Lesson 8 of 12
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1. Representation of Data S1
Types of Data: Qualitative (categorical — names, colours, labels) and Quantitative (numerical). Quantitative data is either discrete (countable, e.g. number of students) or continuous (measurable, any value in a range, e.g. height, time).
Stem-and-Leaf Diagrams
A stem-and-leaf diagram preserves the original data values while showing the distribution. A back-to-back stem-and-leaf compares two datasets.
Box-and-Whisker Plots
Box plot uses five-number summary:
Minimum value | Lower Quartile (Q₁) | Median (Q₂) | Upper Quartile (Q₃) | Maximum value
Interquartile Range (IQR) = Q₃ − Q₁ — measures spread of middle 50%.
Outliers: Values below Q₁ − 1.5×IQR or above Q₃ + 1.5×IQR.
Minimum value | Lower Quartile (Q₁) | Median (Q₂) | Upper Quartile (Q₃) | Maximum value
Interquartile Range (IQR) = Q₃ − Q₁ — measures spread of middle 50%.
Outliers: Values below Q₁ − 1.5×IQR or above Q₃ + 1.5×IQR.
Histograms for Grouped Data
Histogram: For continuous grouped data, the y-axis is frequency density, NOT frequency. This ensures the area of each bar equals the frequency.
Frequency density = Frequency ÷ Class width
Frequency = Frequency density × Class width
Frequency density = Frequency ÷ Class width
Frequency = Frequency density × Class width
📐 Worked Example 1 — Histogram and Frequency Density
Data on journey times (minutes): 0–10: 12, 10–20: 18, 20–40: 22, 40–70: 15. Draw the histogram frequency densities and estimate the number of journeys under 30 minutes.
1
Calculate frequency densities (FD = freq ÷ class width):
0–10: 12/10=1.2 10–20: 18/10=1.8 20–40: 22/20=1.1 40–70: 15/30=0.5
0–10: 12/10=1.2 10–20: 18/10=1.8 20–40: 22/20=1.1 40–70: 15/30=0.5
2
Journeys under 30 min = all of 0–10, all of 10–20, and half of 20–40:
= 12 + 18 + 22/2 = 12 + 18 + 11 = 41 journeys
= 12 + 18 + 22/2 = 12 + 18 + 11 = 41 journeys
Cumulative Frequency Curves
Plot cumulative frequency against the upper class boundary. The curve (ogive) allows estimation of the median (at n/2), quartiles (at n/4 and 3n/4), and percentiles.
2. Measures of Location and Spread S1
Measures of Location
Mean: x̄ = Σx/n or x̄ = Σfx/Σf (frequency distribution)
Median: middle value when data is sorted; average of two middle values if n even
Mode: most frequent value(s)
Measures of Spread — Variance and Standard Deviation
Variance: σ² = Σ(x−x̄)²/n = Σx²/n − x̄² (population)
Variance: s² = Σ(x−x̄)²/(n−1) = [Σx²−(Σx)²/n]/(n−1) (sample)
Standard deviation: σ = √(variance)
Coding: if y=(x−a)/b then ȳ=(x̄−a)/b, σ_y = σ_x/|b|
📐 Worked Example 2 — Mean and Variance
For the data: 3, 7, 7, 8, 10, 12, 14. Find the mean, variance, and standard deviation.
1
n=7, Σx=3+7+7+8+10+12+14=61
x̄ = 61/7 ≈ 8.714
x̄ = 61/7 ≈ 8.714
2
Σx²=9+49+49+64+100+144+196=611
σ² = 611/7 − (61/7)² = 87.286 − 75.918 = 11.368
σ² = 611/7 − (61/7)² = 87.286 − 75.918 = 11.368
3
σ = √11.368 = 3.37 (3 s.f.)
📐 Worked Example 3 — Coding
Data has Σx=360, Σx²=15200, n=20. Using y=x−20, find the mean and standard deviation of x.
