GCE Further Mathematics B (MEI) Y422/01: Statistics major Advanced GCE Mark Scheme for Autumn 2021

Document Content and Description Below

Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y422/01: Statistics major 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 2021Y422/01 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.Y422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 1 (a) 34.711 B1 1.1 ± 1.96 M1 3.3 1.53 50 × M1 1.1 = 34.711 ± 0.424 or (34.287, 35.135) A1 3.4 Allow 34.29 to 35.13 or 35.14 [4] 1 (b) 50 is a sufficiently large sample to apply the CLT which states that for large samples the distribution of the sample mean is approximately Normal B1* *B1 [2] 2.2b 2.4 For mention of central limit theorem For full statement (including CLT) No credit if CLT not mentionedY422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 2 (a) 6 1 1 P( 0) X = = × × 6 6 6 M1 3.1a Allow M1 for 1 1 1 6 6 36 × = 1 = 36 A1 1.1 AG [2] 2 (b) B1 B1 [2] 1.1 1.1 For heights For axes and labels Roughly correct but must have linear scale Do not allow just P on vertical axis 2 (c) The distribution has (slight) negative skew B1 [1] 1.1 Allow ‘roughly symmetrical’ or ‘unimodal’ Not ‘Normal distribution’ 2 (d) DR 1 2 1 2 5 5 E( ) 0 1 2 3 4 5 X = × + × + × + × + × + × 36 36 9 4 9 36 M1 1.1a 105 35 = = = 36 12 2.9166 A1 1.1 Allow fraction or decimal form 2 2 2 2 2 2 2 1 2 1 2 5 5 E( ) 0 1 2 3 4 5 X = × + × + × + × + × + × 36 36 9 4 9 36 371 = = 36 10.3055 M1 1.1 2 Var( ) 10.3055 (2.9166 ) X = −   M1 1.2 259 = = 144 1.80 (1.7986…) A1 1.1 [5] 2 (e) Variance 30 1.7986 1619 = × = 2  (pence2) B1 1.1 [1] 2 (f) Average amount received = 30 × 2.916… = 87.5 B1 3.1a k – 87.5 = 12.5 ⇒ k = 100 B1 1.1 [2] 0.00 0.10 0.20 0.30 0 1 2 3 4 5 P(X) = r rY422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 3 (a) Using B(50, 0.04) M1 3.3 P(X = 2) = 0.276 A1 1.1 BC [2] 3 (b) 0.96 0.04 0.0277 9 × = B1 1.1 Allow 0.028 [1] 3 (c) 0.96 0.442 20 = B1 1.1 [1] 3 (d) Expected value for one misunderstood = 1 25 0.04 = B1 2.1 Must quote probabilities to get full marks Because geometric E1 2.4 For 3 misunderstood expected number = 25 + 25 + 25 = 75 E1 1.1 [3] 3 (e) Require P(2 misunderstood in first 59) × 0.04 B1 3.1a For identifying required probability so using B(59, 0.04) gives P(X = 2) = 0.267 M1 2.2a Use of correct binomial 0.267 × 0.04 = 0.0107 A1 1.1 BC [3] 4 (a) Nuclei decay randomly and decays are independent with constant probability 200 000 1 The number of decays out of 1 000 000 is being counted, so a binomial distribution is appropriate Because n = 1 000 000 is large and p = 200 000 1 is small a Poisson distribution is also appropriate E1 E1 E1 [3] 2.4 2.4 2.4 For partial explanation of binomial For full explanation For explanation of Poisson 4 (b) Po(5) M1 3.3 P( 6) 0.146 X = = A1 1.1 BC P( 6) 1 0.762 0.238 X > = − = A1 [3] 1.1 BC 4 (c) Mean = 10 × 5 = 50 P(at least 60 decays) 1 0.9077 0.0923 = − = B1 B1 [2] 3.3 1.1 BC Allow 0.092Y422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 5 (a) Two A and one B ⁓ N(2 × 3.9 + 7.8, 2 × 0.322 + 0.412) B1 3.3 For N and mean Allow if N stated anywhere in answer SOI N(15.6, 0.3729) M1 1.1 For variance P(≥ 16) = 0. 