Use this diagnostic to find which Unit 2 topics to revisit: one question per topic, two for Topics 2.6, 2.10 and 2.11.
These are original Marlbridge practice questions, not past exam questions, with fictional data. The diagnostic is not calibrated and gives no predicted score.
How to take it. Allow 30 minutes and answer everything before opening any answer. Graphing calculator allowed. N(μ, σ) means a normal distribution with mean μ and standard deviation σ. “Or” includes “both”.
Question 1 (multiple choice · 2.1)
A segmented bar chart shows how pupils travel to two fictional schools: Ashdown (300 pupils) and Brookvale (600). Segment heights:
| School | Walk | Bus | Car |
|---|---|---|---|
| Ashdown | 0.50 | 0.30 | 0.20 |
| Brookvale | 0.25 | 0.30 | 0.45 |
Which statement is supported?
- (A) More Ashdown pupils walk than Brookvale pupils.
- (B) School and travel mode are associated: 0.45 of Brookvale pupils come by car, against 0.20 at Ashdown.
- (C) The same number of pupils at each school come by bus.
- (D) The variables are not associated, because the bus segments match.
Answer and explanation
Answer: (B). The conditional distributions differ, so the variables are associated.
- (A): 0.50 × 300 = 0.25 × 600 = 150, the same.
- (C) confuses proportions with counts: 90 against 180.
- (D): one matching category does not cancel differences elsewhere.
If you missed this: Topic 2.1 study guide.
Question 2 (multiple choice · 2.2)
The fictional Larkspur Hotel recorded 500 guests:
| Booked through | Business | Leisure | Total |
|---|---|---|---|
| Website | 90 | 210 | 300 |
| Agent | 110 | 90 | 200 |
| Total | 200 | 300 | 500 |
What proportion of the business guests booked through the website?
- (A) 0.18
- (B) 0.30
- (C) 0.45
- (D) 0.60
Answer and explanation
Answer: (C). Divide by the Business total: 90 ÷ 200 = 0.45.
- (A) is the joint relative frequency, 90 ÷ 500.
- (B) swaps the direction: 90 ÷ 300.
- (D) is the marginal relative frequency 300 ÷ 500.
If you missed this: Topic 2.2 study guide.
Question 3 (multiple choice · 2.3)
A fictional footballer scores each penalty with probability 0.7, independently. Which simulation correctly estimates P(at least 4 goals in 5 penalties)?
- (A) One digit per penalty, 0–6 = scored; 5 digits per trial; record whether at least 4 are 0–6; repeat many times.
- (B) As (A), but 0–7 = scored.
- (C) As (A), but skip digits that repeat within a trial.
- (D) One digit per trial: 0–6 = “at least 4 scored”.
Answer and explanation
Answer: (A). Seven of ten digits give 0.7, and one trial is 5 penalties.
- (B) gives 0.8.
- (C) is for sampling without replacement.
- (D) gives the whole event probability 0.7; it is about 0.53.
If you missed this: Topic 2.3 study guide.
Question 4 (multiple choice · 2.4)
A fictional café app gives one of three equally likely stamps (Sun, Moon, Star) per visit, independently. Over two visits, what is P(two different stamps)?
- (A) 1/9
- (B) 1/3
- (C) 1/2
- (D) 2/3
Answer and explanation
Answer: (D). Of the 3 × 3 = 9 equally likely ordered pairs, 3 repeat a stamp, so P = 1 − 3/9 = 2/3.
- (A) is one particular pair.
- (B) is P(same stamp), the complement.
- (C) wrongly treats 6 unordered results as equally likely.
If you missed this: Topic 2.4 study guide.
Question 5 (multiple choice · 2.5)
For a random day at a fictional ferry port, P(fog) = 0.12, P(a sailing is cancelled) = 0.05 and P(fog ∩ cancelled) = 0.03. Which statement is correct and correctly justified?
- (A) Not mutually exclusive, because P(fog ∩ cancelled) = 0.03 > 0.
- (B) Mutually exclusive, because fog does not always lead to a cancellation.
- (C) Mutually exclusive, because 0.03 is smaller than both 0.12 and 0.05.
- (D) Not mutually exclusive, because 0.12 + 0.05 is less than 1.
Answer and explanation
Answer: (A). Mutually exclusive events have joint probability 0; here both happen on about 3% of days.
