Topic 13: Statistics and Probability

🟢 Lite — Quick Review (1h–1d)

Statistics is the branch of mathematics concerned with collecting, organising, presenting, analysing, and interpreting data. Probability, meanwhile, measures the likelihood or chance that a particular event will occur, expressed as a number between 0 (impossible) and 1 (certain). In the WASSCE quantitative reasoning paper, questions in this topic assess your ability to handle real data sets and calculate various statistical measures, as well as your understanding of fundamental probability concepts.

Central tendency measures identify a single value that represents the centre of a data set. The three primary measures are: the mean (arithmetic average), the median (the middle value when data is arranged in order), and the mode (the most frequently occurring value). Each has advantages and limitations. The mean uses all data values but is affected by extreme outliers. The median is resistant to outliers but ignores the actual magnitude of values. The mode is useful for categorical data but may not exist or may not be unique.

Key Facts:

• Mean = sum of all values ÷ number of values
• Median: middle value when data is ordered (or average of two middle values for even n)
• Mode: most frequent value(s); a data set may have no mode, one mode (unimodal), or two modes (bimodal)
• Range = highest value - lowest value (measure of spread)
• Probability scale: 0 ≤ P(event) ≤ 1
• P(event) = number of favourable outcomes ÷ total number of possible outcomes
• P(complement) = 1 - P(event)

⚡ Exam Tip: When calculating the mean from a frequency table, remember to multiply each value by its frequency before summing—do not simply average the values.

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🟡 Standard — Regular Study (2d–2mo)

Mean, Median, and Mode

Find the mean, median, and mode of the data set: 4, 7, 3, 8, 7, 2, 7, 9

Mean = (4 + 7 + 3 + 8 + 7 + 2 + 7 + 9) ÷ 8 = 47 ÷ 8 = 5.875

Arranged in order: 2, 3, 4, 7, 7, 7, 8, 9 Median = average of 4th and 5th values = (7 + 7) ÷ 2 = 7

Mode = 7 (appears 3 times, more than any other value)

Grouped Frequency Distributions

For grouped data, the mean is calculated using class midpoints:

Class	Frequency (f)	Midpoint (x)	fx
0-9	5	4.5	22.5
10-19	12	14.5	174
20-29	8	24.5	196
30-39	3	34.5	103.5
40-49	2	44.5	89
Totals	30	—	585

Mean = 585 ÷ 30 = 19.5

Cumulative Frequency and Median

The cumulative frequency is obtained by adding frequencies progressively.

Score	Frequency	Cumulative Frequency
0-9	3	3
10-19	7	10
20-29	12	22
30-39	5	27
40-49	3	30

The median is found at the (n+1)/2 = 15.5th position, which falls in the 20-29 class.

Probability — Basic Concepts

A bag contains 6 red balls and 4 blue balls. Find the probability of drawing a red ball.

P(red) = 6/(6 + 4) = 6/10 = 3/5 = 0.6

Probability of Complementary Events

If P(A) = 0.7, then P(not A) = 1 - 0.7 = 0.3

Probability of Combined Events

For independent events A and B:

• P(A and B) = P(A) × P(B)
• P(A or B) = P(A) + P(B) - P(A and B)

Two dice are thrown. Find the probability of getting a sum of 8.

Total possible outcomes: 6 × 6 = 36 Favourable outcomes (sum = 8): (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes P(sum = 8) = 5/36

Comparison Table: Measures of Central Tendency

Measure	Calculation	Advantage	Disadvantage
Mean	Sum ÷ n	Uses all data	Affected by extreme values
Median	Middle value	Resistant to outliers	Ignores magnitude of values
Mode	Most frequent	Useful for categories	May not exist or be unique

Common Mistakes to Avoid:

1. Forgetting to arrange data before finding the median
2. Mixing up population standard deviation (σ) with sample standard deviation (s)
3. Assuming events are independent when they are not
4. Forgetting to subtract P(both) when calculating P(A or B)
5. Using class boundaries instead of midpoints when calculating mean from grouped data

Problem-Solving Strategy:

1. For statistics: identify what measure is being asked, then apply the correct formula
2. For probability: list all possible outcomes systematically
3. Count favourable outcomes carefully
4. Apply the appropriate probability formula
5. Express your answer as a fraction in simplest form, decimal, or percentage as required

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🔴 Extended — Deep Study (3mo+)

Historical Context: The Birth of Probability Theory

Probability theory emerged from gambling problems in 17th-century France. The Chevalier de Méré approached mathematician Blaise Pascal with questions about dice games. Pascal, collaborating with Pierre de Fermat, developed fundamental counting principles and probability calculations. This correspondence in 1654 is considered the formal beginning of probability theory as a mathematical discipline.

Standard Deviation

Standard deviation measures the spread of data around the mean. For ungrouped data:

σ = √[Σ(x - x̄)² / n]

For the data set: 2, 4, 6, 8, 10 Mean = 6 Variance = [(2-6)² + (4-6)² + (6-6)² + (8-6)² + (10-6)²] ÷ 5 = [16 + 4 + 0 + 4 + 16] ÷ 5 = 40 ÷ 5 = 8 Standard deviation = √8 = 2√2 ≈ 2.83

Measures of Position — Quartiles

Quartiles divide data into four equal parts:

• Q₁ (lower quartile): 25% of data below
• Q₂ (median): 50% of data below
• Q₃ (upper quartile): 75% of data below

Interquartile range (IQR) = Q₃ - Q₁

For the ordered data: 3, 5, 7, 9, 11, 13, 15 Q₂ = 9, Q₁ = 5, Q₃ = 13 IQR = 13 - 5 = 8

Conditional Probability

P(A|B) denotes the probability of event A given that event B has occurred: P(A|B) = P(A and B) / P(B)

In a class of 30 students, 18 play football and 12 play basketball. If 8 students play both sports, what is the probability that a randomly chosen football player also plays basketball?

P(both) = 8/30, P(football) = 18/30 P(basketball|football) = (8/30) / (18/30) = 4/9

Tree Diagrams

For independent events over successive trials, tree diagrams help enumerate outcomes:

The probability of a batsman hitting a boundary is 0.3. Find probabilities for 2 consecutive deliveries:

• HH: 0.3 × 0.3 = 0.09
• HM: 0.3 × 0.7 = 0.21
• MH: 0.7 × 0.3 = 0.21
• MM: 0.7 × 0.7 = 0.49 (Verify: 0.09 + 0.21 + 0.21 + 0.49 = 1.00)

Permutations and Combinations

Permutations (order matters): P(n,r) = n! / (n - r)!

How many ways can 3 prizes be awarded to 10 students if no student can win more than one prize? P(10,3) = 10! / 7! = 10 × 9 × 8 = 720 ways

Combinations (order does not matter): C(n,r) = n! / [r!(n-r)!]

How many ways can a committee of 4 be formed from 7 people? C(7,4) = 7! / (4! × 3!) = (7 × 6 × 5) / (3 × 2 × 1) = 35 ways

WASSCE Examination Patterns:

The WASSCE quantitative reasoning paper typically includes:

1. Mean, median, and mode calculations (Objective)
2. Probability of single and combined events (Objective and Theory)
3. Data interpretation from tables and charts (Objective)
4. Permutations and combinations (Theory)
5. Standard deviation calculations (Theory)

⚡ Pro Exam Tip: In the WASSCE, always simplify probability answers to lowest terms. When using tree diagrams, label each branch with its probability. For combined events, remember that P(A or B) = P(A) + P(B) - P(A and B), and subtract the overlap to avoid double-counting.

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