Statistical Distributions

This lesson introduces statistical distributions at A-Level, focusing on the concept of a random variable and its probability distribution. Understanding distributions is fundamental to modelling real-world phenomena and underpins all of inferential statistics.

Random Variables

A random variable is a variable whose value depends on the outcome of a random event. Random variables are denoted by capital letters (e.g., $X$ ), and their specific values by lowercase letters (e.g., $x$ ).

Type	Description	Example
Discrete	Takes specific, countable values	Number of heads in 10 coin tosses
Continuous	Takes any value in a range	Height of a randomly chosen student

Discrete Probability Distributions

A probability distribution for a discrete random variable $X$ lists all possible values and their probabilities.

Properties:

$0 \leq P(X = x) \leq 1$ for all values $x$ .
$\sum P(X = x) = 1$ (the probabilities of all possible values sum to 1).

Example:

$x$	1	2	3	4
$P(X = x)$	0.1	0.3	0.4	0.2

Check: $0.1 + 0.3 + 0.4 + 0.2 = 1$ ✓

Finding Unknown Probabilities

If a probability is unknown, use the fact that all probabilities sum to 1 to find it.

Example: If $P(X = 1) = 0.2$ , $P(X = 2) = k$ , $P(X = 3) = 2k$ , $P(X = 4) = 0.1$ :

$0.2 + k + 2k + 0.1 = 1 \implies 3k = 0.7 \implies k = \frac{7}{30}$

Expected Value and Variance

Expected Value (Mean)

The expected value of a discrete random variable is the long-run average:

$E(X) = \mu = \sum x \cdot P(X = x)$

Variance

$\text{Var}(X) = E(X^2) - [E(X)]^2$

where $E(X^2) = \sum x^2 \cdot P(X = x)$ .

Standard Deviation

$\sigma = \sqrt{\text{Var}(X)}$

Exam Tip: When calculating $E(X)$ and $\text{Var}(X)$ , set up a table with columns for $x$ , $P(X=x)$ , $xP$ , and $x^2P$ . This organises your working clearly and reduces errors.

Linear Transformations

If $Y = aX + b$ , then:

$E(Y) = aE(X) + b$ $\text{Var}(Y) = a^2 \text{Var}(X)$

Note: Adding a constant $b$ shifts the mean but does not affect the variance. Multiplying by $a$ scales both the mean and the standard deviation.

The Discrete Uniform Distribution

A discrete uniform distribution assigns equal probability to each of $n$ outcomes:

$P(X = x) = \frac{1}{n}$

Example: A fair six-sided die has $P(X = x) = \frac{1}{6}$ for $x = 1, 2, 3, 4, 5, 6$ .

For integers 1 to $n$ : $E(X) = \frac{n + 1}{2}, \quad \text{Var}(X) = \frac{n^2 - 1}{12}$

Summary

A random variable assigns numerical values to outcomes of a random experiment.
A discrete probability distribution lists all values and their probabilities, which must sum to 1.
$E(X) = \sum xP(X=x)$ is the expected (mean) value.
$\text{Var}(X) = E(X^2) - [E(X)]^2$ measures the spread.
For $Y = aX + b$ : $E(Y) = aE(X) + b$ and $\text{Var}(Y) = a^2\text{Var}(X)$ .
The discrete uniform distribution assigns equal probability to each outcome.

Exam Tip: Probability distribution questions often ask you to find an unknown probability by using the fact that all probabilities sum to 1. Always state this condition explicitly in your working.

A-Level Deep Dive: Statistical Distributions

Spec mapping

AQA 7357 specification, Paper 3 — Statistics, Section R: Statistical Distributions. This section requires students to "understand and use simple, finite, discrete probability distributions (defined analytically, in tables or by formula), including the distribution of a random variable, expected value $E(X)$ , variance $Var(X)$ , and the discrete uniform distribution." Binomial $B(n, p)$ and normal $N(\mu, \sigma^2)$ distributions are examined under separate sub-strands and treated in their own lessons; this Deep Dive is restricted to the general discrete framework — probability functions, expectation, variance, and the uniform case. Although the Section R material is concentrated in Paper 3, the language of expectation propagates into Paper 2 Mechanics (impulse and average force as expected change in momentum) and Paper 1 Pure (sigma notation, infinite series convergence). The AQA formula booklet provides $E(X) = \sum x P(X = x)$ and $Var(X) = E(X^2) - [E(X)]^2$ — the second form is the computational form and is the one students are expected to apply.