1
Σy=Σ(x−20)=360−400=−40. ȳ=−40/20=−2
x̄=ȳ+20=18
x̄=ȳ+20=18
2
Σy²=Σ(x−20)²=Σx²−40Σx+400n=15200−14400+8000=8800
σ²_y=Σy²/n−ȳ²=8800/20−4=440−4=436
Since y=x−20, σ_x=σ_y=√436≈20.9
σ²_y=Σy²/n−ȳ²=8800/20−4=440−4=436
Since y=x−20, σ_x=σ_y=√436≈20.9
Choosing the Right Measure
| Situation | Best Measure of Location | Best Measure of Spread |
|---|---|---|
| Symmetric distribution | Mean | Standard deviation |
| Skewed distribution | Median | IQR |
| Outliers present | Median | IQR |
| Categorical data | Mode | — |
| Further calculations needed | Mean | Standard deviation |
3. Probability S1
Fundamental Probability Rules
P(A) = (number of favourable outcomes)/(total outcomes) for equally likely outcomes
0 ≤ P(A) ≤ 1 P(A) + P(A') = 1 P(∅) = 0 P(S) = 1
Addition rule: P(A∪B) = P(A) + P(B) − P(A∩B)
Mutually exclusive: P(A∩B)=0 → P(A∪B)=P(A)+P(B)
Conditional probability: P(A|B) = P(A∩B)/P(B)
Independent events: P(A∩B) = P(A)×P(B) i.e. P(A|B)=P(A)
Multiplication rule: P(A∩B) = P(A|B)×P(B) = P(B|A)×P(A)
📐 Worked Example 4 — Conditional Probability
In a class: P(pass Maths) = 0.7, P(pass Physics) = 0.6, P(pass both) = 0.45. Find: (a) P(pass at least one) (b) P(pass Maths | pass Physics) (c) Are the events independent?
1
(a) P(M∪P) = 0.7+0.6−0.45 = 0.85
2
(b) P(M|P) = P(M∩P)/P(P) = 0.45/0.6 = 0.75
3
(c) If independent: P(M∩P)=P(M)×P(P)=0.7×0.6=0.42≠0.45.
P(M|P)=0.75≠0.7=P(M). Therefore not independent.
P(M|P)=0.75≠0.7=P(M). Therefore not independent.
Tree Diagrams and Venn Diagrams
📐 Worked Example 5 — Tree Diagram (Two-Stage)
A bag has 4 red and 6 blue balls. Two balls are drawn without replacement. Find the probability that both are the same colour.
1
P(RR) = (4/10)×(3/9) = 12/90 = 2/15
P(BB) = (6/10)×(5/9) = 30/90 = 1/3
P(BB) = (6/10)×(5/9) = 30/90 = 1/3
2
P(same colour) = 2/15 + 1/3 = 2/15 + 5/15 = 7/15
Permutations and Combinations
Counting Methods
Permutations (order matters): ⁿPᵣ = n!/(n−r)!
Combinations (order doesn't matter): ⁿCᵣ = n!/[r!(n−r)!]
Arrangements with repeats: n!/p!q!r!... (where p,q,r... are frequencies of repeated items)
Circular arrangements: (n−1)!
📐 Worked Example 6 — Combinations in Probability
A committee of 4 is chosen from 6 men and 5 women. Find the probability that the committee has at least 2 women.
1
Total ways = ¹¹C₄ = 330
2
P(at least 2 women) = P(2W+2M) + P(3W+1M) + P(4W+0M)
= (⁵C₂×⁶C₂ + ⁵C₃×⁶C₁ + ⁵C₄×⁶C₀)/330
= (10×15 + 10×6 + 5×1)/330
= (150+60+5)/330 = 215/330 = 43/66
= (⁵C₂×⁶C₂ + ⁵C₃×⁶C₁ + ⁵C₄×⁶C₀)/330
= (10×15 + 10×6 + 5×1)/330
= (150+60+5)/330 = 215/330 = 43/66
4. Discrete Random Variables S1
Discrete Random Variable (DRV): A variable X that takes specific discrete values with defined probabilities. The probability distribution lists all possible values and their probabilities. Key requirement: ΣP(X=x) = 1.
