CELE Engineering Mathematics — Engineering Data Analysis (Probability and Statistics)Exam Answer Templates

Marks

1

Topic

Descriptive Statistics — Measures of Center

Difficulty

easy

Template Id

T1

Examiner Tip

For 1-mark definition questions, one clean sentence + the formula is sufficient. Do not pad the answer — extra incorrect statements can cost marks.

Model Answer

The arithmetic mean (x̄) is the sum of all observed values divided by the total number of observations: x̄ = (Σx) / n It represents the central (balance) point of the data set.

Question Type

very_short_answer

Answer Structure

• Line 1: State the definition in one precise sentence [0.5 mark]
• Line 2: Write the formula with correct notation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'average of the largest and smallest value' — that is the midrange, not the mean
• Using incorrect notation (e.g., writing μ without specifying it is the population mean)
• Omitting the formula entirely

Key Phrases To Include

• sum of all values
• divided by the number of observations
• x̄ = Σx / n
• central value or balance point

Marks

1

Topic

Probability — Event Types

Difficulty

easy

Template Id

T2

Examiner Tip

The board exam frequently tests whether students confuse these two concepts. Memorize both formulas and contrast them side by side.

Model Answer

Mutually exclusive events cannot occur simultaneously [P(A∩B) = 0], so P(A∪B) = P(A) + P(B). Independent events do not affect each other's probability, so P(A∩B) = P(A)·P(B); they CAN occur simultaneously.

Question Type

very_short_answer

Answer Structure

• Clause 1: Define mutually exclusive with its probability condition [0.5 mark]
• Clause 2: Define independent with its probability condition [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that independent events cannot occur at the same time — this is wrong; only mutually exclusive events cannot
• Confusing the addition rule for ME events with the multiplication rule for independent events

Key Phrases To Include

• cannot occur simultaneously
• P(A∩B) = 0
• do not affect each other
• P(A∩B) = P(A)·P(B)

Marks

2

Topic

Counting — Permutations and Combinations

Difficulty

easy

Template Id

T3

Examiner Tip

Always compute the numerical answer for your example. 'Arranging books' alone gives no credit; 'P(5,3) = 60' shows mastery.

Model Answer

Permutation — arrangement where ORDER MATTERS. Formula: P(n,r) = n! / (n−r)! Example: Arranging 3 of 5 books on a shelf — P(5,3) = 60 ways. Combination — selection where ORDER DOES NOT MATTER. Formula: C(n,r) = n! / [r!(n−r)!] Example: Choosing a committee of 3 from 5 people — C(5,3) = 10 ways. Key distinction: P(5,3) = 60 > C(5,3) = 10 because permutations count re-orderings of the same group as different.

Question Type

short_answer

Answer Structure

• Line 1: Define permutation + write formula [0.5 mark]
• Line 2: Give one correct permutation example with computed value [0.5 mark]
• Line 3: Define combination + write formula [0.5 mark]
• Line 4: Give one correct combination example with computed value [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the definitions (saying order matters for combinations)
• Writing the formula without demonstrating its application in the example
• Using the same example for both — examiners want distinct illustrations

Key Phrases To Include

• order matters
• order does not matter
• P(n,r) = n!/(n-r)!
• C(n,r) = n!/[r!(n-r)!]

Marks

2

Topic

Probability — Addition Rule

Difficulty

medium

Template Id

T4

Examiner Tip

Whenever you see 'OR' in a probability problem, immediately write the addition rule. Then ask: can both happen at once? If yes, compute the overlap and subtract.

Model Answer

GIVEN: Total cards n = 52 Event A: card is a King → P(A) = 4/52 Event B: card is a Heart → P(B) = 13/52 Event A∩B: card is King of Hearts → P(A∩B) = 1/52 (Events are NOT mutually exclusive — the King of Hearts belongs to both.) REQUIRED: P(A∪B) SOLUTION: Using the Addition Rule: P(A∪B) = P(A) + P(B) − P(A∩B) P(A∪B) = 4/52 + 13/52 − 1/52 P(A∪B) = 16/52 ANSWER: P(King or Heart) = 16/52 = 4/13 ≈ 0.308

