Statistics & Probability

From data collection and frequency tables through to hypothesis testing and regression analysis — comprehensive coverage of the Applied Mathematics strand as defined by KICD for Kenya's Competency Based Curriculum. Aligned to KCSE and Senior Secondary assessment frameworks.

Applied Mathematics Junior Secondary → University CBC Aligned KCSE Paper 2
10Sections
35+Worked Examples
3Levels
KNECExam Ready
CBC / KICD Curriculum Context: Statistics and Probability forms a core strand of Applied Mathematics in Kenya's Senior Secondary curriculum (Grades 10–12) under the Competency Based Curriculum (CBC) framework. According to KICD guidelines, students across all three Senior Secondary pathways — STEM, Arts & Sports Science, and Social Sciences — engage with statistical reasoning and probability theory. Mastery of this strand is assessed in KCSE Mathematics Paper 2 (KNEC) and in the Senior Secondary School Assessment (SSSA). The TSC competency framework requires teachers to develop students' ability to collect, organise, analyse, and interpret data as a cross-cutting competency under the BECF (Basic Education Curriculum Framework).
Section 1 · Applied Mathematics

Data Collection & Presentation

Foundational

Statistics begins with data — without carefully collected, organised, and presented data, no analysis is possible. In the CBC curriculum, students first encounter data handling in Grade 7 and progressively deepen their understanding through Senior Secondary. This section covers the essential vocabulary and tools every statistician — and every KCSE student — must master.

Types of Data

All data falls into one of two broad categories depending on its nature. Understanding which type of data you have determines which graphs, calculations, and statistical tests are appropriate.

Qualitative (Categorical) Data
Describes qualities or categories — cannot be measured numerically. Examples: gender, blood type, colour preferences, pass/fail results, county of residence. Further divided into nominal (no order: eye colour) and ordinal (ordered categories: grade A, B, C, D, E).
Quantitative (Numerical) Data
Represents measurable quantities expressed as numbers. Discrete data takes only specific values (countable): number of students, goals scored. Continuous data can take any value in a range (measurable): heights, temperatures, masses, time.
Primary vs Secondary Data
Primary data is collected directly by the researcher through questionnaires, interviews, observation, or experiments. Secondary data comes from existing sources: KNBS reports, government records, published research, internet databases. Primary data is more specific but costly; secondary is cheaper but may be outdated or biased.

Frequency Tables & Grouped Frequency Distributions

A frequency table organises raw data by listing each value (or class interval) and how often it occurs. When data is continuous or has a wide range, we group it into class intervals (also called bins or classes). Each class has a lower class boundary, upper class boundary, class width, and midpoint.

Worked Examples — Frequency Tables
Example 1 Construct a frequency table for: 12, 15, 14, 13, 12, 16, 15, 14, 13, 12, 16, 15, 14, 13, 16
Step 1List distinct values: 12, 13, 14, 15, 16.
Step 2Tally each occurrence: 12 appears 3 times, 13 appears 3 times, 14 appears 3 times, 15 appears 3 times, 16 appears 3 times.
Step 3Record frequencies: Value 12 → f = 3; Value 13 → f = 3; Value 14 → f = 3; Value 15 → f = 3; Value 16 → f = 3. Total Σf = 15.
Σf = 15 ✓ (matches total data points)
Example 2 The masses (kg) of 30 students are recorded. Construct a grouped frequency table with class width 5, starting at 40 kg.
Step 1Choose class intervals: 40–44, 45–49, 50–54, 55–59, 60–64. Width = 5 for each.
Step 2Identify class boundaries (for continuous data): 39.5–44.5, 44.5–49.5, 49.5–54.5, 54.5–59.5, 59.5–64.5.
Step 3Calculate midpoints: 42, 47, 52, 57, 62. These are used when calculating the mean from grouped data: midpoint = (lower boundary + upper boundary) ÷ 2.
Step 4Tally raw data into each interval and record frequencies. Ensure Σf = 30.
Class boundaries and midpoints are critical for grouped data calculations.

Statistical Diagrams: Bar Charts, Histograms, Pie Charts, Stem-and-Leaf

KCSE Key Distinction: A histogram is used for continuous data — bars touch, and area represents frequency (not height, unless class widths are equal). When class widths are unequal, plot frequency density on the y-axis, where Frequency Density = Frequency ÷ Class Width. A bar chart is for discrete or categorical data — bars have gaps and height represents frequency. This distinction is frequently tested in KCSE Paper 2.
Key Formulae — Data Presentation
Frequency Density = Frequency ÷ Class Width
Relative Frequency = f ÷ Σf    (gives proportion, between 0 and 1)
Cumulative Frequency = running total of frequencies
Pie chart angle for class = (f ÷ Σf) × 360°
Worked Example — Histogram with Unequal Class Widths
Example 3 Data: Age 0–9 (f=10, w=10), Age 10–14 (f=15, w=5), Age 15–29 (f=30, w=15), Age 30–49 (f=20, w=20). Calculate frequency densities.
FormulaFrequency Density = Frequency ÷ Class Width
0–9FD = 10 ÷ 10 = 1.0
10–14FD = 15 ÷ 5 = 3.0
15–29FD = 30 ÷ 15 = 2.0
30–49FD = 20 ÷ 20 = 1.0
VerifyCheck: Area = FD × Width. (1.0×10) + (3.0×5) + (2.0×15) + (1.0×20) = 10 + 15 + 30 + 20 = 75 ✔ (equals Σf)
Plot Frequency Density on y-axis; each bar's area equals its class frequency.
Section 2 · Applied Mathematics

