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Additional Mathematics

Statistics and Probability

PDF
Matthew Williams
|May 16, 2026|8 min read
Box PlotsPaper 01Paper 02ProbabilitySection 4Standard DeviationStatisticsStem and LeafVariance

Data types, measures of central tendency and spread, quartiles, variance and standard deviation, stem-and-leaf diagrams, box-and-whisker plots, probability rules, conditional probability, tree diagrams, and Venn diagrams.

Section 4 contributes 5 items to Paper 01 and one 20-mark question to Paper 02. The Paper 02 question typically combines statistical calculation with interpretation and a probability problem using diagrams.

Types of Data

Qualitative (categorical) data describes attributes without numerical value: eye colour, nationality, type of vehicle.

Quantitative data is numerical and divides into two types:

  • Discrete: takes separate countable values (number of siblings, goals scored).
  • Continuous: takes any value within an interval (height, temperature, time).

Measures of Central Tendency

MeasureDefinitionBest used when
MeanSum of all values divided by the countData is symmetric with no extreme outliers
MedianMiddle value when data is orderedData is skewed or has outliers
ModeMost frequent valueIdentifying the most common category

xˉ=∑xn(ungrouped)xˉ=∑fx∑f(grouped/frequency)\bar{x} = \frac{\sum x}{n} \qquad \text{(ungrouped)} \qquad \bar{x} = \frac{\sum fx}{\sum f} \qquad \text{(grouped/frequency)}xˉ=n∑x​(ungrouped)xˉ=∑f∑fx​(grouped/frequency)

Measures of Spread

The simplest measure of spread is the range: range=maximum−minimum\text{range} = \text{maximum} - \text{minimum}range=maximum−minimum. It is easy to compute but sensitive to a single extreme value.

Quartiles and Interquartile Range

Ordered data is divided into four equal parts by the quartiles Q1Q_1Q1​, Q2Q_2Q2​ (median), and Q3Q_3Q3​.

Q1=median of lower half,Q2=median,Q3=median of upper halfQ_1 = \text{median of lower half}, \quad Q_2 = \text{median}, \quad Q_3 = \text{median of upper half}Q1​=median of lower half,Q2​=median,Q3​=median of upper half

IQR=Q3−Q1Semi-IQR=Q3−Q12\text{IQR} = Q_3 - Q_1 \qquad \text{Semi-IQR} = \frac{Q_3 - Q_1}{2}IQR=Q3​−Q1​Semi-IQR=2Q3​−Q1​​

The IQR measures the spread of the middle 50% of the data. Unlike the range, it is not distorted by a single extreme value.

Example

Data (already ordered): 12,15,18,20,22,25,27,30,3212, 15, 18, 20, 22, 25, 27, 30, 3212,15,18,20,22,25,27,30,32.

Median (Q2Q_2Q2​): 5th value =22= 22=22.

Lower half: 12,15,18,2012, 15, 18, 2012,15,18,20. Q1=15+182=16.5Q_1 = \dfrac{15+18}{2} = 16.5Q1​=215+18​=16.5

Upper half: 25,27,30,3225, 27, 30, 3225,27,30,32. Q3=27+302=28.5Q_3 = \dfrac{27+30}{2} = 28.5Q3​=227+30​=28.5

IQR=28.5−16.5=12\text{IQR} = 28.5 - 16.5 = 12IQR=28.5−16.5=12

Percentiles

The pppth percentile is the value below which p%p\%p% of the data lies. Quartiles are specific percentiles: Q1=P25Q_1 = P_{25}Q1​=P25​, Q2=P50Q_2 = P_{50}Q2​=P50​, Q3=P75Q_3 = P_{75}Q3​=P75​.

Variance and Standard Deviation

Variance measures average squared distance from the mean. Standard deviation is its square root, restoring the original units.

S2=∑(x−xˉ)2nS=∑(x−xˉ)2n(ungrouped)S^2 = \frac{\sum(x - \bar{x})^2}{n} \qquad S = \sqrt{\frac{\sum(x-\bar{x})^2}{n}} \qquad \text{(ungrouped)}S2=n∑(x−xˉ)2​S=n∑(x−xˉ)2​​(ungrouped)

S2=∑f(x−xˉ)2∑f(grouped)S^2 = \frac{\sum f(x-\bar{x})^2}{\sum f} \qquad \text{(grouped)}S2=∑f∑f(x−xˉ)2​(grouped)

A small standard deviation means values cluster closely around the mean (consistent). A large standard deviation means values are widely spread (variable).

Exam Tip

Two datasets can have the same mean but very different standard deviations. When comparing distributions, state both the measure of centre and the measure of spread. "Group A has a higher mean but Group B is more consistent because its standard deviation is smaller."

Stem-and-Leaf Diagrams

A stem-and-leaf diagram keeps the original data values while showing the distribution shape.

Constructing one:

  1. Split each value: the stem is all digits except the last; the leaf is the last digit.
  2. Write stems in a column and leaves in ascending order beside each stem.
  3. Include a key.
Example

Data: 17,19,22,31,33,38,44,47,4917, 19, 22, 31, 33, 38, 44, 47, 4917,19,22,31,33,38,44,47,49

1 | 7 9
2 | 2
3 | 1 3 8
4 | 4 7 9

Key: 1 | 7 means 17

A back-to-back stem-and-leaf diagram places two datasets side by side sharing a common stem, allowing direct visual comparison.

Advantages of stem-and-leafDisadvantages
Original values are preservedImpractical for large datasets
Shape of distribution is visibleDifficult to compare non-overlapping ranges

Box-and-Whisker Plots

A box plot summarises a dataset using five values: minimum, Q1Q_1Q1​, median, Q3Q_3Q3​, maximum.

