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

Document Content and Description Below

Oxford Cambridge and RSA Examinations GCE Further Mathematics A Y542/01: Statistics Advanced GCE Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge and RSA) i... s a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2021Y542/01 Mark Scheme October 2021 Annotations and abbreviations Annotation in RM assessor Meaning and  BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Other abbreviations in mark scheme Meaning dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR This question included the instruction: In this question you must show detailed reasoning.Y542/01 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 letters 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 leastY542/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 6 (c) ΣX 2 = 10.5 χ2(4) = 9.488 and 10.5 > 9.488 Reject H0. Significant evidence that data is not consistent with N(100, 152). B1 B1 M1ft A1ft [4] 1.1 1.1 1.1 2.2b Like-with-like comparison needed FT on TS or CV here. Needn’t be stated if next line right FT on TS (but not CV) if method correct. Wrong CV, e.g. 5.991: B1B0M1A0. No ft on H0/H1 6 (d) (i) E.g. Too few in X ≥ 110 or in X ≤ 80, or too many in others, or data truncated, etc B1 [1] 3.5b Any relevant point, needn’t refer to values of X 2 “Divide into 5 minute groups”: B1. “Data discrete”: B0. “The variance” (uncalculated): B0 (ii) B1 B1 [2] 3.3 3.5c Deal with aspect identified in (i) Basically correct, areas roughly same Examples: Uses “data discrete” in (i) More below 100, so translate to left More above 110 so translate to right Divide into 5-minute groups Variance changed, areas not equal Data truncated but worse truncation shown B0 B2 B2 B0 B1 B0 7 (a) H0: Two samples are from identical populations H1: Two samples are from populations with different median ratings. Rm = 1 + 2 + 3 + 4 + 5 + 9 + 10 + 11 (= 45) W = 45 8(8 + 8 + 1) – Rm = 91 Wcrit = 49 Reject H0. Significant evidence that there is a difference in median ratings/opinions have changed B2 M1 A1 B1 B1 M1ft A1ft [8] 1.1 1.1 1.1 1.1 2.1 1.1 1.1 2.2b If no reference to “populations”, maximum B1 Allow H0: “identical population medians”, H1: “not identical populations” or “not identical pop medians” “Pupils’ opinions have not changed”, etc: B2 If omitted, can still get all other marks FT on TS (< 68) or CV FT on TS only. Allow “increased” SC: Sign or paired-sample test, max B2 (hypotheses) 7 (b) Eliminate the difference between individual pupils’ opinions B1 [1] 3.5b “Minimises the difference in tastes” B1 (BOD) Scores arbitrary: B1 (etc). Not “more powerful test”. 7 (c) A paired-sample signed-rank test would have been used B1 3.5c Must mention “paired sample” oe – not just “Wilcoxon Black = PAB version, red = candidate’s versionY542/01 Mark Scheme October 2021 Question Answer Marks AO Guidance [1] signed rank” 7 (d) 0.025 × 12870 M1 3.1a 0.05 × 12870 = 643.5 M1 = 322 A1 [2] 3.2a 321 or 322 or 643 (from 1-tail), must be integerY542/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 8 (a) f(x) = ½ 2 12 0 ∫ a ax x cos( )d = 0.3 2 12 0     sin( ) ax 12 sin(2 ) 0.3 a = a = 0.32175… B1 M1 B1 M1 A1 [5] 3.3 3.1a 1.1 2.1 1.1 Stated or implied, e.g. on diagram ∫f(x) a cos ax dx & equated to 0.3 Correct indefinite integral Correct limits, solve Answer, a.r.t. 0.322 (ignore other answers) 8 (b) F(y) = ½ y [0 ≤ y ≤ 2] M1 A1 3.1a 1.1 Use their f(y) to obtain CDF Correct F(y) (range need not be stated explicitly) P(Y 2 ≤ m) = P(0 < Y ≤ √m) M1 2.1 Find CDF of Y 2, allow m2 instead of √m , or ±√m, here = F(√m) [= ½√m] A1 1.1 Use F(y) correctly ½√P60 = 0.6 M1 1.1 Equate to 0.6 and solve, need √m here P60 = 1.44 A1 [6] 2.2a 1.44 or exact equivalentOCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre d [Show More]

Last updated: 1 year ago

Preview 1 out of 10 pages