Measures of Central Tendency

Foundational / Secondary

A measure of central tendency is a single value that summarises a dataset by describing the centre or most typical value. The three measures — mean, median, and mode — each have distinct advantages and are appropriate in different contexts. Choosing the wrong measure is a common error in KCSE examinations.

Core Formulae — Central Tendency
Mean (ungrouped) x̄ = Σx ÷ n
Weighted / grouped mean x̄ = Σ(f × x) ÷ Σf    (x = midpoint)
Assumed mean method x̄ = A + Σ(f × d) ÷ Σf    where d = x − A
Median (ungrouped, odd n) Middle value at position (n + 1) ÷ 2
Median (ungrouped, even n) Average of values at positions n/2 and (n/2 + 1)
Median (grouped) Md = L + [(n/2 − F) ÷ f] × h
Mode (grouped) Mo = L + [Δ₁ ÷ (Δ₁ + Δ₂)] × h
In the grouped median formula: L = lower class boundary of the median class; n = total frequency; F = cumulative frequency before the median class; f = frequency of the median class; h = class width. In the modal class formula: Δ₁ = difference between modal class frequency and the class before; Δ₂ = difference between modal class frequency and the class after.
Worked Examples — Mean, Median, Mode
Example 1 Find the mean, median, and mode of: 3, 7, 8, 5, 7, 9, 2, 7, 4.
MeanΣx = 3+7+8+5+7+9+2+7+4 = 52. n = 9. x̄ = 52 ÷ 9 ≈ 5.78
MedianSort ascending: 2, 3, 4, 5, 7, 7, 7, 8, 9. Position = (9+1)÷2 = 5th. Median = 7
Mode7 appears 3 times — more than any other value. Mode = 7
Mean ≈ 5.78, Median = 7, Mode = 7
Example 2 Calculate the mean of grouped data: Class 20–29 (f=4), 30–39 (f=9), 40–49 (f=14), 50–59 (f=10), 60–69 (f=3). Total Σf = 40.
Midpointsx: 24.5, 34.5, 44.5, 54.5, 64.5
f × x4×24.5=98, 9×34.5=310.5, 14×44.5=623, 10×54.5=545, 3×64.5=193.5
Σ(fx)98 + 310.5 + 623 + 545 + 193.5 = 1770
Meanx̄ = 1770 ÷ 40 = 44.25
Mean = 44.25
Example 3 — Assumed Mean Method Same data as Example 2. Use A = 44.5 as the assumed mean.
Deviations dd = x − 44.5 → −20, −10, 0, +10, +20
f × d4×(−20)=−80, 9×(−10)=−90, 14×0=0, 10×(+10)=100, 3×(+20)=60
Σ(fd)−80 + (−90) + 0 + 100 + 60 = −10
Meanx̄ = 44.5 + (−10 ÷ 40) = 44.5 − 0.25 = 44.25 ✓
Same answer (44.25) with simpler arithmetic — preferred in KCSE under time pressure.
Example 4 — Grouped Median Using the same distribution (Σf=40), find the median.
Step 1Cumulative frequencies: 4, 13, 27, 37, 40.
Step 2n/2 = 20. The 20th value lies in class 40–49 (cumulative reaches 27 here). Median class = 40–49.
Step 3L = 39.5 (lower boundary), F = 13 (cum. freq. before), f = 14 (median class freq.), h = 10 (width).
ApplyMd = 39.5 + [(20 − 13) ÷ 14] × 10 = 39.5 + (7/14)×10 = 39.5 + 5 = 44.5
Median = 44.5

Which Measure to Use?

Use the Mean when…
Data is roughly symmetrical with no extreme outliers. The mean uses every data value and is ideal for further statistical calculations (variance, z-scores). Best for: test scores, heights, continuous measurements.
Use the Median when…
Data is skewed or contains outliers. The median is resistant to extreme values. Best for: income distributions (where a few millionaires skew the mean), house prices, age distributions in a group with elderly members.
Use the Mode when…
You need the most popular or most frequent category. The only measure usable for nominal (non-ordered categorical) data. Best for: most popular shoe size, most common blood type, most-chosen subject.
Section 3 · Secondary

Measures of Dispersion

Secondary

Two datasets can have the same mean but behave very differently. Measures of dispersion describe how spread out data is around the centre. A small standard deviation means values cluster closely around the mean; a large standard deviation indicates greater spread. In KCSE examinations, standard deviation and variance from grouped data are among the most frequently tested skills.

