Bayes intervention, counting and birthday weeks
MATH40005 Probability and Statistics Resit Preparation
Question 1
1(a). Describe a probability space for a passenger choosing one of three routes A, B, C and then being
either on time or late. (3 marks)
1(b)(i). Passenger route proportions are 0.5, 0.3, 0.2 for A, B, C, and late rates are 0.04, 0.10, 0.05. Find
the probability a randomly selected passenger is late. (5 marks)
1(b)(ii). Given that the passenger is late, find the probability they used route A. (3 marks)
1(b)(iii). A disruption makes every train on routes B and C late, while route A retains late rate 0.04.
Recompute P (A|L). (4 marks)
1(c). How many unordered pairs can be drawn with replacement from eight symbols? (1 marks)
1(d). Assuming independent uniformly distributed birthday weeks, find the probability that at least two
of 12 people share a birthday week. (4 marks)
T otal: 20 marks
17
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 1: solution and marking guide
1(a). One may take Ω = {A, B, C} × {O, L}, F = P(Ω), with probabilities assigned from route propor-
tions and conditional late rates so that P (r, L) = P (r)P (L|r).
3 marks; A:3
1(b)(i). By total probability, P (L) = 0 .5(0.04) + 0.3(0.10) + 0.2(0.05) = 0 .06.
5 marks; A:5
1(b)(ii). P (A|L) = 0 .5(0.04)/0.06 = 1 /3.
3 marks; B:3
1(b)(iii). Now P (L) = 0 .5(0.04) + 0.3 + 0.2 = 0 .52, so P (A|L) = 0 .02/0.52 = 1 /26.
4 marks; D:4
1(c).
8+2−1
2
=
9
2
= 36.
1 marks; B:1
1(d). 1 − (52)12/5212, where (52)12 = 52 · 51 · · · 41.
4 marks; C:4
Question total: 20 marks
20
Joint density, conditional expectation and covariance
MATH40005 Probability and Statistics Resit Preparation
Question 2
2(a). Let f (x, y) = cx on 0 < x < y < 1. Find c. (2 marks)
2(b). Find the marginal densities. (4 marks)
2(c). Find E[Y |X = x]. (4 marks)
2(d). Compute Cov(X, Y ). (5 marks)
2(e). Let R = X/Y . Find its CDF. (5 marks)
T otal: 20 marks
18
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 2: solution and marking guide
2(a). 1 = c
R 1
0 x(1 − x)dx = c/6, so c = 6.
2 marks; A:2
2(b). fX (x) = 6 x(1 − x) for 0 < x < 1; fY (y) =
R y
0 6x dx = 3y2 for 0 < y < 1.
4 marks; A:4
2(c). fY |X=x(y) = 1 /(1 − x) for x < y < 1, so it is uniform on (x, 1) and E[Y |X = x] = ( x + 1)/2.
4 marks; B:2, C:2
2(d). E[X] = 1 /2, E[Y ] = 3 /4, and E[XY ] =
R 1
0
R 1
x 6x2y dy dx = 2 /5. Thus Cov(X, Y ) = 2 /5 − 3/8 =
1/40.
5 marks; A:2, B:2, D:1
2(e). Since 0 < R < 1, for 0 ≤ r ≤ 1, P (R ≤ r) =
R 1
0
R ry
0 6x dx dy = r2. Hence FR(r) = 0 for r < 0, r2
for 0 ≤ r ≤ 1, and 1 for r ≥ 1.
5 marks; B:1, C:1, D:3
Question total: 20 marks
21
Variance inference, multiple testing and p-values
MATH40005 Probability and Statistics Resit Preparation
Question 3
3(a). Ten normal observations have s2 = 9. Find a 95% confidence interval for σ2. Use χ2
9,0.025 = 2.700
and χ2
9,0.975 = 19.023. (4 marks)
3(b). The same sample has ¯x = 51 and s = 3 . Test H0 : µ = 50 versus H1 : µ ̸= 50 and give a p-value.
(5 marks)
3(c). Fifty hypotheses are tested at family-wise level 0.05. What Bonferroni threshold is used, and which
of 0.0007, 0.0012, 0.02 are significant? (4 marks)
3(d). Let T have a continuous CDF F0 under H0. Prove that the upper-tail p-value P = 1 − F0(T ) is
uniform under H0. (4 marks)
3(e). Explain how increasing the significance level affects Type I error and usually affects Type II error.
(3 marks)
T otal: 20 marks
19
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 3: solution and marking guide
3(a). The interval is ((9)(9)/19.023, (9)(9)/2.700) ≈ (4.26, 30.00).
4 marks; A:4
3(b). t = (51 − 50)/(3/
√
10) = 1 .054 with 9 df. The two-sided p-value is about 0.320, so fail to reject at
0.05.
5 marks; A:3, C:1, D:1
3(c). Threshold 0.001. Only 0.0007 is significant.
4 marks; B:2, C:1, D:1
3(d). By the probability integral transform, U = F0(T ) ∼ U (0, 1). For 0 ≤ u ≤ 1,
Pr(P ≤ u) = Pr(1 − U ≤ u) = Pr( U ≥ 1 − u) = u.
Hence P = 1 − F0(T ) ∼ U (0, 1) under H0.
4 marks; B:2, D:2
3(e). The Type I error probability increases because the rejection region expands. For a fixed alternative
and test statistic, Type II error usually decreases, so power increases.
3 marks; A:1, B:1, C:1
Question total: 20 marks
22
Least squares, diagnostics, bootstrap and MLE
MATH40005 Probability and Statistics Resit Preparation
Question 4
4(a). For yi = βx3
i + ei, derive the least-squares estimate of β. (5 marks)
4(b). Interpret the residual plot and suggest a model change.
0 2 4 6 8 10 12 14 16 18 20
−1
−0.5
0
0.5
xi
Residual ˆεi
(3 marks)
4(c). For data {5, 8, 11, 14, 20}, classify {5, 5, 11, 20, 14} and {5, 8, 11, 14} as bootstrap samples or not.
(4 marks)
4(d). Independent observations are uniform on [θ, θ + 2]. Find the MLE set. (5 marks)
4(e). Explain why similar-looking time series can have high correlation without any direct relationship.
(3 marks)
T otal: 20 marks
20
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 4: solution and marking guide
4(a). Differentiate P(yi − bx3
i )2: ˆβ = P x3
i yi/ P x6
i .
5 marks; A:2, B:2, D:1
4(b). The U-shape shows missing curvature. Add a quadratic term or transform the variables; high R2
would not rescue the misspecified model.
3 marks; C:2, D:1
4(c). The first is a bootstrap sample: same size and all values come from the data, with repetition. The
second is not because its size is four rather than five.
4 marks; B:4
4(d). The feasible set is [max xi − 2, min xi]. The likelihood is constant on this set, so every feasible
value is an MLE.
5 marks; A:3, D:2
4(e). A shared time trend or external driver can create correlation. Detrending, domain knowledge and
a causal design are needed before claiming influence.
3 marks; A:3
Question total: 20 marks
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