GCE Further Mathematics B (MEI) Y432/01: Statistics minor Advanced GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y432/01: Statistics minor Advanced GCE Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations OCR (Oxford Cambridg... e and RSA) is 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 2021Y432/01 Mark Scheme October 2021 Annotations and abbreviations Annotation in scoris 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank page Highlighting Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only 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 indicates that the instruction In this question you must show detailed reasoning appears in the question.Y432/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 1 (a) k + 2k + 5k + 10k + 17k = 1 M1 2.4 35k = 1 so 1 k = 35 A1 [2] 1.1 AG 1 (b) The distribution has (strong) negative skew B1 [1] 1.1 1 (c) �(�) = 4 1 7 = 29 7 = 4.143… B1 1.1a BC Accept any equivalent form. Decimal answers should agree to at least 2 significant figures. ���( �) = 1 23 245 = 268 245 = 1.094… B1 [2] 1.1 BC Accept any equivalent form. 1 (d) �(�) = 10 5 7 = 75 7 = 10.714 B1FT 1.1 BC Accept any equivalent form. FT their E(X) from (c) Decimal answers should agree to at least 2 significant figures. ���( �) = 27 17 49 = 1340 49 = 27.347 B1FT [2] 1.1 BC Accept any equivalent form. FT their Var(X) from (c)Y432/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 2 (a) a is the independent variable since the values of a are not subject to random variation B1 [1] 2.4 B1: values of a are controlled B0: d is dependent on a Explanation required 2 (b) d = –1.104a + 197.1 M1 A1 [2] 3.3 1.1 For either –1.104(a) or 197.1 BC y = –1.104x + 197.1 scores M1 A0 2 (c) estimate = 130.9 (m) B1FT [1] 1.1 FT from (b) if the value is plausible from the scatter diagram. Accept 130 as rounded to 2 significant figures. 2 (d) Because this would be extrapolation and it is possible that the relationship is different for young children B1 B1 [2] 2.2b 2.4 For ‘extrapolation’ B1: a 5-year-old child may not be able to read yet B0 for comment about child not being able to drive 2 (e) Residual = 150 – (–1.104… × 40+197.1…) M1 1.1 Subtraction other way around scores M1 only = −3.0 A1FT [2] 1.1 Allow –2.9 (using 1.104 and 197.1) FT from (b) 2 (f) Because the values of a are non-random so it makes no sense to try to predict them. B1 [1] 3.2b Should show understanding of a purpose of a regression line being to make predictions.Y432/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 3 (a) The sample must be random B1 [1] 1.2 3 (b) E8: 23 29 5.5583 120 × = B1 1.1 C13: 2 (28 33.1417) 33.1417 − M1 1.1a = 0.7977 A1 [3] 1.1 3 (c) H0: No association between age and smoking (status) H1: Some association between age & smoking (status) B1 3.4 Both hypotheses needed Use of ‘correlation’ in place of ‘association’ is B0 Degrees of freedom = 2 B1 3.3 Critical value = 5.991 B1 1.1 or �22(6.4801) = 0.9608 or p-value = 0.0392 Test statistic = 3.3642 + 0.6964 + ... + 0.2792 = 6.4801 B1FT 1.1 FT their value of C13 6.4801 > 5.991 M1 2.2b or 0.9608 > 0.95 or 0.0392 < 0.05 Comparing their test and critical values leading to a conclusion. There is sufficient evidence at the 5% level to suggest that there is association between age and smoking (status) A1 [6] 3.5a Correct test and critical values required Use of ‘correlation’ in place of ‘association’ is A0 Conclusion in context 3 (d) For 16-34 year olds the contribution of 3.3642 suggests that more are smokers than would be expected. E1 2.3 Max of 2 marks out of 3 if no contributions are mentioned. Should take each age group in turn and discuss status For 35-59 year olds things are (approximately) as expected if there were no association. E1 3.5a Allow equivalent statements about non-smokers Max 2 marks if done differently For people aged 60 and over the contribution of 1.1775 suggests that fewer are smokers than would be expected. E1 [3] 3.2aY432/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 4 (a) Because (the grouping of points on) the scatter diagram appears to be very roughly elliptical, E1 3.5a For elliptical the distribution may be bivariate Normal. E1 [2] 2.4 For full answer (dep. on first mark) “the data is bivariate Normal” is E0 4 (b) 1 Sxy = − × × = 12317.2 351.9 350.0 0.7 10 M1 1.1a Detailed reasoning required. 1 2 Sxx = − × = 12384.5 351.9 1.139 10 1 2 S yy = − × = 12251.2 350.0 1.2 10 M1 1.1 For either or S S xx yy 0.7 1.139 1.2 xy xx yy S r S S = = × M1 3.3 For general form including square root = 0.60 A1 [4] 1.1 SC2 for correct value without any intermediate calculations (0.598750…) 4 (c) H0: ρ = 0 H1: ρ > 0 B1 3.3 For both hypotheses H0: no correlation in the population where ρ is the population product moment correlation coefficient between x and y B1 2.5 For defining ρ H1: positive correlation in the population For n = 10, 5% critical value (one tailed) = 0.5494 B1 3.4 For critical value scores first B1 Since 0.60 > 0.5494 the result is significant. M1 1.1 For comparison of test statistic and critical value leading to a conclusion There is sufficient evidence (at the 5% level) to suggest that there is positive correlation between directly measured and satellite measured salinity level. A1FT [5] 2.2b FT for conclusion in context 4 (d) It means that one can be more confident that the correlation is genuine, rather than simply the result of random variation. E1 [1] 2.4 4 (e) The test shows that there is almost certainly some real correlation in the population. E1 3.5a However, it is uninformative/of little use since the effect size is so small. E1 [2] 2.2bY432/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 5 (a) (i) P(X ≥ 10) = 1 – 0.9161 M1 3.1b Or P(X ≥ 10) = 1 – P(X ≤ 9) = 0.0839 A1 [2] 1.1 BC (0.083924…) 5 (a) (ii) Poisson (60) M1 3.3 soi P(X < 50) = 0.0844 A1 [2] 1.1 BC (0.084406…) 5 (b) 0.916120 M1 3.1a soi = 0.1733 A1FT [2] 1.1 or 0.1732 from calculator value in (a) 5 (c) Use of Var(X) = E(X) M1 3.1b Var(X) = E(X 2) – (E(X))2 µ = 12 – µ2 M1 2.2a For equation µ = 3 A1 1.1 P(X < 5) = 0.8153 A1 [4] 1.1 BC (0.815263…)Y432/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 6 (a) (i) Uniform distribution on {1, 2, … , n} P( ) X n ≤ = 1 1 4 4 B1 [1] 3.3 6 (a) (ii) P( ) 14 4 1 k X n k ≤ = + M1 2.2a 14 ( 1) n − n = M1 1.1 1 4 n − n = A1 1.1 Single fraction required for A1 [3] 6 (b) E(X) = 51 Var(X) = 850 M1 3.1b For either SD(X) = 29.1 So require P(21.9 < X < 80.1) M1 1.1a For required interval 59 P(22 80) ≤ ≤ = X 101 A1 [3] 1.1OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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