A-Level Mathematics: Statistics

Edexcel · AQA · OCR A-Level Mathematics · Updated March 2026 · 35 min read

The Statistics component of A-Level Mathematics tests your ability to collect, present, and analyse data, reason with probability, and apply formal statistical tests. This guide covers the full Statistics content at both Year 1 and Year 2 level with fully worked exam-style examples.

Exam Board Note

Statistics is tested in Paper 3 (Statistics and Mechanics) for Edexcel, as a standalone Statistics paper for AQA, and in Paper 3 for OCR. Edexcel requires familiarity with the Large Data Set (Camtool weather data). All three boards assess the binomial and normal distributions and hypothesis testing. The critical region approach to hypothesis testing is required by all boards.

1. Statistical Sampling

Sampling Methods

Simple random sampling: Every member has an equal chance of selection. Use a random number generator or lottery method. Advantage: unbiased. Disadvantage: requires a complete sampling frame.

Systematic sampling: Select every $k$th member after a random start. Simple to implement but can introduce bias if there is periodicity in the population.

Stratified sampling: Divide the population into groups (strata) and sample proportionally from each. Ensures representation of subgroups.

Quota sampling: Non-random. Interviewer fills quotas from subgroups. Quick but may introduce bias.

Opportunity (convenience) sampling: Use whoever is available. Simple but highly biased.

A population is the complete set of items being studied. A sample is a subset. A census observes every member of the population; it is expensive but gives exact results.

2. Data Presentation and Interpretation

Measures of Location and Spread

Key Measures

Mean: $\bar{x} = \dfrac{\sum fx}{\sum f}$

Standard deviation: $\sigma = \sqrt{\dfrac{\sum fx^2}{\sum f} - \bar{x}^2}$

Coding: If $y = \dfrac{x - a}{b}$, then $\bar{x} = a + b\bar{y}$ and $\sigma_x = b\sigma_y$.

Outlier: A value more than $1.5 \times \text{IQR}$ below $Q_1$ or above $Q_3$. (Alternative: more than 2 standard deviations from the mean.)

Worked Example 2.1 — Coding to Find Mean and Standard Deviation

Data is coded as $y = \dfrac{x - 50}{10}$. The coded data has $\bar{y} = 2.3$ and $\sigma_y = 1.4$. Find $\bar{x}$ and $\sigma_x$.

Mean: $\bar{x} = 50 + 10 \times 2.3 = 50 + 23 = 73$.

Std dev: $\sigma_x = 10 \times 1.4 = 14$.

Interpreting Diagrams

At A-Level, you must be able to interpret and critique statistical diagrams: box plots, histograms, cumulative frequency curves, and scatter diagrams. Key vocabulary: skewness (positively skewed if mean $>$ median $>$ mode), correlation coefficient, and the distinction between correlation and causation.

3. Probability

Probability Rules

Addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Mutually exclusive: $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.

Independent events: $P(A \cap B) = P(A) \times P(B)$.

Conditional probability: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$

Worked Example 3.1 — Conditional Probability with a Venn Diagram

Events $A$ and $B$ satisfy: $P(A) = 0.6$, $P(B) = 0.45$, and $P(A \mid B) = 0.4$. Find $P(A \cup B)$, and determine whether $A$ and $B$ are independent.

Step 1 $P(A \cap B) = P(A \mid B) \times P(B) = 0.4 \times 0.45 = 0.18$.

Step 2 $P(A \cup B) = 0.6 + 0.45 - 0.18 = 0.87$.

Step 3 Test independence: $P(A) \times P(B) = 0.6 \times 0.45 = 0.27 \neq 0.18 = P(A \cap B)$. So $A$ and $B$ are not independent.

4. Statistical Distributions: Binomial

Binomial Distribution $B(n, p)$

A random variable $X \sim B(n, p)$ if:

There are $n$ independent trials.
Each trial has two outcomes: success (probability $p$) or failure (probability $1-p$).
$p$ is constant across trials.

$$P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$$

Mean: $\mu = np$, $\quad$ Variance: $\sigma^2 = np(1-p)$

Worked Example 4.1 — Binomial Probabilities

A biased coin has $P(\text{Heads}) = 0.6$. It is tossed 8 times. Find: (a) $P(X = 5)$, (b) $P(X \geq 6)$, (c) $E(X)$ and $\text{Var}(X)$.

