Statistics > EXAM REVIEW > AS1101 Probability & Statistics [CT3a], May 2019: Solutions ALL ANSWERS CORRECT-GRADED A. (All)

AS1101 Probability & Statistics [CT3a], May 2019: Solutions ALL ANSWERS CORRECT-GRADED A.

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Solutions to Section A Question 1 (i) Figure 1a is intended to be a histogram, but the x-axis does not have a linear scale and the y-axis has no scale at all. It is clear from looking at figure 1a ... that frequency has been used rather than frequency density. Figure 1b is not a bad attempt at a stem-and-leaf diagram, but it should show a key and there should be a gap between the rows labelled 7 and 4. (ii) The cumulative frequency diagram should be a step function. (iii) The median is 1 hr 19 min, the LQ 22 min and the UQ 2 min 41 min. In all three cases the horizontal line at height n=4 coincides with the graph of the cumulative frequency and the quartile is the mid-point of the area of overlap. [Total: 10 marks] Question 2 (i) Venn diagram should show that B and C are disjoint, that B and D are disjoint and that D is a subset of A. (ii) B \ D is the empty event, which is a subset of everything. Since D is a subset of A, A \ D = D, and clearly C \ D must be a subset of this. (iii) (a) We are looking for P((A[B[C)c) = 1-P(A)-P(B\Ac)-P(C \Ac) = 1-0:6-0:12-0:24 = 0:04: In a 10-week term, consisting of 50 days, I expect to have 2 free days. (b) Let p = P(D \ C) = P(D \ Cc), since P(CjD) = 0:5. Then 0:36 = P(A \ C) ≥ P(C \ D) = p. In addition, p ≤ P(A \ Bc \ Cc) = 0:6 - 0:36 - 0:18 = 0:06. So the largest possible value of p is 0.06. [Total: 10 marks] Question 3 (i) (a) T has an exponential distribution with density 0:5e-0:5t. Therefore we need R23 0:5e-0:5t dt = e-1 - e-1:5 = 0:1447. (b) Let N1 be the number of empty seats in the first two hours, N2 the number in the third hour. Then N1 ∼ P(1), N2 ∼ P(0:5). We want P(N1 = 0 \ N2 > 0) = P(N1 = 0) × P(N2 > 0) = e-1(1 - e-0:5) = 0:1447: AS1101 Probability & Statistics [CT3a], May 2015: Solutions (ii) (a) The two properties F(x) ! 0 as x ! -1 and F(x) ! 1 as x ! 1 are both obvious. We need to find c such that 1-c c2 = 1: it is obvious that c0 = 12. Also, we require F to be non-decreasing: f(x) = dF dx = 2x (1 - x)2 + 2x2 (1 - x)3 = 2x (1 - x)3 > 0 for 0 < x < 1 2 (b) We have E((1 - X)3) = Z0 1=2 (1 -2xx)3 · (1 - x)3 dx = 1 4: [Total: 10 marks] Question 4 Denote by X and Y the number of heads in Andrew and Bob original flips. Therefore A = (X | if X ≥ 2 Y | if X <2 | and | B = ( Y | if X ≥ 2 X | if X <2: Also, note that P(X = i) = P(Y = i) = 1=8 for i = 0; 3 and = 3=8 if i = 1; 2. (i) We have P(A = 3 \ B = 3) = =P(A = 3 \ B = 3 \ (X ≥ 2 [ X < 2)) =P(A = 3 \ B = 3 \ X ≥ 2) + P(A = 3 \ B = 3 \ X < 2) =P(X = 3 \ Y = 3 \ X ≥ 2) + P(X = 3 \ Y = 3 \ X < 2) =P(X = 3 \ Y = 3) + 0 =P(X = 3) × P(Y = [Show More]

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