Question Type

numerical

Answer Structure

• GIVEN block: List n, P(A), P(B), and P(A∩B) with reasoning [0.5 mark]
• State the Addition Rule formula [0.5 mark]
• Substitute values and simplify [0.5 mark]
• Box the final simplified fraction and decimal [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to subtract P(A∩B) — the most common error on this type of question
• Using P(A∩B) = P(A)·P(B) — this multiplication rule applies only to INDEPENDENT events, not here
• Leaving answer as an unsimplified fraction without the decimal

Key Phrases To Include

• P(A∪B) = P(A) + P(B) − P(A∩B)
• NOT mutually exclusive
• King of Hearts (overlap)
• 16/52 = 4/13

Marks

3

Topic

Descriptive Statistics — Variance and Standard Deviation

Difficulty

medium

Template Id

T5

Examiner Tip

Create a three-column table (x, x−μ, (x−μ)²) for every standard deviation problem. It organizes the working, prevents arithmetic errors, and makes it easy for the examiner to award partial marks.

Model Answer

GIVEN: Data set x = {3, 7, 7, 9, 14}; n = 5; population statistics required. REQUIRED: Population mean (μ) and population standard deviation (σ) SOLUTION: Step 1 — Compute the mean: μ = Σx / n = (3 + 7 + 7 + 9 + 14) / 5 = 40 / 5 μ = 8 Step 2 — Compute squared deviations: (3 − 8)² = (−5)² = 25 (7 − 8)² = (−1)² = 1 (7 − 8)² = (−1)² = 1 (9 − 8)² = (1)² = 1 (14 − 8)² = (6)² = 36 Σ(x − μ)² = 25 + 1 + 1 + 1 + 36 = 64 Step 3 — Compute population variance and standard deviation: σ² = Σ(x − μ)² / n = 64 / 5 = 12.8 σ = √12.8 ANSWER: μ = 8 σ = √12.8 ≈ 3.58

Question Type

numerical

Answer Structure

• Step 1: Compute μ = Σx/n with all values shown [1 mark]
• Step 2: Set up and evaluate each squared deviation in a table [1 mark]
• Step 3: Apply σ = √[Σ(x−μ)²/n], compute and state answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Dividing by n−1 = 4 instead of n = 5 (sample vs population confusion) — this changes σ from 3.58 to 4.00
• Squaring before subtracting the mean (computing Σx² / n instead of Σ(x−μ)² / n)
• Not showing individual squared deviations — examiners need to see the working to award partial marks

Key Phrases To Include

• Population standard deviation: divide by n
• μ = Σx/n
• σ² = Σ(x−μ)²/n
• σ = √variance

Marks

2

Topic

Counting — Permutations

Difficulty

easy

Template Id

T6

Examiner Tip

Ask yourself: 'If I swap two people's roles, do I get a different arrangement?' If YES → Permutation. President and VP swapping is a different result, so YES.

Model Answer

GIVEN: n = 10 members; r = 3 positions to fill. REQUIRED: Number of ways to assign 3 distinct positions (Order matters — different positions are different roles). SOLUTION: Since order matters (president ≠ vice-president ≠ secretary), use Permutations: P(n,r) = n! / (n−r)! P(10,3) = 10! / (10−3)! = 10! / 7! P(10,3) = 10 × 9 × 8 = 720 ANSWER: Number of ways = 720

Question Type

numerical

Answer Structure

• Identify n and r; justify that order matters → Permutation [0.5 mark]
• Write the permutation formula [0.5 mark]
• Substitute and compute step-by-step [0.5 mark]
• State the final answer clearly [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using C(10,3) = 120 instead of P(10,3) = 720 — failing to recognize that different offices make order relevant
• Computing 10! fully instead of cancelling — wastes time and risks arithmetic errors
• Not justifying why permutation was chosen

Key Phrases To Include

• order matters
• P(n,r) = n!/(n−r)!
• P(10,3) = 10 × 9 × 8
• 720 ways

Marks

3

Topic

Distributions — Binomial Distribution

Difficulty

medium

Template Id

T7

Examiner Tip

Binomial applies when: (1) fixed number of trials n, (2) only two outcomes, (3) constant probability p, (4) independent trials. State these conditions at the start for full marks on structured questions.

