X$ takes a countable set of values, each with a probability $\mathrm{P}(X = x)$ . The probabilities must satisfy

\mathrm{P}(X = x) \geq 0 \quad\text{for all } x, \qquad \sum_x \mathrm{P}(X = x) = 1.

The second condition (probabilities sum to $1$ ) is what fixes any unknown constant in the distribution.

Expectation

The expectation (mean) is the probability-weighted average of the values:

\mathrm{E}(X) = \sum_x x\,\mathrm{P}(X = x).

It is the long-run average value of $X$ over many repetitions, often denoted $\mu$ .

Variance

The variance measures spread about the mean. Its definition is $\operatorname{Var}(X) = \mathrm{E}\big[(X - \mu)^2\big]$ , but the computational formula is almost always easier:

\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2, \qquad \mathrm{E}(X^2) = \sum_x x^2\,\mathrm{P}(X = x).

The standard deviation is $\sqrt{\operatorname{Var}(X)}$ .

Linear functions of X

For constants $a$ and $b$ ,

\mathrm{E}(aX + b) = a\,\mathrm{E}(X) + b, \qquad \operatorname{Var}(aX + b) = a^2\operatorname{Var}(X).

Expectation is fully linear, but variance scales by $a^2$ and is unaffected by the constant $b$ , because adding a constant shifts the distribution without changing its spread.

Worked example

A random variable $X$ has the distribution $\mathrm{P}(X = 0) = 0.5$ , $\mathrm{P}(X = 1) = 0.3$ , $\mathrm{P}(X = 2) = 0.2$ . Find $\mathrm{E}(X)$ , $\operatorname{Var}(X)$ , and $\operatorname{Var}(2X + 5)$ .

Step 1: Check the distribution

$0.5 + 0.3 + 0.2 = 1$ , so the probabilities are valid.

Step 2: Compute the expectation

\mathrm{E}(X) = 0(0.5) + 1(0.3) + 2(0.2) = 0 + 0.3 + 0.4 = 0.7.

Step 3: Compute E(X squared)

\mathrm{E}(X^2) = 0^2(0.5) + 1^2(0.3) + 2^2(0.2) = 0 + 0.3 + 0.8 = 1.1.

Step 4: Apply the computational formula for variance

\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 1.1 - 0.49 = 0.61.

Step 5: Variance of the linear function

\operatorname{Var}(2X + 5) = 2^2\operatorname{Var}(X) = 4(0.61) = 2.44.

The $+5$ has no effect; only the multiplier $2$ scales the variance, squared.

Examples in context

Example 1. Expected value of a game. A gambling or insurance payoff is a discrete random variable; its expectation is the long-run average gain or loss, the figure that decides whether a bet or premium is fair, and the variance measures the risk.

Example 2. Scaling units. Converting a discrete measurement from one unit to another by $Y = aX + b$ rescales the mean by $a$ and shifts it by $b$ , but the variance scales by $a^2$ , which is why standard deviation (not variance) shares the units of the data.

Try this

Q1. State the computational formula for the variance of a discrete random variable. [1 mark]

Cue. $\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$ .

Q2. If $\mathrm{E}(X) = 5$ , find $\mathrm{E}(2X + 3)$ . [1 mark]

Cue. $2(5) + 3 = 13$ .

Q3. If $\operatorname{Var}(X) = 4$ , find $\operatorname{Var}(3X - 7)$ . [1 mark]

Cue. $3^2(4) = 36$ (the $-7$ has no effect).

Exam-style practice questions

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

Original6 marksA discrete random variable

X

has

\mathrm{P}(X = x) = kx

for

x = 1, 2, 3, 4

. Find

k

, then

\mathrm{E}(X)

and

\operatorname{Var}(X)

Show worked answer →

Probabilities sum to $1$ : $\sum kx = k(1 + 2 + 3 + 4) = 10k = 1$ , so $k = 0.1$ .

The distribution is $\mathrm{P}(X = x) = 0.1x$ : $\mathrm{P}(1) = 0.1$ , $\mathrm{P}(2) = 0.2$ , $\mathrm{P}(3) = 0.3$ , $\mathrm{P}(4) = 0.4$ .

$\mathrm{E}(X) = \sum x\,\mathrm{P}(X = x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3$ .

$\mathrm{E}(X^2) = \sum x^2\,\mathrm{P}(X = x) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10$ .

$\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 10 - 9 = 1$ .

Markers reward finding $k = 0.1$ , the expectation $3$ , $\mathrm{E}(X^2) = 10$ , and the variance $1$ via the computational formula.

Original5 marksA random variable

X

has

\mathrm{E}(X) = 4

and

\operatorname{Var}(X) = 2

. Find

\mathrm{E}(3X - 1)

and

\operatorname{Var}(3X - 1)

Show worked answer →

For a linear function, expectation is linear: $\mathrm{E}(aX + b) = a\,\mathrm{E}(X) + b$ . So

\mathrm{E}(3X - 1) = 3\,\mathrm{E}(X) - 1 = 3(4) - 1 = 11.

Variance scales by the square of the multiplier and ignores the constant:

\operatorname{Var}(aX + b) = a^2\operatorname{Var}(X)

. So

\operatorname{Var}(3X - 1) = 3^2\operatorname{Var}(X) = 9(2) = 18.

Markers reward the linear expectation rule giving $11$ , and the variance rule $a^2\operatorname{Var}(X)$ (with $b$ having no effect) giving $18$ .