Dispersion Formulae
Range Range = Maximum value − Minimum value
Interquartile Range IQR = Q₃ − Q₁
Population Variance σ² = Σ(x − x̄)² ÷ n = Σx² ÷ n − x̄²
Sample Variance s² = Σ(x − x̄)² ÷ (n − 1)
Standard Deviation σ = √[Σ(x − x̄)² ÷ n]   or   σ = √[Σ(fx²) ÷ Σf − x̄²]
Grouped SD (simplified) σ = √[Σ(fd²) ÷ Σf − (Σfd ÷ Σf)²]   where d = x − A
Coefficient of Variation CV = (σ ÷ x̄) × 100%
Outlier boundaries Below Q₁ − 1.5×IQR   or   Above Q₃ + 1.5×IQR
Worked Examples — Standard Deviation
Example 1 — Ungrouped Find the population standard deviation of: 4, 7, 13, 2, 1, 9, 8, 6.
Meanx̄ = (4+7+13+2+1+9+8+6) ÷ 8 = 50 ÷ 8 = 6.25
Deviations²(4−6.25)²=5.0625, (7−6.25)²=0.5625, (13−6.25)²=45.5625, (2−6.25)²=18.0625, (1−6.25)²=27.5625, (9−6.25)²=7.5625, (8−6.25)²=3.0625, (6−6.25)²=0.0625
Σ(x−x̄)²5.0625+0.5625+45.5625+18.0625+27.5625+7.5625+3.0625+0.0625 = 107.5
Varianceσ² = 107.5 ÷ 8 = 13.4375
SDσ = √13.4375 ≈ 3.67
Standard Deviation σ ≈ 3.67
Example 2 — Shortcut Formula Same data: 4, 7, 13, 2, 1, 9, 8, 6. Use σ² = Σx²/n − x̄².
Σx²4²+7²+13²+2²+1²+9²+8²+6² = 16+49+169+4+1+81+64+36 = 420
σ²420 ÷ 8 − (6.25)² = 52.5 − 39.0625 = 13.4375 ✓
SDσ = √13.4375 ≈ 3.67 ✓ — same answer, much faster
The shortcut formula σ² = Σx²/n − x̄² is strongly preferred in KCSE under time pressure.
Example 3 — Grouped Standard Deviation Class: 10–19 (f=5, x=14.5), 20–29 (f=12, x=24.5), 30–39 (f=18, x=34.5), 40–49 (f=10, x=44.5), 50–59 (f=5, x=54.5). Use A = 34.5.
d = x − A−20, −10, 0, +10, +20
fd5×(−20)=−100, 12×(−10)=−120, 18×0=0, 10×10=100, 5×20=100. Σfd = −20
fd²5×400=2000, 12×100=1200, 18×0=0, 10×100=1000, 5×400=2000. Σfd² = 6200
Σf5+12+18+10+5 = 50
σσ = √[6200÷50 − (−20÷50)²] = √[124 − 0.16] = √123.84 ≈ 11.13
σ ≈ 11.13

Box Plots and the Five-Number Summary

A box plot (box-and-whisker diagram) provides a visual summary of distribution using five values. It clearly shows skewness and highlights outliers — skills increasingly assessed in CBC Senior Secondary assessments.

Five-Number Summary
Min, Q₁ (lower quartile), Q₂ (median), Q₃ (upper quartile), Max
Q₁ = value at position (n+1)/4    Q₃ = value at position 3(n+1)/4
For grouped data, use the interpolation formula (same as median formula above).
Outlier fences: Lower = Q₁ − 1.5×IQR  ·  Upper = Q₃ + 1.5×IQR
Section 4 · Secondary

Probability

Secondary

Probability is the mathematical measure of how likely an event is to occur. It was formalised by mathematicians including Kolmogorov in the 20th century, though its roots trace back to 17th-century gambling problems solved by Pascal and Fermat. In the CBC curriculum, probability builds from simple equally-likely outcomes in Grade 8 to conditional probability and Bayes' theorem at university level.