Model Answer

GIVEN: n = 4 trials (bars selected) p = 0.05 (probability of defective bar) q = 1 − p = 0.95 x = 2 (exactly 2 defective) REQUIRED: P(X = 2) using Binomial Distribution SOLUTION: Step 1 — State the binomial formula: P(X = x) = C(n,x) · pˣ · qⁿ⁻ˣ Step 2 — Compute the combination: C(4,2) = 4! / [2! · 2!] = 6 Step 3 — Substitute all values: P(X = 2) = C(4,2) · (0.05)² · (0.95)² P(X = 2) = 6 · (0.0025) · (0.9025) P(X = 2) = 6 × 0.002256 P(X = 2) = 0.013538 ANSWER: P(X = 2) ≈ 0.0135 (or 1.35%)

Question Type

numerical

Answer Structure

• GIVEN block: identify n, p, q, x correctly [0.5 mark]
• State the binomial formula P(X=x) = C(n,x)·pˣ·qⁿ⁻ˣ [0.5 mark]
• Compute C(n,x) and show intermediate values [1 mark]
• Substitute and evaluate to get final probability [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using p = 0.95 (non-defective rate) instead of p = 0.05 for the defective probability
• Forgetting to include C(n,x) — writing only pˣ·qⁿ⁻ˣ
• Using Normal or Poisson instead of Binomial when n and p are both explicitly given

Key Phrases To Include

• P(X = x) = C(n,x) · pˣ · (1−p)ⁿ⁻ˣ
• n = 4, p = 0.05, q = 0.95, x = 2
• C(4,2) = 6
• fixed number of trials, two outcomes

Marks

3

Topic

Distributions — Normal Distribution

Difficulty

medium

Template Id

T8

Examiner Tip

Always DRAW the bell curve on normal distribution problems. Shade the required region, label the z-values and the mean. This sketch organizes your thinking and earns a presentation mark.

Model Answer

GIVEN: μ = 28 MPa, σ = 3 MPa Required range: 25 MPa ≤ X ≤ 34 MPa REQUIRED: P(25 ≤ X ≤ 34) SOLUTION: Step 1 — Standardize both bounds using z = (x − μ) / σ: z₁ = (25 − 28) / 3 = −3/3 = −1.00 z₂ = (34 − 28) / 3 = 6/3 = +2.00 Step 2 — Read z-table values (area from left to z): Φ(−1.00) = 0.1587 Φ(+2.00) = 0.9772 Step 3 — Compute the probability of the interval: P(25 ≤ X ≤ 34) = Φ(2.00) − Φ(−1.00) P(25 ≤ X ≤ 34) = 0.9772 − 0.1587 P(25 ≤ X ≤ 34) = 0.8185 ANSWER: P(25 MPa ≤ X ≤ 34 MPa) = 0.8185 or 81.85% [Bell curve sketch: shade between z = −1 and z = +2; label μ = 28 MPa]

Question Type

numerical

Answer Structure

• GIVEN block with μ, σ, and the two bounds clearly stated [0.5 mark]
• Compute z-scores for both bounds using z = (x−μ)/σ [1 mark]
• Read Φ values from z-table and state them explicitly [0.5 mark]
• Compute P = Φ(z₂) − Φ(z₁) and state final answer as percentage [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing Φ(z₂) + Φ(z₁) instead of subtracting for a two-sided interval
• Getting sign wrong: z = (28−25)/3 = +1 instead of −1 for the lower bound
• Not drawing the bell curve — this loses the presentation mark on 3-mark versions
• Forgetting to convert to percentage when the question says 'express as percentage'

Key Phrases To Include

• z = (x − μ) / σ
• z₁ = −1.00, z₂ = +2.00
• P = Φ(z₂) − Φ(z₁)
• 0.9772 − 0.1587 = 0.8185

Marks

3

Topic

Descriptive Statistics — Sample Variance

Difficulty

medium

Template Id

T9

Examiner Tip

The keyword 'sample' in the problem always signals n−1. Highlight it in your GIVEN block to remind yourself. Graders specifically check the denominator.