Worked example with full mark scheme

Question (8 marks):

The discrete random variable $X$ has probability function

$P(X = x) = \begin{cases} kx & \text{for } x = 1, 2, 3 \\ k(6 - x) & \text{for } x = 4, 5 \\ 0 & \text{otherwise} \end{cases}$

(a) Show that $k = \dfrac{1}{9}$ . (2)

(b) Find $E(X)$ . (2)

Solution with mark scheme:

(a) Step 1 — apply the total-probability axiom.

The probabilities must sum to 1:

$\sum_{x = 1}^{5} P(X = x) = k(1) + k(2) + k(3) + k(2) + k(1) = 9k$

M1 — recognising that $\sum P(X = x) = 1$ over the support and writing the correct sum. The substitution $P(X = 4) = k(6 - 4) = 2k$ and $P(X = 5) = k(6 - 5) = k$ must be evaluated correctly; many candidates double-count or misread the piecewise rule.

A1 — solving $9k = 1 \implies k = \tfrac{1}{9}$ as required.

(b) Step 1 — list the probabilities and apply $E(X) = \sum x P(X = x)$ .

$x$	1	2	3	4	5
$P(X = x)$	$\tfrac{1}{9}$	$\tfrac{2}{9}$	$\tfrac{3}{9}$	$\tfrac{2}{9}$	$\tfrac{1}{9}$

$E(X) = 1 \cdot \tfrac{1}{9} + 2 \cdot \tfrac{2}{9} + 3 \cdot \tfrac{3}{9} + 4 \cdot \tfrac{2}{9} + 5 \cdot \tfrac{1}{9} = \tfrac{1 + 4 + 9 + 8 + 5}{9} = \tfrac{27}{9} = 3$

M1 — correct expectation formula applied with all five terms.

A1 — $E(X) = 3$ . The symmetry of the distribution about $x = 3$ provides a useful check: a symmetric distribution has its expectation at the centre of symmetry.

$E(X^2) = 1 \cdot \tfrac{1}{9} + 4 \cdot \tfrac{2}{9} + 9 \cdot \tfrac{3}{9} + 16 \cdot \tfrac{2}{9} + 25 \cdot \tfrac{1}{9} = \tfrac{1 + 8 + 27 + 32 + 25}{9} = \tfrac{93}{9} = \tfrac{31}{3}$

M1 — applying $E(X^2) = \sum x^2 P(X = x)$ , not $[E(X)]^2$ .

A1 — $E(X^2) = \tfrac{31}{3}$ .

Step 2 — apply the computational form of variance.

$Var(X) = E(X^2) - [E(X)]^2 = \tfrac{31}{3} - 9 = \tfrac{31 - 27}{3} = \tfrac{4}{3}$

M1 — using $Var(X) = E(X^2) - [E(X)]^2$ .

A1 — $Var(X) = \tfrac{4}{3}$ .

Total: 8 marks (M4 A4).

Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): The discrete random variable $Y$ has the discrete uniform distribution on $\{1, 2, 3, 4, 5, 6\}$ , so that $P(Y = y) = \tfrac{1}{6}$ for each $y\in\{1, \ldots, 6\}$ .

(a) Write down $E(Y)$ . (1)

(b) Show that $Var(Y) = \tfrac{35}{12}$ . (3)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1b) — by symmetry of the uniform distribution, $E(Y) = \tfrac{1 + 6}{2} = 3.5$ .

(b)

M1 (AO1.1b) — computing $E(Y^2) = \tfrac{1}{6}(1 + 4 + 9 + 16 + 25 + 36) = \tfrac{91}{6}$ .
M1 (AO1.1a) — applying $Var(Y) = E(Y^2) - [E(Y)]^2 = \tfrac{91}{6} - \tfrac{49}{4}$ .
A1 (AO2.1) — common denominator: $\tfrac{182 - 147}{12} = \tfrac{35}{12}$ as printed.

(c)

B1 (AO1.1b) — $E(W) = 2E(Y) - 5 = 2(3.5) - 5 = 2$ using linearity of expectation.
B1 (AO1.1b) — $Var(W) = 4 Var(Y) = 4 \cdot \tfrac{35}{12} = \tfrac{35}{3}$ using $Var(aY + b) = a^2 Var(Y)$ .

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated question — AQA uses uniform-distribution and linear-transformation questions to test fluency with the algebraic identities for $E$ and $Var$ .

