MATH40005 2026 Practice Paper 8

MATH40005 Probability and Statistics Resit Preparation Question 1: solution and marking guide 1(a). Let D be the condition and + a positive result. By the law of total probability, P(+) = 0 .90(0.04) + 0.06(0.96) = 0 .0936. 3 marks; A:3 1(b). Bayes’ theorem gives P(D | +) = 0.90(0.04) 0.0936 = 5 13 ≈ 0.3846. Sensitivity conditions on already having the condition, whereas the posterior also incorporates the low prevalence and the false positives among the much larger unaffected group. 4 marks; A:1, B:1, D:2 1(c). There are 35 4 possible main-number choices. To match exactly two of the four drawn main numbers, choose two correct numbers and two of the remaining 31 incorrect numbers. The bonus match has probability 1/8. Therefore P = 4 2 31 2 35 4 · 1 8 ≈ 0.00666. 6 marks; A:3, B:1, D:2 1(d). Give two samples to each laboratory first, leaving 8 samples. The number of nonnegative solutions of y1 + · · ·+ y5 = 8 is 8 + 5 − 1 5 − 1 = 12 4 = 495. 3 marks; A:2, C:1 1(e). The complement is that all 16 weeks are distinct. Hence P(at least one shared week ) = 1 − 52 · 51 · · · (52 − 15) 5216 = 1 − (52)16 5216 . 4 marks; A:1, B:2, C:1 Question total: 20 marks 2

MATH40005 Probability and Statistics Resit Preparation Question 2: solution and marking guide 2(a). Normalisation gives 1 = c Z 1 0 Z 1 y y(1 + x) dx dy = c Z 1 0 y 3 2 − y − y2 2 dy = c 7 24 , so c = 24/7. 3 marks; A:3 2(b). For 0 < x < 1, fX (x) = Z x 0 24 7 y(1 + x) dy = 12 7 x2(1 + x). For 0 < y < 1, fY (y) = Z 1 y 24 7 y(1 + x) dx = 24 7 y 3 2 − y − y2 2 . Both densities are zero outside (0, 1). 4 marks; A:2, B:2 2(c). For 0 < y < 1 and y < x < 1, fX|Y =y(x | y) = 1 + x 3 2 − y − y2 2 . Thus E(X | Y = y) = R 1 y x(1 + x) dx 3 2 − y − y2 2 = 5 6 − y2 2 − y3 3 3 2 − y − y2 2 , 0 < y < 1. 5 marks; A:1, B:1, C:1, D:2 2(d). Direct integration gives EX = 27 35 , EY = 18 35 , E(XY ) = 44 105 . Therefore Cov(X, Y ) = 44 105 − 27 35 18 35 = 82 3675 . 5 marks; A:1, C:2, D:2 2(e). No. The joint support is triangular: for 0 < x < y < 1 the joint density is zero while both marginals are positive. Hence the joint density cannot equal fX (x)fY (y) throughout the product of the marginal supports. 3 marks; A:1, B:2 Question total: 20 marks 3

MATH40005 Probability and Statistics Resit Preparation Question 3: solution and marking guide 3(a). The standard error is s/√n = 2/ √ 14. Hence 64.8 ± 2.160 2√ 14 = 64.8 ± 1.1546, so the interval is approximately (63.65, 65.95). 4 marks; A:3, B:1 3(b). The statistic is t = 64.8 − 66 2/ √ 14 ≈ −2.245, with 13 degrees of freedom. The two-sided p-value is 2P(T13 ≥ 2.245) ≈ 0.0428. Since this is below 0.05, reject H0 at the 5% level. 5 marks; A:1, B:1, C:1, D:2 3(c). A p-value is the null probability of obtaining a test statistic at least as extreme as the observed value, with “extreme” defined by the alternative and test. If the null test statistic has a continuous distribution and the p-value is constructed using its exact CDF (with the appropriate tail), then the p-value is exactly uniform on (0, 1) under H0. 3 marks; A:1, B:2 3(d). Under H0, X ∼ HGeom(12, 6, 6). Therefore P(X ≥ 5) = 6 5 6 1 + 6 6 6 0 12 6 = 37 924 ≈ 0.0400. The p-value is below 0.05, so the result is significant at the 5% level. 5 marks; A:2, C:1, D:2 3(e). Increasing α enlarges the rejection region, so the T ype I error probability increases (and equals α for an exact level-α test). Because rejection becomes easier, the T ype II error probability generally decreases for fixed alternatives, so power increases. 3 marks; A:1, B:2 Question total: 20 marks 4

MATH40005 Probability and Statistics Resit Preparation Question 4: solution and marking guide 4(a). The log-likelihood, up to an additive constant, is ℓ(θ) = 2 n log θ − θ X i xi. Thus ℓ′(θ) = 2 n/θ − P i xi, giving bθ = 2nP i xi = 2 ¯x . Since ℓ′′(θ) = −2n/θ2 < 0, this is the unique maximum. 5 marks; A:2, C:2, D:1 4(b). Minimise S(β) = X i (yi − β√xi)2. Then S′(β) = −2 X i √xi(yi − β√xi) = 0 , so, provided P i xi > 0, bβ = P i √xiyiP i xi . The second derivative is 2 P i xi > 0, hence the stationary point is the unique global minimum. 5 marks; A:1, B:1, C:2, D:1 4(c). The spread of the residuals increases withx, indicating heteroscedasticity. The constant-variance assumption is doubtful. Possible remedies include transforming the response (for example a log transformation when scientifically appropriate), using weighted least squares, or explicitly modelling the variance; the residuals should then be rechecked. 3 marks; B:2, D:1 4(d). Generate a large number B of bootstrap samples, each of size n drawn with replacement from the observed data. Compute the median for each sample, obtaining m∗ 1, . . . , m∗ B. Sort these values and take the empirical 2.5% and 97.5% quantiles. The percentile interval is q0.025(m∗), q0.975(m∗) . 4 marks; A:1, B:1, C:1, D:1 4(e). For example, one may model household electricity use (response) as a function of average outdoor temperature (ex- planatory variable). The fitted slope describes the estimated change in expected electricity use per unit change in temperature, subject to the model assumptions. Any similarly clear example earns full credit. 3 marks; A:2, B:1 Question total: 20 marks 5