Probability Formulae
Classical probability P(A) = n(A) ÷ n(S)   (0 ≤ P(A) ≤ 1)
Complement rule P(A') = 1 − P(A)
Addition law (general) P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Mutually exclusive events P(A ∪ B) = P(A) + P(B)   [since P(A ∩ B) = 0]
Multiplication law (dependent) P(A ∩ B) = P(A) × P(B|A)
Independent events P(A ∩ B) = P(A) × P(B)
Conditional probability P(A|B) = P(A ∩ B) ÷ P(B)
Bayes' Theorem P(A|B) = [P(B|A) × P(A)] ÷ P(B)
Worked Examples — Probability
Example 1 A bag contains 5 red, 4 blue, and 3 green marbles. One marble is drawn at random. Find: (a) P(red), (b) P(not blue), (c) P(red or green).
n(S)Total = 5 + 4 + 3 = 12
(a)P(red) = 5/12
(b)P(not blue) = 1 − P(blue) = 1 − 4/12 = 8/12 = 2/3
(c)Red and green are mutually exclusive (a marble can't be both): P(red or green) = 5/12 + 3/12 = 8/12 = 2/3
(a) 5/12   (b) 2/3   (c) 2/3
Example 2 — Tree Diagram / Without Replacement A bag has 6 white and 4 black balls. Two balls are drawn without replacement. Find P(both black) and P(one of each colour).
P(1st black)4/10 = 2/5
P(both black)P(B₁ ∩ B₂) = 4/10 × 3/9 = 12/90 = 2/15
One of eachP(WB) + P(BW) = (6/10 × 4/9) + (4/10 × 6/9) = 24/90 + 24/90 = 48/90 = 8/15
VerifyP(WW) + P(WB or BW) + P(BB) = 6/10×5/9 + 8/15 + 2/15 = 30/90 + 48/90 + 12/90 = 90/90 = 1 ✓
P(both black) = 2/15  ·  P(one of each) = 8/15
Example 3 — Conditional Probability In a class, P(passes Maths) = 0.70, P(passes English) = 0.60, P(passes both) = 0.45. Given a student passes Maths, what is P(also passes English)?
FormulaP(English | Maths) = P(Maths ∩ English) ÷ P(Maths)
Substitute= 0.45 ÷ 0.70 = 0.643 (3 s.f.)
P(English | Maths) ≈ 0.643 ≈ 64.3%
Example 4 — Bayes' Theorem (A-Level / University) A disease affects 1% of the population. A test is 95% accurate (95% sensitivity: P(+|disease) = 0.95) and has a 5% false positive rate (P(+|no disease) = 0.05). A person tests positive. Find P(disease | positive test).
GivenP(D) = 0.01, P(D') = 0.99, P(+|D) = 0.95, P(+|D') = 0.05
P(+)Total probability law: P(+) = P(+|D)×P(D) + P(+|D')×P(D') = 0.95×0.01 + 0.05×0.99 = 0.0095 + 0.0495 = 0.059
BayesP(D|+) = [P(+|D) × P(D)] ÷ P(+) = 0.0095 ÷ 0.059 ≈ 0.161
P(disease | positive test) ≈ 16.1% — Despite 95% accuracy, a positive result only means ~16% chance of disease when prevalence is low. This is why screening programmes must be carefully designed.
Section 5 · Secondary / A-Level

Permutations & Combinations

Secondary

Counting methods — permutations and combinations — form the foundation of probability calculations for complex events. They answer questions like "how many ways can a committee be chosen?" or "how many different arrangements of letters are possible?" The distinction between when order matters (permutations) and when it does not (combinations) is fundamental and frequently tested in KCSE.

Counting Principles & Formulae
Multiplication principle If event A can occur in m ways and B in n ways, both together: m × n ways
Factorial n! = n × (n−1) × (n−2) × … × 2 × 1    (0! = 1)
Permutations (ordered) P(n,r) = ⁿPᵣ = n! ÷ (n−r)!
Permutations with repeats n! ÷ (a! × b! × c! × …)   (if a items identical, b identical, etc.)
Combinations (unordered) C(n,r) = ⁿCᵣ = n! ÷ [r! × (n−r)!]
Symmetry of combinations C(n,r) = C(n, n−r)
Worked Examples — Counting
Example 1 How many 3-letter arrangements can be made from the letters of the word MATHS (no repetition)?
NoteOrder matters (arrangements/permutations). n = 5, r = 3.
P(5,3)= 5! ÷ (5−3)! = 5! ÷ 2! = 120 ÷ 2 = 60 arrangements
60 different 3-letter arrangements
Example 2 A committee of 4 is to be chosen from 9 candidates. How many different committees are possible? How many if the committee must include a specific person, Atieno?
NoteOrder does not matter (committees), so use combinations.
TotalC(9,4) = 9! ÷ (4! × 5!) = (9×8×7×6) ÷ (4×3×2×1) = 3024 ÷ 24 = 126 committees
Include AtienoAtieno's seat is fixed; choose remaining 3 from 8: C(8,3) = 8!÷(3!×5!) = 336÷6 = 56 committees
126 total committees; 56 committees including Atieno
Example 3 How many arrangements of the letters in KENYA are there? How many start with K?
All 5 letters distinctTotal arrangements = 5! = 120
Starting with KFix K in position 1; arrange remaining 4 letters: 4! = 24
120 total; 24 starting with K
Section 6 · Secondary / A-Level

Discrete Probability Distributions

Secondary / A-Level

A probability distribution describes all possible outcomes of a random variable and their corresponding probabilities. For a discrete random variable X, the probability function assigns a probability P(X = x) to each possible value x. This section covers the binomial distribution — the most important discrete distribution in the CBC Senior Secondary syllabus — and introduces the Poisson distribution for events occurring over time or space.