Model Answer

GIVEN: x = {52, 56, 48, 60, 54} kN; n = 5; sample statistics required. REQUIRED: Sample variance s² and sample standard deviation s SOLUTION: Step 1 — Compute the sample mean: x̄ = (52 + 56 + 48 + 60 + 54) / 5 = 270 / 5 = 54 kN Step 2 — Tabulate squared deviations: x | x − x̄ | (x − x̄)² 52 | −2 | 4 56 | +2 | 4 48 | −6 | 36 60 | +6 | 36 54 | 0 | 0 ───────────────────────── Sum 80 Step 3 — Apply sample variance formula (divide by n−1): s² = Σ(x − x̄)² / (n − 1) = 80 / (5 − 1) = 80 / 4 = 20 kN² s = √20 = 4.47 kN ANSWER: Sample variance s² = 20 kN² Sample standard deviation s = √20 ≈ 4.47 kN

Question Type

numerical

Answer Structure

• Step 1: Compute x̄ = 54 kN [0.5 mark]
• Step 2: Table with x, x−x̄, (x−x̄)² and correct sum = 80 [1 mark]
• Step 3: Apply s² = Σ(x−x̄)²/(n−1), show n−1 = 4, compute s [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Dividing by n = 5 instead of n−1 = 4 (population vs sample confusion)
• Not including the unit kN for standard deviation and kN² for variance
• Computing Σx² − (Σx)²/n without showing individual deviations — may lose the table mark

Key Phrases To Include

• sample standard deviation: divide by n−1
• s² = Σ(x − x̄)² / (n−1)
• x̄ = 54 kN
• Σ(x − x̄)² = 80
• s ≈ 4.47 kN

Marks

5

Topic

Distributions — Binomial Distribution, Complement Rule

Difficulty

hard

Template Id

T10

Examiner Tip

For 'at least' problems (X ≥ k), ALWAYS use the complement. Count how many terms are easier to compute below k and subtract from 1. This halves computation time and reduces errors.

Model Answer

GIVEN: n = 4 bids (independent, fixed number of trials) p = 0.40 (probability of winning each bid) q = 1 − p = 0.60 Required: P(X ≥ 2) REQUIRED: P(X ≥ 2) using Binomial Distribution Verify Binomial applies: fixed trials ✓, two outcomes (win/lose) ✓, constant p ✓, independent ✓. SOLUTION: Strategy: Use the complement rule: P(X ≥ 2) = 1 − P(X < 2) = 1 − [P(X = 0) + P(X = 1)] Step 1 — State the Binomial formula: P(X = x) = C(n,x) · pˣ · qⁿ⁻ˣ Step 2 — Compute P(X = 0): P(X = 0) = C(4,0) · (0.40)⁰ · (0.60)⁴ P(X = 0) = 1 · 1 · 0.1296 P(X = 0) = 0.1296 Step 3 — Compute P(X = 1): P(X = 1) = C(4,1) · (0.40)¹ · (0.60)³ P(X = 1) = 4 · 0.40 · 0.216 P(X = 1) = 4 × 0.0864 P(X = 1) = 0.3456 Step 4 — Apply the complement: P(X ≥ 2) = 1 − [0.1296 + 0.3456] P(X ≥ 2) = 1 − 0.4752 P(X ≥ 2) = 0.5248 Verification (optional, earns bonus credit): P(X=2) = C(4,2)·(0.4)²·(0.6)² = 6·0.16·0.36 = 0.3456 P(X=3) = C(4,3)·(0.4)³·(0.6)¹ = 4·0.064·0.6 = 0.1536 P(X=4) = C(4,4)·(0.4)⁴·(0.6)⁰ = 1·0.0256·1 = 0.0256 Sum = 0.3456 + 0.1536 + 0.0256 = 0.5248 ✓ ANSWER: P(X ≥ 2) = 0.5248 or 52.48%

Question Type

long_answer

Answer Structure

• GIVEN block: n, p, q, and clear statement of requirement [0.5 mark]
• Verify binomial conditions are met [0.5 mark]
• State complement strategy: P(X≥2) = 1 − [P(X=0) + P(X=1)] [0.5 mark]
• Write and apply binomial formula for P(X=0) with full computation [1 mark]
• Write and apply binomial formula for P(X=1) with full computation [1 mark]
• Apply complement rule and state final answer [1 mark]
• Optional verification by direct computation of P(X=2)+P(X=3)+P(X=4) [0.5 bonus]

Scoring Breakdown

Common Mark Deductions

• Computing P(X=2)+P(X=3)+P(X=4) directly without using the complement — wastes time and risks more arithmetic errors; no mark deduction but inefficient
• Using P(X≥2) = 1 − P(X=0) only (forgetting P(X=1))
• Mixing up p and q in the formula — using (0.60)ˣ instead of (0.40)ˣ
• Not showing the binomial formula before substituting values

Key Phrases To Include

• P(X ≥ 2) = 1 − P(X < 2)
• complement rule
• P(X = x) = C(n,x)·pˣ·qⁿ⁻ˣ
• n = 4, p = 0.40, q = 0.60
• 0.5248

Marks

2

Topic

Distributions — Normal Distribution, Empirical Rule

Difficulty

easy

Template Id

T11

Examiner Tip

Memorize the rule as '1-2-3 sigmas for 68-95-99.7 percent.' In the PRC exam, this rule often appears as a quick 1–2 mark question requiring direct application, not table lookup.