Synoptic links

Connects to:

Binomial distribution $B(n, p)$ (own lesson): the binomial is derived by treating $n$ independent Bernoulli trials. Its mean $np$ and variance $np(1 - p)$ are special cases of the general $E$ and $Var$ formulae applied to a sum of indicator variables. The general framework here is the foundation; the binomial is one named example.
Normal distribution $N(\mu, \sigma^2)$ (own lesson): the parameters $\mu$ and $\sigma^2$ are precisely $E(X)$ and $Var(X)$ for a continuous random variable. The discrete formulae generalise to integrals: $E(X) = \int x f(x) \, dx$ . Understanding the discrete case first is the conceptual ladder to the continuous case.
Mechanics — impulse and average force: impulse equals change in momentum; over many small collisions, the expected impulse per collision drives the average force. Thinking of force as an expectation links Section R to Newtonian mechanics in a way A* candidates exploit in modelling questions.
Pure — series and sigma notation: $E(X) = \sum x P(X = x)$ is a finite series. Manipulating these sums uses the $\sum\$ -notation rules from Pure: $\sum (a x + b) P(X = x) = a E(X) + b$ . Linearity of expectation is a sigma-notation identity in disguise.
Modelling and expected value: in financial-decision contexts (insurance premiums, expected utility), $E(X)$ is the long-run average payoff. Variance $Var(X)$ measures risk. The economics intuition — "high variance = high risk" — is the same object as the AQA computational definition.

Mark-scheme literacy

Statistical-distribution questions on AQA 7357 split AO marks heavily toward AO1:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–75%	Computing $\sum P(X = x) = 1$ , applying $E(X) = \sum x P(X = x)$ , computing $E(X^2)$ , applying $Var(X) = E(X^2) - [E(X)]^2$
AO2 (reasoning / interpretation)	15–30%	Recognising symmetry to short-circuit $E(X)$ , justifying use of linearity $E(aX + b) = aE(X) + b$ , interpreting variance as spread
AO3 (problem-solving)	5–15%	Modelling a real-world scenario as a discrete distribution, choosing whether uniform/binomial/normal applies, justifying assumptions

Examiner-rewarded phrasing: "since probabilities must sum to 1, we have $\sum P(X = x) = 1$ "; "using $Var(X) = E(X^2) - [E(X)]^2$ "; "by linearity of expectation". Phrases that lose marks: writing $E(X^2)$ when meaning $[E(X)]^2$ or vice versa (a costly notational slip); leaving $Var(X)$ as a decimal when the question asks for an exact fraction; failing to verify that probabilities sum to 1 before computing $E(X)$ .

A specific AQA pattern to watch: questions phrased "show that $k = \ldots$ " demand that you derive $k$ from the total-probability axiom and state the conclusion explicitly. Computing $k$ but not writing the final equality $k = \tfrac{1}{9}$ can lose the A1.

Grade-band model answers

3-mark question

Question: The discrete random variable $X$ has probability distribution given by $P(X = x) = \tfrac{x}{10}$ for $x = 1, 2, 3, 4$ . Find $E(X)$ .

Grade C response (~170 words):

$E(X) = 1 \cdot \tfrac{1}{10} + 2 \cdot \tfrac{2}{10} + 3 \cdot \tfrac{3}{10} + 4 \cdot \tfrac{4}{10} = \tfrac{1 + 4 + 9 + 16}{10} = \tfrac{30}{10} = 3$ .

So $E(X) = 3$ .

Examiner commentary: Full marks (3/3). The candidate applies the expectation formula correctly, evaluates the sum cleanly, and gives the answer as an exact integer. Working is brief but every step is verifiable. A typical Grade C answer for a procedural question — efficient and correct. One minor point: the candidate could have first verified $\sum P(X = x) = \tfrac{10}{10} = 1$ to confirm the distribution is valid. On longer questions where validity is not given, this check is the difference between full and partial marks. Here it is implicit and the marker accepts the omission.

Grade A response (~210 words):*

First verify the distribution is valid: $\sum_{x = 1}^{4} P(X = x) = \tfrac{1 + 2 + 3 + 4}{10} = \tfrac{10}{10} = 1$ . Confirmed.

Apply the definition of expectation:

Statistical Distributions

Statistical Distributions

Random Variables

Discrete Probability Distributions

Finding Unknown Probabilities

Expected Value and Variance

Expected Value (Mean)

Variance

Standard Deviation

Linear Transformations

The Discrete Uniform Distribution

Summary

A-Level Deep Dive: Statistical Distributions

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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