- (B) and (C) ignore the test: a joint probability above 0 means both can happen.
- (D) has the right verdict for the wrong reason.
If you missed this: Topic 2.5 study guide.
Question 6 (multiple choice · 2.6)
At a fictional college, 40% of students take Spanish, and 25% of these also take French. What is P(a randomly chosen student takes both)?
- (A) 0.25
- (B) 0.10
- (C) 0.625
- (D) 0.65
Answer and explanation
Answer: (B). General multiplication rule: P(S ∩ F) = P(S) · P(F | S) = 0.40 × 0.25 = 0.10.
- (A) is the conditional probability P(F | S).
- (C) divides, 0.25 ÷ 0.40.
- (D) adds the two values.
If you missed this: Topic 2.6 study guide.
Question 7 (short answer · 2.6)
A fictional cycle-hire scheme has 60% e-bikes and 40% standard bikes. After a hire, 5% of e-bikes and 15% of standard bikes need a repair. One returned bike is chosen at random.
(a) Find the probability that it needs a repair. (b) It needs a repair. Find the probability that it is an e-bike, and explain why this is not 0.60.
Answer and explanation
(a) Multiply along tree branches, then add: P(repair) = 0.60 × 0.05 + 0.40 × 0.15 = 0.03 + 0.06 = 0.09.
(b) P(e-bike | repair) = 0.03 ÷ 0.09 ≈ 0.3333. Standard bikes need repairs three times as often, so they cause most repairs.
If you missed this: Topic 2.6 study guide, Worked example 2.
Question 8 (multiple choice · 2.7)
For a randomly chosen household in a fictional town, P(has a garden) = 0.70, P(has a pet) = 0.40 and P(both) = 0.28. Which statement is correct?
- (A) Independent; P(garden or pet) = 1.10
- (B) Not independent, because P(both) is not 0; P(garden or pet) = 0.82
- (C) Independent; P(garden or pet) = 0.82
- (D) Independent; P(garden or pet) = 0.54
Answer and explanation
Answer: (C). 0.70 × 0.40 = 0.28 = P(both), so the events are independent; P(G ∪ P) = 0.70 + 0.40 − 0.28 = 0.82.
- (A) forgets the overlap; it exceeds 1.
- (B) confuses independent with mutually exclusive.
- (D) subtracts the overlap twice.
If you missed this: Topic 2.7 study guide.
Question 9 (multiple choice · 2.8)
X = the number of a fictional ski resort’s 3 lifts closed on a random morning. Cumulative distribution:
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X ≤ x) | 0.55 | 0.85 | 0.97 | 1.00 |
Find P(at least 2 lifts closed).
- (A) 0.12
- (B) 0.15
- (C) 0.45
- (D) 0.97
Answer and explanation
Answer: (B). P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − 0.85 = 0.15.
- (A) is P(X = 2) only, leaving out X = 3.
- (C) is P(X ≥ 1).
- (D) reads P(X ≤ 2), the wrong direction.
If you missed this: Topic 2.8 study guide.
Question 10 (multiple choice · 2.9)
A fictional raffle ticket costs $2. It wins $50 with probability 0.01, $10 with probability 0.05 and nothing otherwise. What is the expected net gain per ticket?
- (A) −$1.00
- (B) $1.00
- (C) $18.00
- (D) −$2.00
Answer and explanation
Answer: (A). Net gains are $48, $8 and −$2 with probabilities 0.01, 0.05 and 0.94: E(X) = 0.48 + 0.40 − 1.88 = −$1.00. In the long run, buyers lose about $1 per ticket.
- (B) ignores the $2 cost.
- (C) averages 48, 8 and −2 without weighting.
- (D) is the most likely value, not the mean.
If you missed this: Topic 2.9 study guide.
Question 11 (multiple choice · 2.10)
A fictional player makes each free throw with probability 0.75, independently. In 8 throws, what is P(at least 7 made)?
- (A) 0.1001
- (B) 0.2670
- (C) 0.3671
- (D) 0.6329
Answer and explanation
Answer: (C). X ~ B(8, 0.75). P(X ≥ 7) = ₈C₇(0.75)⁷(0.25) + (0.75)⁸ = 0.2670 + 0.1001 = 0.3671.