Model Answer

Empirical Rule (68-95-99.7 Rule): • 68% of data falls within μ ± 1σ • 95% of data falls within μ ± 2σ • 99.7% of data falls within μ ± 3σ (Applies to normally distributed data.) Application: For 95%: range = μ ± 2σ Lower bound = 150 − 2(20) = 150 − 40 = 110 kPa Upper bound = 150 + 2(20) = 150 + 40 = 190 kPa ANSWER: Approximately 95% of measurements fall between 110 kPa and 190 kPa.

Question Type

short_answer

Answer Structure

• State all three tiers of the empirical rule (68%, 95%, 99.7%) [0.5 mark]
• Identify the correct tier (95% → ±2σ) and apply to given μ and σ [0.5 mark]
• Compute both bounds with correct unit (kPa) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using ±1σ instead of ±2σ for the 95% case
• Computing the range as 2σ = 40 only, not applying to both sides
• Omitting the unit kPa from the answer

Key Phrases To Include

• 68-95-99.7 rule
• 95% → μ ± 2σ
• 110 kPa and 190 kPa
• normally distributed

Marks

2

Topic

Probability — Classical Probability, Sample Space

Difficulty

medium

Template Id

T12

Examiner Tip

For dice problems, list outcomes systematically by ascending sum. Never rely on memory alone — write out the pairs to avoid missing ordered pairs like (5,6) vs (6,5).

Model Answer

GIVEN: Two fair six-sided dice; total outcomes = 6 × 6 = 36. REQUIRED: P(sum > 9) = P(sum = 10, 11, or 12) SOLUTION: Favorable outcomes (sum > 9): Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes Sum = 11: (5,6), (6,5) → 2 outcomes Sum = 12: (6,6) → 1 outcome Total favorable = 3 + 2 + 1 = 6 Probability: P(sum > 9) = 6 / 36 = 1/6 ANSWER: P(sum > 9) = 6/36 = 1/6 ≈ 0.167

Question Type

numerical

Answer Structure

• State total sample space = 36 [0.5 mark]
• Enumerate ALL favorable outcomes for sums 10, 11, 12 in an organized list [1 mark]
• Compute probability = 6/36 and simplify [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Counting only sum = 10 and missing 11 and 12
• Miscounting ordered pairs — (4,6) and (6,4) are DIFFERENT outcomes
• Using total outcomes = 12 (adding dice faces) instead of 36 (multiplying)

Key Phrases To Include

• total outcomes = 6 × 6 = 36
• sum > 9 means sum = 10, 11, or 12
• favorable outcomes = 6
• P = 6/36 = 1/6

Marks

3

Topic

Counting — Combinations with Constraints

Difficulty

medium

Template Id

T13

Examiner Tip

When one or more members are FIXED (required), remove them from n and choose the remaining r − (number fixed) from the rest. This reduces the problem to a simpler combination.

Model Answer

GIVEN: n = 8 faculty members; committee size r = 3; order does NOT matter (a committee is a group, not ranked positions). Part (a) — Any 3 members from 8: Use Combination: C(n,r) = n! / [r!(n−r)!] C(8,3) = 8! / [3! · 5!] = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56 ways Part (b) — Committee MUST include the department chair: The chair is fixed. Remaining 2 spots are filled from the other 7 members. C(7,2) = 7! / [2! · 5!] = (7 × 6) / (2 × 1) = 42 / 2 = 21 ways ANSWER: (a) 56 committees can be formed. (b) 21 committees include the department chair.