Expectation and Variance of a DRV
E(X) = Σ x P(X=x) (expected value / mean)
E(X²) = Σ x² P(X=x)
Var(X) = E(X²) − [E(X)]²
E(aX+b) = aE(X) + b Var(aX+b) = a²Var(X)
E(X+Y) = E(X) + E(Y) (always, even if not independent)
Var(X+Y) = Var(X) + Var(Y) (only when X and Y are independent)
📐 Worked Example 7 — Probability Distribution Table
A DRV X has P(X=x) = k(x+1) for x = 0, 1, 2, 3. Find k, E(X), and Var(X).
1
ΣP=1: k(1)+k(2)+k(3)+k(4)=10k=1 → k=1/10
2
E(X)=0×(1/10)+1×(2/10)+2×(3/10)+3×(4/10)=0+2/10+6/10+12/10=20/10=2
3
E(X²)=0×(1/10)+1×(2/10)+4×(3/10)+9×(4/10)=0+2/10+12/10+36/10=50/10=5
Var(X)=E(X²)−[E(X)]²=5−4=1
Var(X)=E(X²)−[E(X)]²=5−4=1
Geometric Distribution
Geometric Distribution: Models the number of trials until the first success, where each trial has probability p of success (independent trials).
X ~ Geo(p): P(X=r) = (1−p)^(r−1) p for r = 1, 2, 3, ...
E(X) = 1/p Var(X) = (1−p)/p²
X ~ Geo(p): P(X=r) = (1−p)^(r−1) p for r = 1, 2, 3, ...
E(X) = 1/p Var(X) = (1−p)/p²
📐 Worked Example 8 — Geometric Distribution
A biased coin has P(head) = 0.3. The coin is tossed until a head appears. Find: (a) P(X=4) (b) P(X≤3) (c) E(X) and Var(X)
1
(a) P(X=4) = (0.7)³×(0.3) = 0.343×0.3 = 0.1029
2
(b) P(X≤3) = P(X=1)+P(X=2)+P(X=3)
= 0.3 + 0.7×0.3 + 0.7²×0.3 = 0.3+0.21+0.147 = 0.657
Alternative: P(X≤3) = 1−P(X>3) = 1−(0.7)³ = 1−0.343 = 0.657 ✓
= 0.3 + 0.7×0.3 + 0.7²×0.3 = 0.3+0.21+0.147 = 0.657
Alternative: P(X≤3) = 1−P(X>3) = 1−(0.7)³ = 1−0.343 = 0.657 ✓
3
(c) E(X) = 1/0.3 = 10/3 ≈ 3.33
Var(X) = 0.7/0.09 = 70/9 ≈ 7.78
Var(X) = 0.7/0.09 = 70/9 ≈ 7.78
5. Binomial Distribution S1
Binomial Distribution: X ~ B(n, p) where n = number of trials, p = probability of success in each trial. Conditions: fixed number of trials, independent trials, constant probability, only two outcomes (success/failure).
Binomial Distribution Formulae
P(X=r) = ⁿCᵣ pʳ (1−p)ⁿ⁻ʳ for r = 0, 1, 2, ..., n
E(X) = np Var(X) = np(1−p) = npq where q = 1−p
P(X≤r) = Σ ⁿCₖ pᵏ qⁿ⁻ᵏ for k=0 to r (use tables or calculator)
📐 Worked Example 9 — Binomial Probabilities
X~B(10, 0.35). Find: (a) P(X=4) (b) P(X≥3) (c) E(X) and Var(X) (d) P(X=k) is maximum at k=?
1
(a) P(X=4) = ¹⁰C₄×(0.35)⁴×(0.65)⁶
= 210×0.01501×0.07542 ≈ 0.2377
= 210×0.01501×0.07542 ≈ 0.2377
2
(b) P(X≥3) = 1−P(X≤2) = 1−[P(0)+P(1)+P(2)]
P(0)=(0.65)¹⁰≈0.01346; P(1)=10×0.35×(0.65)⁹≈0.07249; P(2)=45×(0.35)²×(0.65)⁸≈0.17567
P(X≤2)≈0.26162. P(X≥3)≈0.7384
P(0)=(0.65)¹⁰≈0.01346; P(1)=10×0.35×(0.65)⁹≈0.07249; P(2)=45×(0.35)²×(0.65)⁸≈0.17567
P(X≤2)≈0.26162. P(X≥3)≈0.7384
3
(c) E(X)=10×0.35=3.5. Var(X)=10×0.35×0.65=2.275
4
(d) Modal value: find k where P(X=k) ≥ P(X=k±1). Generally k = floor((n+1)p).
