What is not checking the distribution sums to one?

Always verify (or impose) sum P(X = x) = 1 before computing the mean; an unnormalised distribution gives wrong answers.

A variable X has P(X = x) = 14 for x = 1, 2, 3, 4. Find E(X). [2 marks]

If Var(X) = 4, find Var(2X + 5). [2 marks]

OCR OCR 2019-style practice: A discrete random variable X has P(X = x) = kx for x = 1, 2, 3, 4. Find k, then E(X).

Probabilities sum to 1 (M1): k(1) + k(2) + k(3) + k(4) = 10k = 1, so k = 110 (A1). Expectation E(X) = sum x\,P(X = x) (M1): = 1·110 + 2·210 + 3·310 + 4·410 (A1). Evaluate: = 1 + 4 + 9 + 1610 = 3010 = 3 (A1). Markers reward setting the total probability to 1, the value of k, the expectation formula, and the value 3.

OCR OCR 2022-style practice: A discrete random variable X has E(X) = 2 and Var(X) = 1.5. Find E(3X - 1) and Var(3X - 1), and find E(X^2).

Expectation is linear (M1): E(3X - 1) = 3E(X) - 1 = 3(2) - 1 = 5 (A1). Variance ignores the additive constant and squares the multiplier (M1): Var(3X - 1) = 3^2Var(X) = 9(1.5) = 13.5 (A1). Use Var(X) = E(X^2) - [E(X)]^2 (M1): 1.5 = E(X^2) - 2^2, so E(X^2) = 1.5 + 4 = 5.5 (A1). Markers reward the linearity of expectation, the a^2 rule for variance, the computational variance formula, and the value of E(X^2).

EnglandFurther MathsSyllabus dot point

How do you describe a discrete random variable and find its expectation and variance?

Discrete random variables, the probability distribution, expectation and variance, and the effect of a linear transformation aX + b on the mean and variance.

A focused answer to the OCR A-Level Further Mathematics A Statistics option content on discrete random variables, covering the probability distribution and the condition that probabilities sum to one, the expectation E(X) and variance Var(X), the computational formula for variance, and the effect of a linear transformation aX plus b on the mean and variance.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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Quick answer

A discrete random variable $X$ takes values with probabilities that must sum to one: $\sum \mathrm{P}(X = x) = 1$ . The expectation (mean) is $\mathrm{E}(X) = \sum x\,\mathrm{P}(X = x)$ , and the variance is $\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$ , where $\mathrm{E}(X^2) = \sum x^2\,\mathrm{P}(X = x)$ . Under a linear transformation, the mean transforms fully but the variance ignores the shift and squares the scale: $\mathrm{E}(aX + b) = a\mathrm{E}(X) + b$ and $\mathrm{Var}(aX + b) = a^2\mathrm{Var}(X)$ . More generally $\mathrm{E}(g(X)) = \sum g(x)\,\mathrm{P}(X = x)$ for any function $g$ .

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What this dot point is asking
The probability distribution
Expectation
Variance
Linear transformations
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What this dot point is asking

OCR's Statistics option wants you to describe a discrete random variable by its probability distribution (checking the probabilities sum to one), to calculate its expectation $\mathrm{E}(X)$ and variance $\mathrm{Var}(X)$ using the standard formulae, and to find the effect of a linear transformation $aX + b$ on the mean and variance.

The probability distribution

A discrete random variable is fully described by listing its possible values and their probabilities. The one constraint is that these probabilities form a valid distribution: each is between $0$ and $1$ , and they sum to $1$ . This sum condition is how you find an unknown constant in a distribution.

Expectation

The expectation, or mean, is the long-run average value, computed as a probability-weighted sum of the values.

Variance

The variance measures spread about the mean. The computational formula "mean of the squares minus the square of the mean" is almost always quicker than the definition.

Linear transformations

Scaling and shifting a random variable affects the mean and variance differently. The mean follows the transformation exactly, but the variance is unchanged by the additive shift (which moves the whole distribution without changing its spread) and is multiplied by the square of the scale factor. The reason is geometric: adding a constant slides every value along by the same amount, so distances from the mean, and hence the spread, are untouched; multiplying by $a$ stretches all those distances by $a$ , and because variance is measured in squared units, it grows by $a^2$ . The standard deviation, being the square root of the variance, therefore scales by $|a|$ , which is often the more natural quantity to report because it shares the units of $X$ .

Discrete random variables are the foundation for the named distributions (Poisson, geometric) and parallel the continuous case, where sums become integrals.

Try this

Q1. A variable $X$ has $\mathrm{P}(X = x) = \tfrac{1}{4}$ for $x = 1, 2, 3, 4$ . Find $\mathrm{E}(X)$ . [2 marks]

Cue. $\tfrac{1}{4}(1 + 2 + 3 + 4) = \tfrac{10}{4} = 2.5$ .

Q2. If $\mathrm{Var}(X) = 4$ , find $\mathrm{Var}(2X + 5)$ . [2 marks]

Cue. $2^2 \times 4 = 16$ (the $+5$ has no effect).

Exam-style practice questions

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

OCR 20195 marksA discrete random variable

X

has

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

for

x = 1, 2, 3, 4

. Find

k

, then

\mathrm{E}(X)

Show worked answer →

Probabilities sum to $1$ (M1): $k(1) + k(2) + k(3) + k(4) = 10k = 1$ , so $k = \tfrac{1}{10}$ (A1).

Expectation $\mathrm{E}(X) = \sum x\,\mathrm{P}(X = x)$ (M1): $= 1\cdot\tfrac{1}{10} + 2\cdot\tfrac{2}{10} + 3\cdot\tfrac{3}{10} + 4\cdot\tfrac{4}{10}$ (A1).

Evaluate: $= \tfrac{1 + 4 + 9 + 16}{10} = \tfrac{30}{10} = 3$ (A1).

Markers reward setting the total probability to $1$ , the value of $k$ , the expectation formula, and the value $3$ .

OCR 20226 marksA discrete random variable

X

has

\mathrm{E}(X) = 2

and

\mathrm{Var}(X) = 1.5

. Find

\mathrm{E}(3X - 1)

and

\mathrm{Var}(3X - 1)

, and find

\mathrm{E}(X^2)

Show worked answer →

Expectation is linear (M1): $\mathrm{E}(3X - 1) = 3\mathrm{E}(X) - 1 = 3(2) - 1 = 5$ (A1).

Variance ignores the additive constant and squares the multiplier (M1): $\mathrm{Var}(3X - 1) = 3^2\mathrm{Var}(X) = 9(1.5) = 13.5$ (A1).

Use $\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$ (M1): $1.5 = \mathrm{E}(X^2) - 2^2$ , so $\mathrm{E}(X^2) = 1.5 + 4 = 5.5$ (A1).

Markers reward the linearity of expectation, the $a^2$ rule for variance, the computational variance formula, and the value of $\mathrm{E}(X^2)$ .

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Sources & how we know this

OCR A Level Further Mathematics A (H245) specification — OCR (2017)