256 (0.25622…) A1 3.4 BC [3] 5 (b) Four B – one C ⁓ N(4×7.8 – 30.2, 4×0.412 + 0.642) B1 3.3 For N and mean Allow -1 for mean Allow if N stated anywhere in answer SOI N(1, 1.082) M1 1.1 For variance P(within 1 unit) = 0.473 (0.47274…) A1 3.4 BC [3] 5 (c) DR H0: μ = 30.2 H1: μ ≠ 30.2 B1 3.3 Hypotheses in words only must include “population” where μ is the population mean capacitance B1 1.2 For definition in context Sample mean = 29.96 B1 1.1 Est. population variance = − 1 299.6 9(8981.0 10 2 ) M1 1.1 = 0.5538 A1 1.1 Or sd = 0.7442 Test statistic 29.96 30.2 0.5538 10 − = M1 3.3 FT their mean and/or sd = −1.020 A1 1.1 BC Refer to t9 M1 3.4 No FT if not t9 Critical value (2-tailed) at 5% level is 2.262 A1 1.1 Or P(t < ‒1.020) = 0.1672 −1.020 > –2.262 so not significant (do not reject H0) M1 2.2b Or 1.020 < 2.262 Or 0.1672 > 0.025 Insufficient evidence to suggest that the capacitance of the batch is different from 30.2 E1 3.5a Answer must be in context [11]Y422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 6 (a) Mean = 1.725 B1 1.1 Or 345 200 Variance = 1.768 B1 1.1 Condone 1.759 (using divisor n) The variance is reasonably close to the mean so this does support the suitability of a Poisson model E1 2.2b Dep on mean and variance correct [3] 6 (b) Cell C3 = 0.3106 B1 3.4 Do not allow 0.311 Cell D3 = 62.1124 B1FT 2.2a 200 × their C3 (62.12 if use 0.3106) Allow 62.2 from 0.311 Cell E3 = 2 (65 62.1224) 62.1224 − M1FT 1.1a Must show working to get M1 = 0.1342 A1 1.1 Allow 0.126 from 62.2 [4] 6 (c) Because otherwise some expected frequencies would be less than 5 so too small for the test to be valid E1 [1] 3.5b For ‘less than 5 so invalid’ 6 (d) H0: Poisson model is a good fit H1: Poisson model is not a good fit B1 2.5 X 2 = 2.43 B1FT 1.1 FT Their value of E3 Refer to 2 χ5 B1 3.4 For degrees of freedom = 5 soi Critical value at 5% level = 11.07 B1 1.1 2.43 < 11.07 so result is not significant M1 1.1 For comparison with critical value Allow M1 (not A1) for comparison with any chi squared critical value eg 1.145 or 5.991 There is insufficient evidence to suggest that the Po(1.7) model is not a good fit. A1 2.2b Conclusion in context [6]Y422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 7 (a) The pairing will eliminate any differences in grip strengths between different people and so will only compare the grip strengths of the dominant and nondominant hands E1 E1 [2] 2.2b 2.2b Give 1 mark for any valid comment For 2 marks must include pairing 7 (b) The parent population of differences must be Normally distributed E1 E1 [2] 1.1 1.2 For Normally distributed For full answer including ‘differences’ 7 (c) It does because the confidence interval contains 2 E1 [1] 3.5a 7 (d) (i) Sample mean difference = 2.39 B1 1.1 SD 0.45 1.96 100 = × M1 3.1b Sample SD = 2.30 (2.2959…) A1 1.1 [3] 7 (d) (ii) The sample must be random B1 3.2b since only a random sample enables proper inference about the population to be undertaken B1 2.4 Do not allow eg a random sample is less likely to be biased [2]Y422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 8 (a) (i) Predicted = 50.5 B1 1.1 Do not allow answer to more than 2dp [1] 8 (a) (ii) Although this point