- (A) is P(X = 8), from the off-by-one 1 − P(X ≤ 7).
- (B) is P(X = 7) only.
- (D) is P(X ≤ 6), the complement.
If you missed this: Topic 2.10 study guide.
Question 12 (short answer · 2.10)
On a fictional airline, 6% of booked passengers do not show up. The airline sells 124 tickets for a flight with 120 seats. Let X = the number of no-shows.
(a) State the binomial conditions in context. Which is most open to doubt? (b) Find and interpret the mean of X, and find its standard deviation. (c) The flight is over-full if fewer than 4 passengers fail to show. Find this probability.
Answer and explanation
(a) Two outcomes (show or not); n = 124 fixed; p = 0.06 for each; independent passengers. Independence is most doubtful: groups travel together.
(b) μ = 124 × 0.06 = 7.44 no-shows: over many such flights, no-shows average about 7.44 per flight. σ = √[124 × 0.06 × 0.94] ≈ 2.64 no-shows.
(c) “Fewer than 4” is X ≤ 3. X ~ B(124, 0.06), so P(X ≤ 3) = 0.0561.
If you missed this: Topic 2.10 study guide.
Question 13 (multiple choice · 2.11)
The lifetimes of a fictional battery are approximately N(40, 3) hours. Using the empirical rule, about what percentage last more than 34 hours?
- (A) 2.5%
- (B) 84%
- (C) 95%
- (D) 97.5%
Answer and explanation
Answer: (D). 34 = μ − 2σ. Half of the 5% outside μ ± 2σ lies below it, so 97.5% lies above.
- (A) is the area below 34 hours.
- (B) is the area above μ − σ = 37 hours.
- (C) is the area between 34 and 46 hours.
If you missed this: Topic 2.11 study guide.
Question 14 (short answer · 2.11)
At a fictional swimming club, 100 m times are approximately N(68, 2.4) seconds.
(a) Find the probability that a randomly chosen swim takes less than 64 seconds. (b) The fastest 5% of times earn a squad place. Find the qualifying time.
Answer and explanation
(a) z = (64 − 68) ÷ 2.4 ≈ −1.67. P(T < 64) = normalcdf(lower = −10⁹⁹, upper = 64, μ = 68, σ = 2.4) = 0.0478 (table: 0.0475).
(b) Fastest means lowest times, so the area to the left is 0.05: z = −1.6449, and 68 + (−1.6449)(2.4) ≈ 64.05 s. Using 0.95 gives 71.95 s, the slowest 5%.
If you missed this: Topic 2.11 study guide.
Question 15 (multiple choice · 2.12)
Sixteen fictional volunteers were randomly assigned, 8 to a mint drink and 8 to water, before a typing test. The mint group’s mean was 1.5 words per minute higher. In 1,000 random reallocations of the 16 scores, 229 differences (mint − water) were 1.5 or more. Best conclusion?
- (A) The reallocations prove that mint has no effect.
- (B) The randomization distribution is centred near 1.5, so the result is typical.
- (C) The mint group typed faster, so mint improves typing.
- (D) Chance assignment alone often gives a difference this large, so there is no convincing evidence that mint helps.
Answer and explanation
Answer: (D). Reallocating moves only the labels, showing what happens if mint makes no difference; 229 ÷ 1,000 ≈ 0.23 is not small.
- (A) is too strong: a typical result is not proof.
- (B) is wrong: it is centred near 0.
- (C) ignores chance variation from random assignment.
If you missed this: Topic 2.12 study guide.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 2.1 | 1 | Two-way tables |
| 2.2 | 2 | Two-way summary statistics |
| 2.3 | 3 | Simulation |
| 2.4 | 4 | Probability basics |
| 2.5 | 5 | Mutually exclusive events |
| 2.6 | 6, 7 | Conditional probability |
| 2.7 | 8 | Independence and unions |
| 2.8 | 9 | Random variables |
| 2.9 | 10 | Mean and standard deviation |
| 2.10 | 11, 12 | Binomial distribution |
| 2.11 | 13, 14 | Normal distribution |
| 2.12 | 15 | Sampling distributions |
How to use your result
- Mark each question right, partly right or wrong.
- Fix weak topics in order: 2.6 underpins 2.7, and both feed 2.10.
- Read the study guide, then do its practice set.
- Then try the Unit 2 mixed review.