Quick key: 1 A, 2 C, 3 D, 4 B, 5 C, 6 A, 7 B, 8 D, 9 C, 10 A, 11 B, 12 D, 13 B, 14 C, 15 A, 16 D, 17 B, 18 C, 19 D, 20 A, 21 C, 22 B, 23 D, 24 A, 25 B, 26 C, 27 D, 28 A, 29 B, 30 C, 31 D, 32 A, 33 B, 34 C, 35 D, 36 A, 37 B, 38 C, 39 D, 40 A.
| Marks (out of 40) |
What it means |
| 36 to 40 |
Exam-ready; keep drilling the rapid review in Part 8 |
| 30 to 35 |
Close; redo the parts for the questions you missed |
| Below 30 |
Go back through Parts 1 to 5 in order, then retake this paper |
Data and descriptive statistics
Q1 (A): the four categories have a natural order but unequal gaps, so the level is ordinal.
Q2 (C): a number computed from a sample is a statistic. A parameter describes the whole population.
Q3 (D): width = range ÷ classes = (12.0 − 3.6)/6 = 1.4, and the course rounds a class width up to a convenient whole number, 2.
Q4 (B): relative frequency = 29/60 = 0.483. Choice (D) 0.517 is the share outside the class.
Q5 (C): shares of a whole are shown with a pie chart (or a bar chart); a histogram is for a quantitative variable and a line chart is for time.
Q6 (A): (12 + 15 + 11 + 18 + 14)/5 = 70/5 = 14.
Q7 (B): n = 6, so the median is the average of the 3rd and 4th ordered values: (15 + 16)/2 = 15.5.
Q8 (D): the mean is 10, squared deviations are 16, 4, 0, 4, 16 (sum 40), and the sample variance is 40/(5 − 1) = 10, so s = √10 = 3.16. The population SD would be √(40/5) = 2.83 (choice A), but the data are a sample.
Q9 (C): CV = (12/80) × 100 = 15%.
Q10 (A): the mean is pulled above the median and the mode (mode < median < mean), which means a long right tail, so right-skewed.
Q11 (B): z = (85 − 70)/10 = 1.5.
Q12 (D): n = 8, i = 0.25 × 8 = 2, which is a whole number, so Q1 is the average of the 2nd and 3rd values: (109 + 114)/2 = 111.5.
Probability and Bayes
Q13 (B): independent events satisfy P(A ∩ B) = P(A)P(B), and 0.4 × 0.3 = 0.12.
Q14 (C): 0.5 + 0.4 − 0.2 = 0.7.
Q15 (A): P(A | B) = 0.2 ÷ 0.4 = 0.5.
Q16 (D): P(BT | AC) = P(BT ∩ AC)/P(AC) = 0.2/0.7 = 0.286. Choice (A) is the reverse conditional.
Q17 (B): mutually exclusive, so there is no overlap term: 0.3 + 0.5 = 0.8.
Q18 (C): general multiplication law with the conditional given online: 0.60 × 0.25 = 0.15.
Q19 (D): independent, so 0.04 × 0.04 = 0.0016.
Q20 (A): joints: defective and flagged 0.02 × 0.90 = 0.018; good and flagged 0.98 × 0.05 = 0.049; total flagged 0.067; P(defective | flagged) = 0.018/0.067 = 0.269.
Q21 (C): joints: A 0.70 × 0.01 = 0.007; B 0.30 × 0.04 = 0.012; total defective 0.019; P(B | defective) = 0.012/0.019 = 0.632.
Q22 (B): the given group is the 40 managers, 10 of whom are female: 10/40 = 0.25. (Choice D, 0.40, is the share of all employees who are female: 80/200.)
Discrete distributions
Q23 (D): only (D) has all probabilities between 0 and 1 and a sum of exactly 1. (A) sums to 1.1, (B) has a negative probability, (C) sums to 1.2.
Q24 (A): 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7.
Q25 (B): (0 − 1.7)²(0.1) + (1 − 1.7)²(0.3) + (2 − 1.7)²(0.4) + (3 − 1.7)²(0.2) = 0.289 + 0.147 + 0.036 + 0.338 = 0.81. The SD would be 0.9.
Q26 (C): C(5, 2)(0.3)²(0.7)³ = 10 × 0.09 × 0.343 = 0.3087. Choice (A) forgets the 10 ways.
Q27 (D): μ = np = 12 × 0.25 = 3; σ = √(npq) = √(12 × 0.25 × 0.75) = √2.25 = 1.5. Choice (A) forgets the square root.
Q28 (A): at least one is the complement of none: 1 − (0.9)¹⁰ = 1 − 0.3487 = 0.651.
Q29 (B): 4² e⁻⁴ ÷ 2! = 16 × 0.01832 ÷ 2 = 0.1465. Choice (D) 0.238 is P(2 or fewer).
Q30 (C): 20 minutes is one third of an hour, so λ = 3 × 1/3 = 1 and P(0) = e⁻¹ = 0.368. Choice (A) 0.050 uses λ = 3, the wrong interval.
Continuous distributions
Q31 (D): (30 − 22)/(30 − 10) = 8/20 = 0.40.
Q32 (A): μ = (0 + 12)/2 = 6; σ = 12/√12 = 3.46.
Q33 (B): the mean is 1/λ = 5, so λ = 0.2 per minute and P(x > 10) = e⁻⁰·²×10 = e⁻² = 0.1353.
Q34 (C): λ = 6/60 = 0.1 per minute; P(x ≤ 15) = 1 − e⁻⁰·¹×15 = 1 − e⁻¹·⁵ = 1 − 0.2231 = 0.7769. Choice (A) is the "more than 15 minutes" probability.
Q35 (D): z = (115 − 100)/15 = 1.00, tail = 0.5 − 0.3413 = 0.1587.
Q36 (A): z = (46 − 50)/8 = −0.5 and z = (58 − 50)/8 = 1.0. Opposite sides, so add: 0.1915 + 0.3413 = 0.5328.
Q37 (B): z = (35 − 50)/10 = −1.5; lower tail = 0.5 − 0.4332 = 0.0668.
Q38 (C): the area from the mean to the cut-off is 0.45, so z = 1.645 and x = 100 + 1.645(15) = 124.7. Choice (B) uses z = 1.96, which is the cut-off for the top 2.5%.
Q39 (D): waiting time between arrivals is exponential; the count of arrivals would be Poisson.
Q40 (A): z = (550 − 500)/100 = 0.5 and z = (700 − 500)/100 = 2.0. Same side of the mean, so subtract: 0.4772 − 0.1915 = 0.2857.