Question Type

numerical

Answer Structure

• Identify that order does not matter → Combinations [0.5 mark]
• Part (a): Apply C(8,3) with working = 56 [1 mark]
• Part (b): Recognize chair is fixed, reduce to C(7,2) with working = 21 [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using P(8,3) = 336 instead of C(8,3) for an unranked committee
• For part (b), computing C(8,3) − C(7,3) = 56 − 35 = 21 (also correct, but students who use this rarely show clear logic)
• For part (b), computing C(7,2) = 21 but forgetting that the chair counts as the 3rd member

Key Phrases To Include

• order does not matter → Combination
• C(8,3) = 56
• chair is fixed; choose remaining 2 from 7
• C(7,2) = 21

Marks

3

Topic

Distributions — Poisson Distribution

Difficulty

medium

Template Id

T14

Examiner Tip

Use Poisson when the problem gives a RATE (events per unit time/area) and asks for a specific count, with no fixed n. The trigger word is 'on average' or 'mean rate of λ per [unit]'.

Model Answer

GIVEN: λ (mean rate) = 3 vehicles per minute x = 5 (desired number of arrivals) Distribution: Poisson (rare/random events over a fixed interval) REQUIRED: P(X = 5) SOLUTION: Step 1 — State the Poisson formula: P(X = x) = (e^−λ · λˣ) / x! Step 2 — Substitute values: P(X = 5) = (e^−3 · 3⁵) / 5! Step 3 — Evaluate each component: e^−3 = 0.049787 3⁵ = 243 5! = 120 Step 4 — Compute: P(X = 5) = (0.049787 × 243) / 120 P(X = 5) = 12.098 / 120 P(X = 5) = 0.10082 ANSWER: P(X = 5) ≈ 0.1008 or 10.08%

Question Type

numerical

Answer Structure

• GIVEN block: identify λ and x; state Poisson distribution applies [0.5 mark]
• Write the Poisson formula P(X=x) = e^−λ · λˣ / x! [0.5 mark]
• Compute e^−3, 3⁵, and 5! individually and show values [1 mark]
• Multiply/divide to final answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Binomial instead of Poisson when only the mean rate (λ) is given and there is no fixed n
• Computing e^3 (positive exponent) instead of e^−3
• Using 3! or 5 instead of 5! = 120

Key Phrases To Include

• Poisson distribution
• λ = 3
• P(X = x) = e^−λ · λˣ / x!
• e^−3 = 0.04979
• 0.1008

Marks

5

Topic

Descriptive Statistics — Measures of Center and Spread

Difficulty

easy

Template Id

T15

Examiner Tip

For 5-mark descriptive statistics questions, allocate ~1 mark per measure. ALWAYS sort first — the median and range both require ordered data. Write the sorted sequence prominently at the top of your solution.

Model Answer

GIVEN: Data set x = {18, 21, 19, 25, 17, 23, 22} kN; n = 7 REQUIRED: Mean, Median, Mode, Range SOLUTION: Step 1 — Arrange in ascending order: 17, 18, 19, 21, 22, 23, 25 Step 2 — (a) MEAN: x̄ = Σx / n = (17 + 18 + 19 + 21 + 22 + 23 + 25) / 7 x̄ = 145 / 7 x̄ = 20.71 kN Step 3 — (b) MEDIAN: n = 7 (odd) → median is the 4th value in the ordered set Ordered: 17, 18, 19, [21], 22, 23, 25 Median = 21 kN Step 4 — (c) MODE: Each value appears exactly once → No mode (or: the data set has no mode). Step 5 — (d) RANGE: Range = Maximum − Minimum = 25 − 17 = 8 kN ANSWER: (a) Mean x̄ = 20.71 kN (b) Median = 21 kN (c) Mode = No mode (d) Range = 8 kN

Question Type

long_answer

Answer Structure

• Arrange data in ascending order (prerequisite step) [0.5 mark]
• (a) Mean: show Σx = 145, divide by 7, state x̄ = 20.71 kN [1 mark]
• (b) Median: identify position for odd n = 7 as (n+1)/2 = 4th term, state 21 kN [1 mark]
• (c) Mode: examine frequencies, correctly state 'no mode' [1 mark]
• (d) Range: state max − min = 25 − 17 = 8 kN [1 mark]
• Present all answers in an organized summary block with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Finding the median without first sorting the data (computing the 4th element of the UNSORTED list gives 25, which is wrong)
• Stating 'no mode' when the question expects the student to identify it — some boards expect the explicit statement
• Computing range as max + min (addition) instead of max − min
• Omitting the unit kN from each answer

Key Phrases To Include

• Arrange in ascending order
• x̄ = Σx/n
• median: middle value for odd n
• mode: most frequent value
• range = max − min
• no mode