(n+1)p=11×0.35=3.85 → floor=3. Check: P(3)≈0.2522>P(4)≈0.2377 ✓
Modal value = 3
(n+1)p=11×0.35=3.85 → floor=3. Check: P(3)≈0.2522>P(4)≈0.2377 ✓
Modal value = 3
Normal Approximation to Binomial
When n is large and p is not too close to 0 or 1 (rule of thumb: np>5 and nq>5), the binomial B(n,p) can be approximated by Normal(np, npq). Apply a continuity correction: P(X≤k) becomes P(Y≤k+0.5) where Y~N(np, npq).
6. Poisson Distribution S1
Poisson Distribution: X ~ Po(λ) models the number of events occurring in a fixed interval of time or space, where events occur randomly, independently, at a constant average rate λ.
Poisson Distribution
P(X=r) = e^(−λ) λʳ / r! for r = 0, 1, 2, ...
E(X) = Var(X) = λ (mean equals variance — key property)
If X~Po(λ₁) and Y~Po(λ₂) are independent: X+Y~Po(λ₁+λ₂)
Poisson approximation to Binomial: when n large, p small, np=λ moderate
📐 Worked Example 10 — Poisson Distribution
Calls arrive at a helpdesk at a rate of 4 per hour. Find: (a) P(exactly 3 calls in an hour) (b) P(at most 2 calls in 30 minutes) (c) P(at least 1 call in 15 minutes)
1
(a) λ=4. P(X=3) = e⁻⁴×4³/3! = e⁻⁴×64/6 ≈ 0.01832×10.667 ≈ 0.1954
2
(b) 30 min → λ=2. P(X≤2) = P(0)+P(1)+P(2)
= e⁻²(1+2+2) = 5e⁻² ≈ 5×0.1353 = 0.6767
= e⁻²(1+2+2) = 5e⁻² ≈ 5×0.1353 = 0.6767
3
(c) 15 min → λ=1. P(X≥1) = 1−P(X=0) = 1−e⁻¹ ≈ 1−0.3679 = 0.6321
Choosing the Right Distribution:
Binomial B(n,p): Fixed n trials, constant p, independent, count successes.
Geometric Geo(p): Count trials until first success, independent, constant p.
Poisson Po(λ): Count events in interval, random/independent, constant rate λ, E=Var=λ.
Normal N(μ,σ²): Continuous data, symmetric bell curve (Lesson 9).
Binomial B(n,p): Fixed n trials, constant p, independent, count successes.
Geometric Geo(p): Count trials until first success, independent, constant p.
Poisson Po(λ): Count events in interval, random/independent, constant rate λ, E=Var=λ.
Normal N(μ,σ²): Continuous data, symmetric bell curve (Lesson 9).
📝 Exam Practice Questions
Q1 [4 marks] — Data: 12, 15, 18, 18, 20, 22, 25, 28, 30, 35. Find the median, quartiles, IQR, and identify any outliers using the 1.5×IQR rule.
n=10. Median = (20+22)/2 = 21
Q₁ = median of lower half {12,15,18,18,20} = 18
Q₃ = median of upper half {22,25,28,30,35} = 28
IQR = 28−18 = 10
Lower fence: 18−15=3. Upper fence: 28+15=43. No values outside [3,43].
No outliers.
Q₁ = median of lower half {12,15,18,18,20} = 18
Q₃ = median of upper half {22,25,28,30,35} = 28
IQR = 28−18 = 10
Lower fence: 18−15=3. Upper fence: 28+15=43. No values outside [3,43].
No outliers.
Q2 [3 marks] — For 50 values: Σx=480, Σx²=5200. Find the mean, variance, and standard deviation.
x̄ = 480/50 = 9.6
σ² = 5200/50 − 9.6² = 104 − 92.16 = 11.84
σ = √11.84 ≈ 3.44
σ² = 5200/50 − 9.6² = 104 − 92.16 = 11.84
σ = √11.84 ≈ 3.44
Q3 [4 marks] — Events A and B satisfy P(A)=0.4, P(B)=0.5, P(A∪B)=0.7. (a) Find P(A∩B). (b) Find P(A|B). (c) Are A and B independent? (d) Are A and B mutually exclusive?