Min    Q₁    Q₂    Q₃    Max
 |------[======|======]------|

The box spans from Q1Q_1Q1​ to Q3Q_3Q3​ and contains the middle 50% of the data. The whiskers extend to the minimum and maximum.

Reading Skewness from a Box Plot

PatternSkew
Q3−Q2=Q2−Q1Q_3 - Q_2 = Q_2 - Q_1Q3​−Q2​=Q2​−Q1​ and whiskers roughly equalSymmetric
Q3−Q2>Q2−Q1Q_3 - Q_2 > Q_2 - Q_1Q3​−Q2​>Q2​−Q1​ or right whisker longerPositive skew (tail to right)
Q3−Q2<Q2−Q1Q_3 - Q_2 < Q_2 - Q_1Q3​−Q2​<Q2​−Q1​ or left whisker longerNegative skew (tail to left)

For a positively skewed distribution: Mode <<< Median <<< Mean. For a negatively skewed distribution: Mean <<< Median <<< Mode.

Probability Theory

An experiment produces outcomes. The set of all possible outcomes is the sample space SSS. An event is a subset of the sample space.

P(A)=number of outcomes in Atotal number of outcomes in SP(A) = \frac{\text{number of outcomes in } A}{\text{total number of outcomes in } S}P(A)=total number of outcomes in Snumber of outcomes in A​

This is classical probability, valid when all outcomes are equally likely.

Relative frequency gives an experimental estimate of probability when theoretical equally-likely outcomes cannot be assumed:

P(A)≈number of times A occurredtotal number of trialsP(A) \approx \frac{\text{number of times } A \text{ occurred}}{\text{total number of trials}}P(A)≈total number of trialsnumber of times A occurred​

As the number of trials increases, the relative frequency approaches the true probability.

Basic Laws

  • 0≤P(A)≤10 \leq P(A) \leq 10≤P(A)≤1 for any event AAA.
  • ∑P(all outcomes)=1\sum P(\text{all outcomes}) = 1∑P(all outcomes)=1.
  • P(A′)=1−P(A)P(A') = 1 - P(A)P(A′)=1−P(A), where A′A'A′ is the complement of AAA (the event that AAA does not occur).

The Addition Rule

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)P(A∪B)=P(A)+P(B)−P(A∩B)

For mutually exclusive events (AAA and BBB cannot both occur): P(A∩B)=0P(A \cap B) = 0P(A∩B)=0, so:

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)P(A∪B)=P(A)+P(B)

Example

A card is drawn from a standard deck. P(K)=4/52P(K) = 4/52P(K)=4/52 and P(A)=4/52P(A) = 4/52P(A)=4/52 where KKK = King and AAA = Ace. Since these are mutually exclusive:

P(K∪A)=452+452=852=213P(K \cup A) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}P(K∪A)=524​+524​=528​=132​

Conditional Probability

The conditional probability of AAA given BBB has occurred:

P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}P(A∣B)=P(B)P(A∩B)​

Rearranging: P(A∩B)=P(A∣B)⋅P(B)P(A \cap B) = P(A|B) \cdot P(B)P(A∩B)=P(A∣B)⋅P(B).

Independent Events

Events AAA and BBB are independent if knowing that BBB occurred gives no information about AAA:

P(A∣B)=P(A)equivalentlyP(A∩B)=P(A)⋅P(B)P(A|B) = P(A) \qquad \text{equivalently} \qquad P(A \cap B) = P(A) \cdot P(B)P(A∣B)=P(A)equivalentlyP(A∩B)=P(A)⋅P(B)

Exam Tip

Do not confuse "mutually exclusive" with "independent." Mutually exclusive events (P(A∩B)=0P(A \cap B) = 0P(A∩B)=0) cannot both occur. Independent events can both occur, but neither influences the other's probability. Two events with non-zero probabilities cannot be both mutually exclusive and independent.

Probability Diagrams

Possibility Space Diagrams

A possibility space (sample space) diagram lists all outcomes in a grid. Useful for two-stage experiments like rolling two dice.

Example

Two fair dice are rolled. The sample space has 6×6=366 \times 6 = 366×6=36 equally likely outcomes.

P(sum=7)=636=16P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}P(sum=7)=366​=61​ (the six pairs that sum to 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)).

Tree Diagrams

Tree diagrams show sequential outcomes. Branches show each possible outcome at each stage; multiply along branches for intersection probabilities.

The syllabus restricts tree diagrams to two initial branches.

Example

A bag contains 3 red and 5 blue balls. A ball is drawn, its colour noted, then a second ball is drawn without replacement.

P(both red) =38×27=656=328= \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}=83​×72​=566​=283​

P(one of each) =38×57+58×37=1556+1556=3056=1528= \dfrac{3}{8} \times \dfrac{5}{7} + \dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56} + \dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28}=83​×75​+85​×73​=5615​+5615​=5630​=2815​

Venn Diagrams

Venn diagrams with two sets partition the sample space into four regions: AAA only, BBB only, A∩BA \cap BA∩B, and neither. The sum of all regions must equal P(S)=1P(S) = 1P(S)=1 (or n(S)n(S)n(S) for counts).

The syllabus restricts Venn diagrams to two sets.

Example

In a class of 30 students, 18 study Physics, 15 study Chemistry, and 8 study both.

P(Physics only) =18−830=1030= \dfrac{18-8}{30} = \dfrac{10}{30}=3018−8​=3010​

P(at least one) =18+15−830=2530=56= \dfrac{18+15-8}{30} = \dfrac{25}{30} = \dfrac{5}{6}=3018+15−8​=3025​=65​

P(neither) =30−2530=530=16= \dfrac{30-25}{30} = \dfrac{5}{30} = \dfrac{1}{6}=3030−25​=305​=61​

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