Engineering Data Analysis (Probability and Statistics) - Exam Answer Templates
Proper answer writing is the single most controllable factor in PRC board exam performance. A candidate who knows the correct answer but writes it poorly may lose 30–50% of available marks due to missing formulas, skipped steps, or unclear final answers. These templates show you EXACTLY how to structure responses — from one-liner definitions to full five-step numerical solutions — so that every mark-earning element is visible to the examiner. In Engineering Mathematics, partial credit is awarded for correct method even when the final numerical answer is wrong; these templates teach you how to capture every point. Study the answer structure, memorize the key phrases, and practice writing at the speed required by the board exam (roughly 1 minute per mark).
Answer templates
Define the arithmetic mean of a data set. [1 mark]
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.
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
Criteria
Correct definition including both the concept of summation and division by n, or correct formula with notation
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
State the difference between mutually exclusive events and independent events in probability. [1 mark]
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.
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
Criteria
Correct contrast of both terms with their respective probability rules
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)
Distinguish between a permutation and a combination. Give one example of each. [2 marks]
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.
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
Criteria
Correct definition and formula for permutation with a valid worked example
Criteria
Correct definition and formula for combination with a valid worked example
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)!]
A card is drawn at random from a standard deck of 52 cards. Find the probability that the card is a king OR a heart. [2 marks]
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
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
Criteria
Correctly identifying the overlap P(A∩B) = 1/52 and citing the general addition rule
Criteria
Correct substitution and simplification to 16/52 = 4/13 ≈ 0.308
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
Compute the population mean and population standard deviation of the data set: 3, 7, 7, 9, 14. [3 marks]
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
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
Criteria
Correct mean μ = 8 shown with working
Criteria
Correct squared deviations summing to 64, shown individually
Criteria
Correct application of population variance formula (dividing by n, not n−1) and σ ≈ 3.58
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
In how many ways can a president, vice-president, and secretary be selected from a club with 10 members? [2 marks]
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
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
Criteria
Correct identification of permutation (order matters) and statement of formula P(n,r) = n!/(n−r)!
Criteria
Correct substitution and evaluation: P(10,3) = 720
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
A quality control inspector finds that 5% of steel bars produced by a plant are defective. If 4 bars are randomly selected, find the probability that exactly 2 are defective. Use the binomial distribution. [3 marks]
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%)
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
Criteria
Correct identification of binomial conditions (n, p, q, x) and statement of formula
Criteria
Correct calculation of C(4,2) = 6 and (0.05)² = 0.0025, (0.95)² = 0.9025
Criteria
Correct final probability ≈ 0.0135
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
The compressive strength of concrete cylinders in a project is normally distributed with a mean of 28 MPa and a standard deviation of 3 MPa. Find the probability that a randomly selected cylinder has a compressive strength between 25 MPa and 34 MPa. [3 marks]
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]
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
Criteria
Correct z-scores: z₁ = −1.00 and z₂ = +2.00 with working shown
Criteria
Correct Φ values cited: Φ(−1) = 0.1587, Φ(+2) = 0.9772
Criteria
Correct subtraction and final answer 0.8185 (81.85%)
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
A structural engineer tests five steel wire samples and records their tensile strength (kN): 52, 56, 48, 60, 54. Compute the sample variance and sample standard deviation. [3 marks]
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
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
Criteria
Correct mean x̄ = 54 kN with working
Criteria
Correct squared deviations table summing to 80
Criteria
Correct use of n−1 = 4 denominator giving s² = 20 kN² and s ≈ 4.47 kN
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
A construction firm wins each bid independently with a probability of 0.40. Find the probability that the firm wins at least 2 out of the next 4 bids. [5 marks]
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%
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
Criteria
Correct setup: GIVEN, Binomial conditions verified, complement strategy identified
Criteria
Correct P(X=0) = 0.1296 with full working shown
Criteria
Correct P(X=1) = 0.3456 with full working shown
Criteria
Correct application of complement rule: 1 − 0.4752
Criteria
Correct final answer P(X≥2) = 0.5248 or 52.48% clearly stated and boxed
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
State the 68-95-99.7 empirical rule for normal distributions and apply it: A set of soil bearing capacity tests has μ = 150 kPa and σ = 20 kPa. Between what two values do approximately 95% of measurements fall? [2 marks]
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.
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
Criteria
Correct statement of the 95% empirical rule as μ ± 2σ
Criteria
Correct computation: lower = 110 kPa, upper = 190 kPa
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
Two dice are rolled. What is the probability of getting a sum greater than 9? [2 marks]
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
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
Criteria
Correct total sample space (36) and complete enumeration of favorable outcomes (6 outcomes)
Criteria
Correct probability 6/36 = 1/6 ≈ 0.167
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
A civil engineering department has 8 faculty members. In how many ways can a committee of 3 be formed if: (a) any 3 members may be selected, and (b) the committee must include the department chair? [3 marks]
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.
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
Criteria
Correct use of combination formula C(8,3) = 56 for part (a)
Criteria
Correct logic for part (b): chair is fixed → choose 2 from remaining 7 = C(7,2)
Criteria
Correct evaluation C(7,2) = 21 with working shown
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
The number of vehicles arriving at a tollbooth follows a Poisson distribution with a mean of 3 vehicles per minute. Find the probability that exactly 5 vehicles arrive in a given minute. [3 marks]
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%
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
Criteria
Correct identification of Poisson distribution and statement of formula
Criteria
Correct evaluation of e^−3 = 0.04979, 3⁵ = 243, 5! = 120
Criteria
Correct final probability ≈ 0.1008
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
The breaking loads (kN) of seven rope specimens are: 18, 21, 19, 25, 17, 23, 22. Find the (a) mean, (b) median, (c) mode, and (d) range. [5 marks]
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
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
Criteria
Correct mean x̄ = 145/7 ≈ 20.71 kN with Σx shown
Criteria
Correct median = 21 kN with the ordered data and position clearly identified
Criteria
Correct statement that there is no mode (each value unique)
Criteria
Correct range = 8 kN = max − min
Criteria
Correct ascending order shown and all answers stated with kN unit
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
Mark wise strategy
State the definition precisely and write the formula. For numerical 1-mark questions, show only the key substitution and the answer. Do not over-explain — every extra wrong statement is a risk.