(a) P(A∩B)=P(A)+P(B)−P(A∪B)=0.4+0.5−0.7=0.2
(b) P(A|B)=P(A∩B)/P(B)=0.2/0.5=0.4
(c) P(A|B)=0.4=P(A) ✓ → A and B are independent
(Also: P(A∩B)=0.2=P(A)×P(B)=0.4×0.5=0.2 ✓)
(d) P(A∩B)=0.2≠0 → not mutually exclusive
(b) P(A|B)=P(A∩B)/P(B)=0.2/0.5=0.4
(c) P(A|B)=0.4=P(A) ✓ → A and B are independent
(Also: P(A∩B)=0.2=P(A)×P(B)=0.4×0.5=0.2 ✓)
(d) P(A∩B)=0.2≠0 → not mutually exclusive
Exam Tip: Independence means P(A∩B)=P(A)P(B). Mutually exclusive means P(A∩B)=0. Two events with positive probability CANNOT be both independent and mutually exclusive simultaneously.
Q4 [4 marks] — A DRV X has the distribution: P(X=1)=0.1, P(X=2)=0.3, P(X=3)=k, P(X=4)=0.2, P(X=5)=0.1. Find k, E(X), Var(X), and E(3X−2).
k=1−(0.1+0.3+0.2+0.1)=0.3
E(X)=1(0.1)+2(0.3)+3(0.3)+4(0.2)+5(0.1)=0.1+0.6+0.9+0.8+0.5=2.9
E(X²)=1(0.1)+4(0.3)+9(0.3)+16(0.2)+25(0.1)=0.1+1.2+2.7+3.2+2.5=9.7
Var(X)=9.7−2.9²=9.7−8.41=1.29
E(3X−2)=3E(X)−2=8.7−2=6.7
E(X)=1(0.1)+2(0.3)+3(0.3)+4(0.2)+5(0.1)=0.1+0.6+0.9+0.8+0.5=2.9
E(X²)=1(0.1)+4(0.3)+9(0.3)+16(0.2)+25(0.1)=0.1+1.2+2.7+3.2+2.5=9.7
Var(X)=9.7−2.9²=9.7−8.41=1.29
E(3X−2)=3E(X)−2=8.7−2=6.7
Q5 [3 marks] — X~B(15, 0.4). Find P(X=6), P(4≤X≤7), and the most likely value of X.
P(X=6)=¹⁵C₆×(0.4)⁶×(0.6)⁹=5005×0.004096×0.010078≈0.2066
P(4≤X≤7)=P(4)+P(5)+P(6)+P(7)
≈0.1268+0.1859+0.2066+0.1771≈0.6964
Modal value: (n+1)p=16×0.4=6.4 → floor=6. Mode=6
P(4≤X≤7)=P(4)+P(5)+P(6)+P(7)
≈0.1268+0.1859+0.2066+0.1771≈0.6964
Modal value: (n+1)p=16×0.4=6.4 → floor=6. Mode=6
Q6 [4 marks] — Flaws in a fabric occur at a rate of 2 per metre. (a) State the distribution of X = number of flaws in 3 metres. (b) Find P(X=5). (c) Find P(X>8). (d) In 0.5 m, find P(at least one flaw).
(a) X~Po(6) (λ=2×3=6)
(b) P(X=5)=e⁻⁶×6⁵/5!=e⁻⁶×7776/120≈0.002479×64.8≈0.1606
(c) P(X>8)=1−P(X≤8). Using cumulative Poisson tables or calculation:
P(X≤8)≈0.8472. P(X>8)≈0.1528
(d) 0.5m: λ=1. P(X≥1)=1−e⁻¹≈0.6321
(b) P(X=5)=e⁻⁶×6⁵/5!=e⁻⁶×7776/120≈0.002479×64.8≈0.1606
(c) P(X>8)=1−P(X≤8). Using cumulative Poisson tables or calculation:
P(X≤8)≈0.8472. P(X>8)≈0.1528
(d) 0.5m: λ=1. P(X≥1)=1−e⁻¹≈0.6321