Discrete Probability Distribution — Basics

Properties of a Discrete Probability Distribution
0 ≤ P(X = x) ≤ 1 for all x
Σ P(X = x) = 1   (probabilities sum to 1)
Expected value: E(X) = μ = Σ[x × P(X = x)]
Variance: Var(X) = E(X²) − [E(X)]² = Σ[x² × P(X=x)] − μ²
Worked Example — Probability Distribution Table
Example 1 X takes values 1, 2, 3, 4 with probabilities 0.1, k, 0.3, 0.2 respectively. Find k, E(X), and Var(X).
Find kΣ P = 1: 0.1 + k + 0.3 + 0.2 = 1 → k = 1 − 0.6 = 0.4
E(X)= 1(0.1) + 2(0.4) + 3(0.3) + 4(0.2) = 0.1 + 0.8 + 0.9 + 0.8 = 2.6
E(X²)= 1²(0.1) + 2²(0.4) + 3²(0.3) + 4²(0.2) = 0.1 + 1.6 + 2.7 + 3.2 = 7.6
Var(X)= E(X²) − [E(X)]² = 7.6 − (2.6)² = 7.6 − 6.76 = 0.84
k = 0.4; E(X) = 2.6; Var(X) = 0.84

Binomial Distribution B(n, p)

The binomial distribution models the number of successes X in n independent trials, each with a constant probability of success p. Conditions required: (1) fixed number of trials n, (2) only two outcomes per trial (success/failure), (3) trials are independent, (4) constant probability p.

Binomial Distribution
Probability function P(X = r) = C(n,r) × pʳ × (1−p)ⁿ⁻ʳ
Mean μ = E(X) = np
Variance σ² = Var(X) = np(1−p) = npq   where q = 1−p
Standard Deviation σ = √(npq)
Worked Examples — Binomial Distribution
Example 2 A fair coin is tossed 8 times. Find: (a) P(exactly 5 heads), (b) P(at most 2 heads), (c) Mean and SD of heads.
SetupX ~ B(8, 0.5). n = 8, p = 0.5, q = 0.5.
(a)P(X=5) = C(8,5) × (0.5)⁵ × (0.5)³ = 56 × (0.5)⁸ = 56 ÷ 256 = 0.2188
(b)P(X ≤ 2) = P(0) + P(1) + P(2).
P(0) = (0.5)⁸ = 1/256 ≈ 0.0039
P(1) = C(8,1)(0.5)⁸ = 8/256 ≈ 0.0313
P(2) = C(8,2)(0.5)⁸ = 28/256 ≈ 0.1094
P(X≤2) ≈ 0.0039+0.0313+0.1094 = 0.1445
(c)Mean = np = 8×0.5 = 4. SD = √(npq) = √(8×0.5×0.5) = √2 ≈ 1.414
(a) 0.2188   (b) 0.1445   (c) Mean = 4, SD ≈ 1.414
Example 3 In a multiple-choice test with 10 questions, each with 4 options, a student guesses all answers randomly. Find P(at least 3 correct).
SetupX ~ B(10, 0.25). P(correct by guessing) = 1/4.
ComplementP(X ≥ 3) = 1 − P(X ≤ 2) = 1 − [P(0) + P(1) + P(2)]
P(0)= (0.75)¹⁰ ≈ 0.0563
P(1)= C(10,1)(0.25)¹(0.75)⁹ = 10×0.25×0.0751 ≈ 0.1877
P(2)= C(10,2)(0.25)²(0.75)⁸ = 45×0.0625×0.1001 ≈ 0.2816
AnswerP(X≥3) = 1 − (0.0563+0.1877+0.2816) = 1 − 0.5256 = 0.4744
P(at least 3 correct) ≈ 0.474 ≈ 47.4%

Poisson Distribution Po(λ)

Poisson Distribution — for rare events over a fixed time/space
Probability function P(X = k) = e⁻λ × λᵏ ÷ k!
Mean = Variance E(X) = Var(X) = λ
Conditions Events occur independently, at a constant average rate λ, singly (not simultaneously)
Worked Example — Poisson Distribution
Example 4 Customers arrive at a bank at an average rate of 3 per minute. Find P(exactly 5 arrive in a given minute) and P(fewer than 2 arrive).
SetupX ~ Po(3). λ = 3, e⁻³ ≈ 0.0498.
P(X=5)= e⁻³ × 3⁵ ÷ 5! = 0.0498 × 243 ÷ 120 = 0.0498 × 2.025 ≈ 0.1008
P(X<2)= P(0) + P(1) = e⁻³×3⁰/0! + e⁻³×3¹/1! = 0.0498 + 0.1494 = 0.1992
P(X=5) ≈ 0.101; P(X<2) ≈ 0.199
Section 7 · A-Level / University

The Normal Distribution N(μ, σ²)

Advanced

The normal distribution is the most important probability distribution in statistics. Its characteristic bell-shaped curve arises naturally in many biological, physical, and social measurements — heights of people, errors in measurements, exam marks. The Central Limit Theorem (discussed in Section 8) guarantees that sample means follow a normal distribution, making it fundamental to statistical inference.

