IGCSE Mathematics: Statistics & Probability

Cambridge IGCSE & Edexcel IGCSE · Updated March 2026 · 25 min read

Statistics and Probability questions appear on every IGCSE Mathematics paper and test your ability to interpret data, calculate measures, and reason about chance. This guide covers all required techniques with fully worked examples.

Exam Board Note

This guide covers Cambridge IGCSE Mathematics (0580) and Edexcel International GCSE Mathematics (4MA1). Histograms with unequal class widths (frequency density), cumulative frequency curves, and conditional probability are Extended tier topics on Cambridge; Edexcel assesses them at all levels.

1. Data Collection and Sampling

Types of Data

Qualitative (categorical): Data described in words, e.g., colour, gender.

Quantitative: Numerical data. Discrete data can only take specific values (e.g., number of siblings). Continuous data can take any value in a range (e.g., height).

Common sampling methods include random sampling (each member of the population has an equal chance of being chosen), stratified sampling (the population is divided into groups and a proportional sample is taken from each), and systematic sampling (every $n$th member is chosen).

Worked Example 1.1 — Stratified Sampling

A school has 300 Year 10 students and 200 Year 11 students. A stratified sample of 50 students is required. How many should come from each year?

Step 1 Total students $= 500$. Year 10 proportion $= \dfrac{300}{500} = 0.6$.

Step 2 Year 10 sample: $0.6 \times 50 = 30$. Year 11 sample: $0.4 \times 50 = 20$.

2. Averages and Measures of Spread

Averages

Mean: $\bar{x} = \dfrac{\sum fx}{\sum f}$ where $f$ is the frequency and $x$ is the midpoint of each class.

Median: Middle value when data is ordered. For $n$ values, the median is the $\dfrac{n+1}{2}$th value.

Mode: Most frequently occurring value (or modal class for grouped data).

Range: Maximum $-$ Minimum.

Interquartile Range (IQR): $Q_3 - Q_1$, a measure of spread that is not affected by outliers.

Worked Example 2.1 — Mean from a Frequency Table

Calculate the estimated mean from the grouped frequency table below.

Time (minutes)	Frequency $f$	Midpoint $x$	$fx$
$0 \leq t < 10$	4	5	20
$10 \leq t < 20$	11	15	165
$20 \leq t < 30$	9	25	225
$30 \leq t < 40$	6	35	210
Total	30		620

Result Estimated mean $= \dfrac{\sum fx}{\sum f} = \dfrac{620}{30} \approx 20.7$ minutes.

3. Histograms

For histograms with unequal class widths, the vertical axis shows frequency density, not frequency:

$$\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}$$

The area of each bar equals the frequency: $\text{Frequency} = \text{Frequency density} \times \text{Class width}$.

Worked Example 3.1 — Reading a Histogram

A histogram has a bar from $20$ to $30$ (width $= 10$) with frequency density $3.5$. Another bar from $30$ to $50$ (width $= 20$) has frequency density $2.0$. How many values are in the range $20 \leq x < 50$?

Step 1 Frequency for $20$–$30$: $3.5 \times 10 = 35$.

Step 2 Frequency for $30$–$50$: $2.0 \times 20 = 40$.

Total $35 + 40 = 75$ values.

4. Cumulative Frequency and Box Plots

A cumulative frequency curve (or ogive) plots cumulative frequency against the upper class boundary. It allows you to estimate the median, quartiles, and percentiles.

Reading from a Cumulative Frequency Curve

For a dataset of $n$ values:

Median $Q_2$: at cumulative frequency $\dfrac{n}{2}$
Lower quartile $Q_1$: at cumulative frequency $\dfrac{n}{4}$
Upper quartile $Q_3$: at cumulative frequency $\dfrac{3n}{4}$
IQR $= Q_3 - Q_1$

Worked Example 4.1 — Cumulative Frequency and Box Plot

80 students took a test. The cumulative frequency at scores 30, 40, 50, 60, 70 is 8, 24, 52, 68, 80. Estimate the median, IQR, and describe the distribution.

Step 1 Median: at CF $= 40$ (half of 80). From the curve, the median score $\approx 50$.

Step 2 $Q_1$: at CF $= 20$. From the curve, $Q_1 \approx 43$.

Step 3 $Q_3$: at CF $= 60$. From the curve, $Q_3 \approx 57$.

Step 4 IQR $= 57 - 43 = 14$.

Step 5 The box plot has: minimum $\approx 30$, $Q_1 \approx 43$, median $\approx 50$, $Q_3 \approx 57$, maximum $= 70$. The distribution is roughly symmetric with a slight positive skew.

5. Scatter Diagrams and Correlation

A scatter diagram plots pairs of values to investigate the relationship between two variables. The pattern of points indicates the type of correlation:

Positive correlation: As one variable increases, so does the other.
Negative correlation: As one variable increases, the other decreases.
No correlation: No clear pattern.

The correlation may also be described as strong (points close to a line) or weak (points scattered more widely).

A line of best fit (by eye) should pass through the mean point $(\bar{x}, \bar{y})$ with roughly equal numbers of points on each side. Do not force the line through the origin unless the data demands it.

6. Basic Probability

Probability Rules

Probability always satisfies $0 \leq P(A) \leq 1$.

$$P(A') = 1 - P(A)$$

Addition rule (mutually exclusive events): $P(A \cup B) = P(A) + P(B)$

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

Multiplication rule (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 6.1 — Conditional Probability

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that the second ball is red given the first was red.

