OCR GCE Further Mathematics A Y542/01: Statistics Advanced GCE Mark Scheme for Autumn 2021

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Mark Scheme October 2021 Question Answer Marks AO Guidance 1 (a) y = 52.7 + 0.251x B1* B1* depB1 [3] 1.1 1.1 1.1 a in range [0.250, 0.251] b correct to 3 SF Completely correct including let... ters SC: Correct formulae used for a and b: M1(A1)A1 1 (b) This quantity is minimised to find best-fit line B1 [1] 2.4 Need “minimised” or “this is its minimum value” OE 1 (c) y′ = 11.5 + 0.139x [y′ = 95 × (their a – 32) + 95 × their b] M1 A1ft [2] 1.1 1.1 Apply inverse formula at least once All correct, any letters, ft on their y 2 E(D) = 2×0.1 + 4×0.3 + 6×0.2 = 2.6 M1 A1 2.1 1.1 NB: a is not needed by this method Or change 0, 2, 4, 6 to 4, 10, 16, 22 E(D2) = 22×0.1 + 42×0.3 + 62×0.2 [= 12.4] M1 1.1 Or Σ(x – µ)2p(x) and find a Var(D) = 12.4 – 2.62 = 5.64 M1 A1 1.1 1.1 Σp2d oe gets max M1A1M0M1M1 Var(3D + 4) = 9×Var(D) = 50.76 M1 A1 [7] 3.1a 3.4 Allow even if their Var(D) < 0 SC: Σ(x – µ)2p(x): M1A1, a = 0.4 M1 M1(use this formula), A1M1A1 3 (a) (i) P(X ≥ 5) – P(X ≥ 11) = 0.74 – 0.710 M1 3.1b Allow 1 term wrong at either end Or pq4 + … + pq9 = 0.212 A1 [2] 3.4 awrt 0.212 (ii) 0.7n–1 < ⅓, or 0.103 > 0.1 > 0.072 M1 2.1 Solve 0.3 × 0.7n–1 = 0.1 or < 0.1, allow inequality error nmin = 5 A1 [2] 1.1 5 only SC: 5 without sufficient justification: B1 3 (b) 2 1 p 42 − p = ⇒ 42p2 + p – 1 = 0 M1 A1 3.1a 1.1 Equate correct variance formula to 42 Correct simplified quadratic equation p = 17 A1 2.2a Explicitly reject p = − 16 A1 2.3 SC: if – 17 and 16 , allow A1 for explicitly rejecting – 17 E(X) = 7 A1 2.1Y542/01 Mark Scheme October 2021 Question Answer Marks AO Guidance [5] 4 (a) (i) µˆ = = x 16.8 B1 [1] 1.1 Or exact equivalent (ii) 48398 2 16.8 160 − [= 20.2475] M1 1.1 If single formula used, full marks if correct; M0M1 if wrong but divisor 159 seen anywhere × 160 159 M1 1.1 = 20.3748… A1 [3] 1.1 Awrt 20.4, www 4 (b) 2 x z ± σ /160 M1 3.3 Any z from Φ–1, 160 needed, allow √ errors z = 2.576 A1 1.1 Or better, e.g. 2.575829 (15.88, 17.72) A1 [3] 3.4 Both, 4 sf required by question, www (NB: σ 2 = 20.2475 gives same end-points to 4 SF but this gets M1A1A0) 4 (c) (i) Not needed in (a) as E(X) and Var(X) are independent of the distribution B1 [1] 2.4 Mention at least one of E(X) and Var(X) explicitly, or “not relevant to X ” (ii) Needed in (b) as parent distribution not stated to be normal B1 [1] 2.4 Must make it clear that two distributions are involved. “n is large” etc: B0Y542/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 5 (a) The value of Pearson’s pmcc would be changed by (most) such changes. The value of Spearman’s rs would not be changed as the ranks remain unchanged. B1 B1 [2] 2.5 2.5 Explain effect on Pearson, or not known bivariate normal or not testing for linear correlation Explain why no effect on Spearman (not “not likely to be affected”, or “not much affected” or “association not correlation” 5 (b) H0: no association between ranks of numbers of items H1: (positive) association between ranks B1 1.1 Don’t insist on “population” here, but allow use of ρs in both, even if no explanation (not just rs). Context needed, but don’t worry about 1- or 2-tailed here Ranks 1 2 3 4 5 6 7 8 9 4 1 3 2 8 5 9 7 6 M1 1.1 Σd 2 = 38 A1 1.1 rs = 2 2 6 1 9(9 1) Σd − − M1 1.2 = 0.683 A1 1.1 < 0.700 B1 1.1 Compare TS (–1 ≤ TS ≤ 1) with 0.7, independent Do not reject H0. Insufficient evidence of association between rankings of the items M1ft A1ft [8] 1.1 2.2b ft on TS provided correct formula used, or on CV 0.600 In context, not too positive. FT on TS only SC: 0.600 (2-tailed): B0 M1A0 6 (a) H0: Data consistent with N(100, 152) H1: Data not consistent with N(100, 152) B1 [1] 1.1 Allow: “follows N(100, 152)” or “can be modelled by”. Parameters not needed. No other alternatives seen! 6 (b) P(100 ≤ X < 110) = 0.2475 BC Expected frequency = 500 × 0.2475 [= 123.754] B1 3.4 Probability needs to be seen (129 123.754)2 123.754 − [= 0.222…, AG] M1 A1 [3] 2.1 2.2a Sufficient working to justify AG, needs 123.754 at least [Show More]

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