lies within the data (interpolation), the points do not lie too close to the line and the value of r2 is not too close to 1 so the estimate is only moderately reliable B1 B1 2.2a 3.5b Mention of 1 of the three points Mention of at least 2 points with correct conclusion [2] 8 (a) (iii) Coordinates (47.3, 48.7) B1 1.1 [1] 8 (a) (iv) This is the point with coordinates which are the means of the x- and y-values respectively B1 [1] 1.1 Allow ‘This is the centroid’ 8 (b) (i) The scatter diagram is very roughly elliptical and E1 3.5a so the distribution may be bivariate Normal E1 2.4 [2] 8 (b) (ii) 1 Svt = − × × 3886.53 80.37 970.86 20 (= −14.87…) M1 1.1a Numerical evaluations are not required at this stage 1 2 Stt = − × 324.71 80.37 20 (= 1.743…) M1 1.1 For either Stt or Svv 1 2 Svv = − × 47829.24 970.86 20 (= 700.78…) 14.87 1.743 700.78 tv tt vv S r S S − = = × M1 3.3 For general form including sq. root = −0.4255 A1 1.1 BC [4] 8 (b) (iii) H0: ρ = 0, H1: ρ < 0 B1 3.3 For both hypotheses Do not allow r in place of ρ Hypotheses in words only get B1 unless population mentioned where ρ is the population pmcc between t and v B1 2.5 For defining ρ For n = 20, the 5% critical value is 0.3783 B1 3.4 For correct critical value Since |−0.4255| > 0.3783 the result is significant, so there is sufficient evidence to reject H0 M1 1.1 For comparison and conclusion Allow −0.4255 < −0.3783 There is sufficient evidence at the 5% level to suggest that there is negative correlation between marathon time and VO2max A1FT 2.2b FT for conclusion in words Answer must be in context [5]Y422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 9 (a) ( ) 12 12 ( 1) P 2 1 n X n n + > = + M1 M1 3.1a 1.1 For correct denominator For correct numerator 1 2(2 1) n + n = + A1 1.1 [3] 9 (b) (2n + 1) values so Var( ) (2 1) 1 X n = + − 12 1 [ 2 ] M1 3.1a Var of sum of 10 values = × + − 10 (2 1) 1 12 1 [ ] n 2 M1 1.1 Allow M1 for 10 attempt at variance × any 10 10 2 = + 3 3 n n A1 1.1 [3] 10 (a) 104 P( 56) 0.208 500 T ≤ = = B1 1.1 253 P( 61) 1 0.494 500 T > = − = B1 1.1 [2] 10 (b) E(T) = 25 + 28 + 5 + 3 = 61 1 1 2 2 Var( ) 10 6 4 16 T = × + × + + 12 12 94 = 3 (= 31.333) B1 M1 A1 3.1a 1.1 1.1 W ⁓ N(61, 31.333) so P(W ≤ 56) = 0.186 P(W > 61) = 0.5 B1 B1 [5] 3.3 1.1 BC 10 (c) Because the mean is 61 and both the uniform and Normal distributions are symmetrical so you would expect the simulated probability to be very close to 0.5 E1 E1 [2] 2.2b 2.4 For second mark must mention symmetricalY422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 11 (a) 2 3 2 2 0 2 F(3) 1 d (3 ) d 1 = ⇒ + − = ∫ ∫ ax x b x x M1 3.1a 8 1 ⇒ + = 3 3 a b 1 A1 1.1 2 3 3 2 0 2 E( ) 2 d (3 ) d 2 X ax x bx x x = ⇒ + − = ∫ ∫ M1 3.1a 34 ⇒ + = 4 2 a b A1 1.1 18 a b = = , 2 A1 1.1 [5] 11 (b) 2 2 0 1 1 F(2) d = = ∫ 8 3 x x B1 3.1a 2 2 16 2(3 ) d m ⇒ − = ∫ x x M1 2.2a 2 2 1 3 ⇒ − − + = 3 3 6 (3 ) m 3 3 ⇒ − = ⇒ = (3 ) m m 4 2.09 (2.0914…) A1 1.1 Or m = − 3 3 3 4 [3] 11 (c) Using N 2, ( ) 0.2 50 M1 3.1a For use of Normal distribution N(2, 0.004) M1 1.1a For correct values Estimate P(Mean < 1.9) = 0.0569 A1 1.1 [3]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

Last updated: 3 years ago

Preview 1 out of 13 pages