(a) $P(X=5) = \dbinom{8}{5}(0.6)^5(0.4)^3 = 56 \times 0.07776 \times 0.064 = 0.2787$

(b) $P(X \geq 6) = P(6) + P(7) + P(8)$

$P(X=6) = \dbinom{8}{6}(0.6)^6(0.4)^2 = 28 \times 0.046656 \times 0.16 = 0.2090$

$P(X=7) = \dbinom{8}{7}(0.6)^7(0.4)^1 = 8 \times 0.027994 \times 0.4 = 0.0896$

$P(X=8) = (0.6)^8 = 0.01680$

$P(X \geq 6) = 0.2090 + 0.0896 + 0.0168 = 0.3154$

5. Statistical Distributions: Normal

Normal Distribution $N(\mu, \sigma^2)$

If $X \sim N(\mu, \sigma^2)$, the distribution is bell-shaped and symmetric about $\mu$.

Standardise: $Z = \dfrac{X - \mu}{\sigma}$ where $Z \sim N(0, 1)$.

Use statistical tables or a calculator to find $P(Z < z)$.

Key percentage points of $N(0,1)$:

$P(Z < 1.645) = 0.95$ (90% two-tail)
$P(Z < 1.960) = 0.975$ (95% two-tail)
$P(Z < 2.326) = 0.99$ (98% two-tail)

STAT·1 — Standard normal distribution $Z \sim N(0,1)$. Green: $P(-1 \lt Z \lt 1) \approx 68.3\%$. Red tails: $P(|Z| \gt 1.96) \approx 5\%$ in total (2.5% each side), giving the 95% two-tailed critical value $z = \pm 1.96$.

Worked Example 5.1 — Normal Distribution

The heights of students follow $X \sim N(168, 64)$. Find: (a) $P(X < 175)$, (b) $P(160 < X < 175)$, (c) the value of $h$ such that $P(X > h) = 0.1$.

(a) $P(X < 175) = P\!\left(Z < \dfrac{175-168}{8}\right) = P(Z < 0.875) \approx 0.8092$.

(b) $P(160 < X < 175) = P(-1 < Z < 0.875) = \Phi(0.875) - \Phi(-1) = 0.8092 - (1 - 0.8413) = 0.8092 - 0.1587 = 0.6505$.

Worked Example 5.2 — Finding $\mu$ and $\sigma$

$X \sim N(\mu, \sigma^2)$ with $P(X < 10) = 0.2$ and $P(X > 20) = 0.15$. Find $\mu$ and $\sigma$.

Step 1 From tables: $P(Z < -0.842) = 0.2$ and $P(Z < 1.036) = 0.85$ (since $P(X>20)=0.15$).

$\dfrac{10 - \mu}{\sigma} = -0.842$ and $\dfrac{20 - \mu}{\sigma} = 1.036$.

Step 2 Subtract the first equation from the second:

$\dfrac{10}{\sigma} = 1.878 \Rightarrow \sigma = 5.325$.

Step 3 $\mu = 10 + 0.842 \times 5.325 = 14.48$. So $\mu \approx 14.5$ and $\sigma \approx 5.33$.

Normal Approximation to the Binomial

When $n$ is large and $p$ is not too close to 0 or 1, $B(n,p)$ is approximately $N(np, np(1-p))$. Apply a continuity correction: $P(X \leq k)$ becomes $P\!\left(X < k + \frac{1}{2}\right)$.

6. Hypothesis Testing

Hypothesis Testing Framework

A hypothesis test uses sample data to assess a claim about a population parameter.

Null hypothesis $H_0$: the statement we assume to be true.

Alternative hypothesis $H_1$: what we accept if we reject $H_0$.

The significance level $\alpha$ is the maximum acceptable probability of incorrectly rejecting $H_0$ (Type I error).

One-tailed test: $H_1$ states a direction (e.g. $p > p_0$ or $p < p_0$).

Two-tailed test: $H_1: p \neq p_0$. Each tail has significance level $\alpha/2$.

Worked Example 6.1 — One-Tailed Binomial Hypothesis Test

A factory claims that 20% of items it produces are defective. A quality controller suspects the true proportion is higher. She tests a random sample of 15 items and finds 6 defective. Test the factory's claim at the 5% significance level.

Step 1 Let $p$ be the probability that an item is defective.

$H_0: p = 0.2$ (accept the factory's claim).

$H_1: p > 0.2$ (one-tailed, since we suspect the proportion is higher).

Step 2 Assume $H_0$: $X \sim B(15, 0.2)$. Under $H_0$, the observed value is $x = 6$.

Compute $P(X \geq 6) = 1 - P(X \leq 5)$.

$P(X \leq 5) = \sum_{r=0}^{5} \binom{15}{r}(0.2)^r(0.8)^{15-r}$

Using a calculator or tables: $P(X \leq 5) \approx 0.9389$.

$P(X \geq 6) = 1 - 0.9389 = 0.0611$.

Step 3 Compare with significance level: $0.0611 > 0.05$.