Normal Distribution Formulae
Probability density function f(x) = (1/σ√(2π)) × e^[−(x−μ)²/(2σ²)]
Standardisation (Z-score) Z = (X − μ) ÷ σ    (Z ~ N(0,1))
68-95-99.7 Empirical Rule P(μ−σ < X < μ+σ) = 68.27%, P(μ−2σ < X < μ+2σ) = 95.45%, P(μ−3σ < X < μ+3σ) = 99.73%
Standard normal tables Φ(z) = P(Z ≤ z)   (read from tables or use symmetry)
Symmetry P(Z > z) = 1 − Φ(z)  ·  P(Z < −z) = 1 − Φ(z)
Worked Examples — Normal Distribution
Example 1 Exam scores follow N(65, 10²). Find: (a) P(score > 80), (b) P(50 < score < 75), (c) The score at the 90th percentile.
Setupμ = 65, σ = 10
(a)Z = (80−65)/10 = 1.5. P(X>80) = P(Z>1.5) = 1 − Φ(1.5) = 1 − 0.9332 = 0.0668 (6.68%)
(b)Z₁ = (50−65)/10 = −1.5; Z₂ = (75−65)/10 = 1.0.
P(−1.5 < Z < 1.0) = Φ(1.0) − Φ(−1.5) = 0.8413 − (1−0.9332) = 0.8413 − 0.0668 = 0.7745
(c)From tables, Φ(1.2816) = 0.90. So Z = 1.2816.
X = μ + Zσ = 65 + 1.2816×10 = 65 + 12.816 ≈ 77.8
(a) 6.68%   (b) 77.45%   (c) ≈ 77.8 marks
Example 2 — Normal Approximation to Binomial X ~ B(100, 0.4). Use normal approximation to find P(X > 45).
Conditionsnp = 40 > 5 and nq = 60 > 5 ✓ — normal approximation is valid.
Parametersμ = np = 40. σ = √(npq) = √(100×0.4×0.6) = √24 ≈ 4.899
Continuity correctionP(X > 45) ≈ P(X > 45.5) with continuity correction.
Z-scoreZ = (45.5 − 40) ÷ 4.899 ≈ 1.122. P(Z>1.122) = 1 − Φ(1.12) ≈ 1 − 0.8686 = 0.1314
P(X > 45) ≈ 0.131 ≈ 13.1%
Section 8 · University

Sampling, Estimation & the Central Limit Theorem

University

We rarely have access to an entire population. Instead, we use a sample to make inferences about population parameters. The Central Limit Theorem is the theoretical backbone that makes this possible — it guarantees that sample means will be approximately normally distributed for sufficiently large samples, regardless of the population's shape.

Sampling Methods

Simple Random Sampling
Every member of the population has an equal probability of being selected. Can be achieved using a random number table or computer-generated random numbers. Eliminates selection bias but requires a complete sampling frame.
Stratified Sampling
Population divided into non-overlapping strata (e.g. by gender, county, school year). Sample taken proportionally from each stratum. Ensures representation of subgroups — particularly important in educational surveys.
Systematic Sampling
Select every kth member after a random start. If population size N and sample size n, then k = N/n. Simple to implement but can introduce bias if there is periodicity in the population.
Cluster Sampling
Population divided into natural clusters (e.g. schools, villages). A random selection of whole clusters is surveyed. Cost-effective for geographically dispersed populations but less precise than stratified sampling.

The Central Limit Theorem (CLT)

Central Limit Theorem & Sampling Distribution of the Mean
CLT Statement For large n (≥30), X̄ ~ N(μ, σ²/n) regardless of the population distribution.
Standard Error of the Mean SE = σ ÷ √n    (or s ÷ √n for sample SD)
Z-score for sample mean Z = (X̄ − μ) ÷ (σ/√n)
95% Confidence Interval for μ CI = X̄ ± 1.96 × (σ/√n)
99% Confidence Interval for μ CI = X̄ ± 2.576 × (σ/√n)
Worked Example — Confidence Interval
Example 1 A sample of 50 students has a mean score of 72 marks with SD = 8. Construct a 95% confidence interval for the population mean.
Givenn = 50, X̄ = 72, s = 8. Use Z = 1.96 for 95% CI.
SE= 8 ÷ √50 = 8 ÷ 7.071 ≈ 1.131
Margin of Error= 1.96 × 1.131 ≈ 2.217
CI72 ± 2.22 → (69.78, 74.22)
We are 95% confident the population mean lies between 69.78 and 74.22 marks.
Section 9 · University

Hypothesis Testing

University

Hypothesis testing provides a systematic, objective framework for using sample data to make decisions about population parameters. It is the engine of scientific research, medical trials, quality control, and policy evaluation. A hypothesis test never proves a hypothesis — it either provides evidence against the null hypothesis or fails to do so.