Do
- Write the formula or definition on the first line
- Use standard notation (x̄, μ, σ, P(A), C(n,r))
- State the final numerical answer clearly
- Include the unit if the quantity is physical (e.g., kN, MPa)
Don't
- Do not write a full paragraph for a 1-mark answer
- Do not show lengthy step-by-step working — just the key step and answer
- Do not leave blanks — even a formula with no substitution earns partial credit
Use the GIVEN → REQUIRED → SOLUTION → ANSWER format. The formula line and the final answer line are the two most critical lines. Show the substitution explicitly between them.
Do
- Start with a GIVEN block listing all known values
- Write the formula before substituting
- Show ONE clear computation step between the formula and the answer
- Box the final answer with its unit
Don't
- Do not skip the formula and go straight to numbers
- Do not present a 2-mark answer as a one-liner calculation — the examiner needs to see method
- Do not forget to simplify fractions (e.g., 16/52 → 4/13)
Three marks typically correspond to three distinct steps. Identify what each step earns and write it clearly. A tabular format for variance/probability enumeration earns clarity marks and prevents arithmetic errors.
Do
- Clearly label each step (Step 1, Step 2, Step 3)
- Use a table for variance problems (x, x−x̄, (x−x̄)²)
- Draw and label a bell curve for normal distribution problems
- Justify your choice of formula (why binomial? why permutation?)
Don't
- Do not compress three distinct computations into one line
- Do not forget to interpret the answer (e.g., '0.8185 or 81.85% of specimens fall in this range')
- Do not assume the examiner knows your intermediate values — write them all out
Plan before writing. A 5-mark question has five distinct mark-earning elements. Use the complement rule, complement probabilities, or part-by-part structure to maximize marks. Always include a verification step if time allows — it demonstrates mastery and can earn the final 0.5 mark.
Do
- Write a complete GIVEN block at the start
- State which distribution/formula/approach applies and WHY
- Show ALL intermediate computations (each component is a potential mark)
- Use the complement rule for 'at least' probability problems
- Include a verification step or reasonableness check at the end
- Present a clean summary block with all answers at the end
Don't
- Do not attempt a 5-mark problem in 3 minutes — allocate proper time
- Do not skip justifications — 'I used binomial because there are fixed independent trials with two outcomes' earns a setup mark
- Do not leave a 5-mark question blank — even a correct formula with no numbers earns 1 mark
- Do not write in paragraph form — use numbered steps and labels
General answer writing tips
- Always follow the four-part format for numerical problems: GIVEN → REQUIRED → SOLUTION → ANSWER. Examiners scan for this structure first before reading the working.
- Write the governing formula on its own line BEFORE substituting numbers. This earns the formula mark even if your arithmetic later goes wrong.
- Box or underline your final numerical answer with the correct SI unit. An unboxed answer buried in working may be missed by the examiner.
- For probability answers, always express the result as both a fraction and a decimal (e.g., 6/36 = 0.167). Board exam choices are often given as decimals.
- State whether you are using population (divide by n) or sample (divide by n−1) standard deviation at the beginning of every statistics problem — this one line prevents a common deduction.
- In counting problems (permutations/combinations), explicitly justify your choice: write 'Order matters → Permutation' or 'Order does not matter → Combination' before applying the formula.
- For normal distribution problems, always draw and label the bell curve with the z-score and shaded region before reading the z-table. This diagram earns a presentation mark and prevents sign errors.
- Check reasonableness: probabilities must be between 0 and 1; standard deviation must be non-negative; the number of combinations must be a positive integer. Write a one-line check at the end.
Diagrams
Probability and Statistics Problem-Solving Decision Tree
A comprehensive flowchart that guides students through the decision process for every problem type in this chapter. Starting from the nature of the question, the tree branches into Descriptive Statistics (with population vs sample fork), Probability (with OR/AND/mutually exclusive forks), Counting (with order matters fork), and Distributions (Binomial, Poisson, Normal). Students can use this as a first-step tool before selecting their formula.
Engineering Data Analysis — Master Formula Map
A hierarchical mind map covering all four major topic areas: Descriptive Statistics (mean, variance, standard deviation, median, mode, range), Probability (classical, addition, multiplication, complement), Counting (permutation and combination with their formulas), and Distributions (Binomial, Normal, Poisson with key parameters). Ideal as a one-page formula reference before the board exam.
Standard Deviation Calculation Procedure
A step-by-step flowchart of the standard deviation computation procedure. The critical decision node between population (÷n) and sample (÷n−1) is highlighted as the most common source of error. This diagram serves as a procedural checklist students can follow during the exam to avoid skipping steps.
Distribution of Engineering Data Analysis Question Types in PRC Board Exams
A pie chart showing the approximate distribution of question types for Engineering Data Analysis in PRC Civil Engineer board examinations based on historical patterns. Descriptive Statistics and Probability (Classical + Binomial + Normal) together account for ~85% of questions. Students should allocate study time proportionally, prioritizing Normal and Binomial distributions which together represent 40% of questions.