Since the $p$-value exceeds the significance level, we fail to reject $H_0$. There is insufficient evidence at the 5% level to conclude that the proportion of defective items exceeds 20%.

Worked Example 6.2 — Two-Tailed Test Using Critical Region

A coin is tossed 20 times. Test at the 5% significance level whether the coin is biased, given 14 heads are observed. Use the critical region approach.

Step 1 $H_0: p = 0.5$, $H_1: p \neq 0.5$ (two-tailed). $X \sim B(20, 0.5)$.

Step 2 Find critical regions. Each tail should have probability $\leq 0.025$.

Lower tail: Find largest $c_1$ with $P(X \leq c_1) \leq 0.025$.

$P(X \leq 5) = 0.0207 \leq 0.025$, $P(X \leq 6) = 0.0577 > 0.025$. So $c_1 = 5$.

Upper tail: Find smallest $c_2$ with $P(X \geq c_2) \leq 0.025$.

$P(X \geq 15) = 0.0207 \leq 0.025$, $P(X \geq 14) = 0.0577 > 0.025$. So $c_2 = 15$.

Step 3 Critical region: $X \leq 5$ or $X \geq 15$.

Observed value: $x = 14$. Since $14 \notin$ critical region, fail to reject $H_0$. There is insufficient evidence that the coin is biased.

7. Practice Problems

Problem 1

A class of 200 students has 80 in Year 12 and 120 in Year 13. A stratified sample of 30 is required. How many come from each year?

Show Solution

Year 12: $\dfrac{80}{200} \times 30 = 12$ students. Year 13: $\dfrac{120}{200} \times 30 = 18$ students.

Problem 2

$X \sim B(12, 0.35)$. Find: (a) $P(X = 4)$, (b) $P(X \leq 3)$, (c) $E(X)$, (d) $\text{Var}(X)$.

Show Solution

(a) $P(X=4) = \dbinom{12}{4}(0.35)^4(0.65)^8 = 495 \times 0.015006 \times 0.031941 \approx 0.2367$

(b) $P(X \leq 3) = P(0)+P(1)+P(2)+P(3)$. Using a calculator: $\approx 0.3467$.

(c) $E(X) = 12 \times 0.35 = 4.2$.

(d) $\text{Var}(X) = 12 \times 0.35 \times 0.65 = 2.73$.

Problem 3

$X \sim N(50, 25)$. Find $P(45 < X < 60)$.

Show Solution

$\sigma = 5$. Standardise: $P\!\left(\dfrac{45-50}{5} < Z < \dfrac{60-50}{5}\right) = P(-1 < Z < 2)$

$= \Phi(2) - \Phi(-1) = 0.9772 - 0.1587 = 0.8185$.

Problem 4

A supermarket claims that 30% of customers use self-checkout. In a sample of 25 customers, 4 used self-checkout. Test at the 10% significance level whether the true proportion is less than 30% ($H_1: p < 0.3$).

Show Solution

$H_0: p = 0.3$, $H_1: p < 0.3$. $X \sim B(25, 0.3)$ under $H_0$.

$P(X \leq 4) = \sum_{r=0}^{4}\dbinom{25}{r}(0.3)^r(0.7)^{25-r}$.

Using cumulative binomial tables: $P(X \leq 4) \approx 0.0905$.

Since $0.0905 < 0.10$ (significance level), reject $H_0$.

There is sufficient evidence at the 10% level that the true proportion is less than 30%.

Problem 5

$X \sim N(\mu, \sigma^2)$ with $P(X < 15) = 0.3$ and $P(X < 25) = 0.8$. Find $\mu$ and $\sigma$.

Show Solution

$P(Z < z_1) = 0.3 \Rightarrow z_1 = -0.524$. $P(Z < z_2) = 0.8 \Rightarrow z_2 = 0.842$.

$\dfrac{15 - \mu}{\sigma} = -0.524$ and $\dfrac{25 - \mu}{\sigma} = 0.842$.

Subtract: $\dfrac{10}{\sigma} = 1.366 \Rightarrow \sigma = 7.32$.

$\mu = 15 + 0.524 \times 7.32 = 18.84$. So $\mu \approx 18.8$, $\sigma \approx 7.32$.

Problem 6

Events $A$ and $B$ have $P(A) = 0.5$, $P(B) = 0.4$, $P(A \cup B) = 0.7$. (a) Find $P(A \cap B)$. (b) Find $P(A \mid B)$. (c) Are $A$ and $B$ independent?

Show Solution

(a) $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.7 = 0.2$.

(b) $P(A \mid B) = \dfrac{0.2}{0.4} = 0.5$.

(c) $P(A) \times P(B) = 0.5 \times 0.4 = 0.2 = P(A \cap B)$. So $A$ and $B$ are independent.