The Five-Step Procedure for Hypothesis Testing

  1. State the hypotheses: The null hypothesis H₀ is the default assumption (no effect, no difference, equals the claimed value). The alternative hypothesis H₁ (or Hₐ) is what you are trying to find evidence for. H₁ determines whether the test is one-tailed or two-tailed.
  2. Choose the significance level α: Typically α = 0.05 (5%) or α = 0.01 (1%). This is the probability of rejecting H₀ when it is actually true (Type I error rate).
  3. Select the test and calculate the test statistic: For means with known σ: Z-test. For means with unknown σ and small n: t-test. For proportions: Z-test. For goodness of fit / independence: Chi-square test.
  4. Find the p-value or critical value: The p-value is the probability of observing a test statistic at least as extreme as the one computed, assuming H₀ is true.
  5. Make a decision: If p-value < α (or test statistic falls in rejection region): Reject H₀. If p-value ≥ α: Fail to reject H₀. Never say "accept H₁" — the test cannot prove H₁, only provide evidence for it.
Test Statistics
One-sample Z-test (σ known) Z = (X̄ − μ₀) ÷ (σ ÷ √n)
One-sample t-test (σ unknown) T = (X̄ − μ₀) ÷ (s ÷ √n) ~ t(n−1)
Z-test for proportion Z = (p̂ − p₀) ÷ √[p₀(1−p₀)/n]
Chi-square goodness of fit χ² = Σ [(O − E)² ÷ E] ~ χ²(df)
Chi-square independence df = (rows − 1) × (cols − 1)    Expected: E = (row total × col total) ÷ grand total
Worked Examples — Hypothesis Testing
Example 1 — Z-test A teacher claims the average score for a large standardised test is 70. A sample of 36 students has x̄ = 73.2, σ = 9.6. Test at α = 0.05 (two-tailed).
H₀ and H₁H₀: μ = 70  ·  H₁: μ ≠ 70 (two-tailed)
Critical valuesAt α = 0.05 two-tailed: Z_crit = ±1.96
Test statisticZ = (73.2 − 70) ÷ (9.6 ÷ √36) = 3.2 ÷ (9.6 ÷ 6) = 3.2 ÷ 1.6 = 2.00
Decision|Z| = 2.00 > 1.96. Reject H₀. There is sufficient evidence at 5% significance that the mean score differs from 70.
Reject H₀. The sample mean of 73.2 is significantly different from 70.
Example 2 — Chi-Square Test of Independence A survey of 200 students asks whether they prefer Science or Arts, and records their gender. Observed: (Male: Science=60, Arts=40), (Female: Science=50, Arts=50). Test at α = 0.05 whether subject preference is independent of gender.
H₀ and H₁H₀: Subject preference is independent of gender. H₁: They are associated.
Row/Col totalsMale=100, Female=100. Science=110, Arts=90. Grand total=200.
Expected valuesE(M,Sci) = 100×110/200 = 55; E(M,Arts) = 100×90/200 = 45; E(F,Sci) = 55; E(F,Arts) = 45.
χ²= (60−55)²/55 + (40−45)²/45 + (50−55)²/55 + (50−45)²/45
= 25/55 + 25/45 + 25/55 + 25/45
= 0.455 + 0.556 + 0.455 + 0.556 = 2.02
Decisiondf = (2−1)(2−1) = 1. Critical value χ²(1, 0.05) = 3.841. Since 2.02 < 3.841, Fail to reject H₀. No significant association at 5% level.
Fail to reject H₀. Insufficient evidence of association between gender and subject preference.
Type I and Type II Errors:
Type I Error (False Positive): Rejecting H₀ when it is actually true. Probability = α (the significance level you set).
Type II Error (False Negative): Failing to reject H₀ when H₁ is actually true. Probability = β.
Power of the test = 1 − β = probability of correctly rejecting a false H₀.
Reducing α decreases Type I error but increases Type II error — there is always a trade-off.
Section 10 · University

Regression & Correlation

University

Correlation measures the strength and direction of a linear relationship between two variables. Regression goes further — it models the relationship as an equation that allows prediction. Both tools are essential in economics, medicine, education research, and engineering. A critical principle: correlation does not imply causation. Two variables may be correlated because they share a common cause, not because one causes the other.