Step 1 After one red ball is removed: 4 red and 3 blue balls remain (7 total).

Result $P(\text{2nd red} \mid \text{1st red}) = \dfrac{4}{7}$.

7. Tree Diagrams

Tree diagrams show all possible outcomes of a multi-stage experiment. Multiply along branches (AND) and add across branches (OR).

Worked Example 7.1 — Tree Diagram Without Replacement

A box contains 4 green and 6 yellow counters. Two counters are drawn without replacement. Find the probability that exactly one is green.

Step 1 Draw the tree. First draw: $P(G) = \dfrac{4}{10}$, $P(Y) = \dfrac{6}{10}$.

Second draw given first was G: $P(G) = \dfrac{3}{9}$, $P(Y) = \dfrac{6}{9}$.

Second draw given first was Y: $P(G) = \dfrac{4}{9}$, $P(Y) = \dfrac{5}{9}$.

Step 2 Exactly one green occurs in two ways: GY or YG.

$$P(GY) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$$

$$P(YG) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$$

Step 3 $P(\text{exactly one green}) = \dfrac{24}{90} + \dfrac{24}{90} = \dfrac{48}{90} = \dfrac{8}{15}$.

8. Venn Diagrams

Venn diagrams represent sets and their relationships. The universal set $\xi$ contains all elements. The complement $A'$ is everything not in $A$.

Worked Example 8.1 — Two-Set Venn Diagram

In a class of 30 students, 18 study French ($F$), 14 study Spanish ($S$), and 6 study both. Find: (a) $P(F \cup S)$, (b) $P(F \text{ only})$, (c) $P(S \mid F)$.

Fill the Venn diagram:

Both $F$ and $S$: 6
$F$ only: $18 - 6 = 12$
$S$ only: $14 - 6 = 8$
Neither: $30 - 12 - 6 - 8 = 4$

(a) $P(F \cup S) = \dfrac{12 + 6 + 8}{30} = \dfrac{26}{30} = \dfrac{13}{15}$

(b) $P(F \text{ only}) = \dfrac{12}{30} = \dfrac{2}{5}$

9. Practice Problems

Problem 1

Find the mean, median, and mode of: $3, 7, 7, 9, 11, 13, 13, 13, 15$.

Show Solution

Mean: $\dfrac{3+7+7+9+11+13+13+13+15}{9} = \dfrac{91}{9} \approx 10.1$

Median: 9 values, median is the 5th: $11$.

Mode: $13$ (appears 3 times).

Problem 2

A histogram bar for the class $10 \leq x < 15$ has frequency density $6$. How many values are in this class?

Show Solution

Class width $= 15 - 10 = 5$.

Frequency $= 6 \times 5 = 30$.

Problem 3

A fair coin is flipped and a fair die is rolled. Find the probability of getting a head and a prime number.

Show Solution

Primes on a die: $\{2, 3, 5\}$, so $P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$.

Events are independent: $P(H \cap \text{prime}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$.

Problem 4

60 people were surveyed. 35 own a cat ($C$), 28 own a dog ($D$), and 12 own both. Find the probability that a randomly chosen person owns neither.

Show Solution

$|C \cup D| = 35 + 28 - 12 = 51$. Neither $= 60 - 51 = 9$.

$P(\text{neither}) = \dfrac{9}{60} = \dfrac{3}{20}$.

Problem 5

A bag contains 3 red, 4 blue, and 5 green balls. Two balls are drawn without replacement. Find the probability that both are the same colour.

Show Solution

Total balls $= 12$.

$P(RR) = \dfrac{3}{12} \times \dfrac{2}{11} = \dfrac{6}{132}$

$P(BB) = \dfrac{4}{12} \times \dfrac{3}{11} = \dfrac{12}{132}$

$P(GG) = \dfrac{5}{12} \times \dfrac{4}{11} = \dfrac{20}{132}$

$P(\text{same colour}) = \dfrac{6+12+20}{132} = \dfrac{38}{132} = \dfrac{19}{66}$

Problem 6

The table gives the number of goals scored by a football team in 30 matches: 0 goals (8 matches), 1 goal (10 matches), 2 goals (7 matches), 3 goals (4 matches), 4 goals (1 match). Calculate the mean number of goals per match.

Show Solution

$\sum fx = 0(8) + 1(10) + 2(7) + 3(4) + 4(1) = 0 + 10 + 14 + 12 + 4 = 40$

Mean $= \dfrac{40}{30} = \dfrac{4}{3} \approx 1.33$ goals per match.

Problem 7

In a class, $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cap B) = 0.3$. Are events $A$ and $B$ independent? Justify your answer.

Show Solution

For independence, we need $P(A \cap B) = P(A) \times P(B)$.

$P(A) \times P(B) = 0.6 \times 0.5 = 0.30$.

Since $P(A \cap B) = 0.3 = 0.30$, events $A$ and $B$ are independent.

Problem 8

The IQR of a dataset is 18 and $Q_1 = 24$. Find $Q_3$ and state the range within which values are not considered outliers (using the rule: outlier if below $Q_1 - 1.5 \times \text{IQR}$ or above $Q_3 + 1.5 \times \text{IQR}$).

Show Solution

$Q_3 = Q_1 + \text{IQR} = 24 + 18 = 42$.

Lower fence: $24 - 1.5 \times 18 = 24 - 27 = -3$.

Upper fence: $42 + 1.5 \times 18 = 42 + 27 = 69$.

Values between $-3$ and $69$ are not considered outliers.