Correlation & Regression Formulae
Pearson's r (correlation coefficient) r = Σ(xᵢ−x̄)(yᵢ−ȳ) ÷ [√Σ(xᵢ−x̄)² × √Σ(yᵢ−ȳ)²]
Computational form of r r = [nΣxy − ΣxΣy] ÷ √{[nΣx² − (Σx)²][nΣy² − (Σy)²]}
Regression line y on x ŷ = a + bx   (use for predicting y from x)
Gradient b b = [nΣxy − ΣxΣy] ÷ [nΣx² − (Σx)²]
Intercept a a = ȳ − b×x̄    (line always passes through (x̄, ȳ))
Coefficient of determination r² = proportion of variance in y explained by x
Spearman's rank correlation rₛ = 1 − [6Σd²] ÷ [n(n²−1)]    (d = difference in ranks)
Interpreting r: r = +1 (perfect positive linear correlation) · r = −1 (perfect negative linear correlation) · r = 0 (no linear correlation) · |r| > 0.8 (strong) · 0.5 < |r| < 0.8 (moderate) · |r| < 0.5 (weak). Note: r measures only linear relationships; two variables can have r = 0 but still be strongly related in a non-linear way.
Worked Example — Correlation and Regression
Example 1 Five students: x = hours studied (2, 4, 6, 8, 10) and y = test score (45, 55, 65, 72, 80). Find r and the regression equation.
Sumsn=5, Σx=30, Σy=317, Σx²=220, Σy²=20383, Σxy=2070
Meansx̄ = 30/5 = 6; ȳ = 317/5 = 63.4
r numeratornΣxy − ΣxΣy = 5×2070 − 30×317 = 10350 − 9510 = 840
r denominator√{[5×220−900][5×20383−100489]} = √{[1100−900][101915−100489]} = √{200 × 1426} = √285200 ≈ 534.04
r= 840 ÷ 534.04 ≈ 0.972 — strong positive correlation ✓
b= 840 ÷ 200 = 4.2
a & equationa = 63.4 − 4.2×6 = 63.4 − 25.2 = 38.2. Regression line: ŷ = 38.2 + 4.2x
r ≈ 0.972 (very strong positive). Predicted score for 7 hours: ŷ = 38.2 + 4.2×7 = 67.6
Example 2 — Spearman's Rank Correlation Five judges rank five contestants A–E in two categories. Ranks: Category 1: A=1, B=3, C=2, D=5, E=4. Category 2: A=2, B=1, C=4, D=3, E=5. Find rₛ.
Differences dA: 1−2=−1; B: 3−1=2; C: 2−4=−2; D: 5−3=2; E: 4−5=−1
d²1, 4, 4, 4, 1 → Σd² = 14
rₛ= 1 − 6×14 ÷ [5×(25−1)] = 1 − 84 ÷ 120 = 1 − 0.7 = 0.3
rₛ = 0.3 — weak positive rank correlation between the two judges.

Common Errors in Statistics

✗ WrongUsing mean when data has extreme outliers
✓ RightUse median for skewed distributions or data with outliers
Example: One student scores 2, all others score 80–90. Mean is misleading; median better represents the group.
✗ WrongIn histogram, plotting frequency on y-axis when class widths are unequal
✓ RightPlot frequency density = frequency ÷ class width
The area of each bar, not the height, must equal the frequency.
✗ WrongP(A and B) = P(A) + P(B) for any two events
✓ RightP(A ∩ B) = P(A) × P(B) only when A and B are independent
Always check independence or use P(A ∩ B) = P(A) × P(B|A).
✗ WrongDividing by n when calculating sample variance/SD
✓ RightFor a sample: divide by (n−1) to get unbiased estimate
σ² = Σ(x−x̄)²/n (population); s² = Σ(x−x̄)²/(n−1) (sample estimate).
✗ Wrong"Correlation proves causation" — claiming X causes Y because r is high
✓ RightCorrelation only indicates that X and Y move together linearly
Ice cream sales and drowning rates are positively correlated — both driven by hot weather, not by each other.
✗ WrongAfter failing to reject H₀, concluding "H₀ is true" or "H₀ is accepted"
✓ Right"There is insufficient evidence to reject H₀ at the α level"
Hypothesis tests cannot prove a null hypothesis; they can only fail to disprove it given the sample size.

Quick Reference Formula Sheet

Central Tendency
x̄ = Σx/n
x̄ grouped = Σ(fx)/Σf
Md = L + [(n/2−F)/f]×h
Mo = L + [Δ₁/(Δ₁+Δ₂)]×h
Dispersion
σ² = Σx²/n − x̄²
σ grouped = √[Σfd²/Σf − (Σfd/Σf)²]
IQR = Q₃ − Q₁
CV = (σ/x̄) × 100%
Probability
P(A∪B) = P(A)+P(B)−P(A∩B)
P(A|B) = P(A∩B)/P(B)
P(A∩B) = P(A)×P(B|A)
P(A') = 1 − P(A)
Distributions
Binomial: P(r) = C(n,r)pʳqⁿ⁻ʳ
B(n,p): μ=np, σ²=npq
Poisson: P(k) = e⁻λλᵏ/k!
Po(λ): μ = Var = λ
Normal Distribution
Z = (X − μ)/σ
68-95-99.7% rule
P(Z>z) = 1 − Φ(z)
CI: X̄ ± z × σ/√n
Regression
b = [nΣxy−ΣxΣy]/[nΣx²−(Σx)²]
a = ȳ − bx̄
r = Sxy/√(Sxx×Syy)
rₛ = 1 − 6Σd²/[n(n²−1)]
Hypothesis Tests
Z = (X̄−μ₀)/(σ/√n)
T = (X̄−μ₀)/(s/√n)
χ² = Σ(O−E)²/E
SE = σ/√n
Counting
ⁿPᵣ = n!/(n−r)!
ⁿCᵣ = n!/[r!(n−r)!]
Repeats: n!/(a!b!c!…)
C(n,r) = C(n,n−r)