GCE Mathematics A H240/02: Pure Mathematics and Statistics Advanced GCE Mark Scheme for November 2020

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GCE Mathematics A H240/02: Pure Mathematics and Statistics Advanced GCE Mark Scheme for November 2020 Oxford Cambridge and RSA Examinations GCE Mathematics A H240/02: Pure Mathematics and Stat ... istics Advanced GCE Mark Scheme for November 2020Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge 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 2020H240/02 Mark Scheme November 2020 2 Text Instructions 1. 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.H240/02 Mark Scheme November 2020 3 2. Subject-specific Marking Instructions for A Level Mathematics A a Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^) is sufficient, but not required. For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. Award NR (No Response) - if there is nothing written at all in the answer space and no attempt elsewhere in the script - OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’) - OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question. Note: Award 0 marks only for an attempt that earns no credit (including copying out the question). If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt unless it is ambiguous for which part it is intended. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. If you are in any doubt whatsoever you should contact your Team Leader.H240/02 Mark Scheme November 2020 4 c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. e The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.H240/02 Mark Scheme November 2020 5 • When a value is not given in the paper accept any answer that agrees with the correct value to 3 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. NB for Specification B (MEI) the rubric is not specific about the level of accuracy required, so this statement reads “2 s.f”. Follow through should be used so that only one mark in any question is lost for each distinct accuracy error. Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. g Rules for replaced work and multiple attempts: • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others. • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete. • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the command words “Show” or “Determine”. Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. j If in any case the scheme operates with considerable unfairness consult your Team Leader.H240/02 Mark Scheme November 2020 6 Question Answer Mark AO Guidance 1 (a) (i) 14(2x + 3)6 oe M1 1.1 M1 for k(2x + 3)6 or 14u6 seen A1 1.1 A1 all correct ISW [2] 1 (a) (ii) 3x2ln x + x2 oe M1 1.1 Attempt use product rule; allow incorrect sign; allow lnx 3x2 + x2 A1 1.1 One term correct A1 1.1 All correct ISW eg factorised incorrectly [3] 1 (b) 15 sin 5x (+ c) M1 1.1 M1 for sin5x seen NOT –5sin5x A1 1.1 A1 all correct. Allow without "+ c" [2] 1 (c) (y =) 3x2 – 5x + 5 M1 1.1 M1 for attempt integrate 6x – 5 A1 1.1 A1 all correct, including y = [2] 2 Numerator ≡ (x + 1)(x – 2)(2x + 3) M1 3.1a Attempt factorise numerator into 3 linear factors Denominator ≡ (x + 1)(x – 2) M1 1.1 Attempt factorise denominator into 2 linear factors M1 1.1 "cancel" two common factors in num & denom Ans: 2x + 3 A1 1.1 Allow no mention of x ≠ −1or x ≠ 2 conditions. NB correct answer with no working or partial working: 4 marks [4] SC: Answer x + 3 2 B3 Alternative method 2 3 2 3 2 2 2 2 3 2 2 7 6 2 2 4 3 3 6 3 3 6 x x x x x x x x x x x x x + − − + − − − − − − − − − 2x + 3 M1 A1 A1 A1 Attempt long division by x2 – x – 2 or by x + 1 or by x – 2 Obtain "2x" in quotient Obtain "+ 3" in quotient Answer 2x + 3 clear (not just in the division sum)H240/02 Mark Scheme November 2020 7 Question Answer Mark AO Guidance 3 (a) (i) 1 + (-2)(-x) + ( 2)( 3) − − 2! (−x)2 + ( 2)( 3)( 4) − − − 3! (−x)3 M1 1.1 Correct expressions for at least three terms. May be implied ≡ 1 + 2x + 3x2 + 4x3 A1 1.1 cao [2] 3 (a) (ii) (n + 1) xn B1 2.2a Allow xn = (n + 1) xn [1] 3 (b) 1−1x oe B1 1.1 [1] 3 (c) 2 + 3x + 4x2 + 5x3 + ... = 1 + x + x2 + x3 + ... + 1 + 2x + 3x2 + 4x3 + ... M1 3.1a = 1 1−x + 1 2 (1 ) −x = ( ) 2 1 1 (1 ) x x − + − M1 3.1a Their (b)(i) + 1 2 (1 ) −x and attempt single term = 2 2 (1 ) x − x − A1 1.1 cao Unsupported answer, no marks [3] (a – x)(1 – x)–2 M1 a + 2ax + 3ax2 + 4ax3 + ... – (x + 2x2 + 3x3 + 4x4 + .....) M1 a = 2 2 2 (1 ) x − x − Justification for all terms up to infinity A1 NB other correct methods existH240/02 Mark Scheme November 2020 8 Question Answer Mark AO Guidance 4 DR 3sin4 ϕ + sin2 ϕ – 4 = 0 (3sin2 ϕ + 4)(sin2 ϕ – 1) = 0 B1 2.1 Attempt to solve QE in sin2 ϕ or QE in u with u = sin2 ϕ soi Must see method sin2 ϕ = – 4 3 or sin2 ϕ = 1 (or sin ϕ = 1) B1 1.1 May be implied from x = sin2 ϕ and x = – 4 3 or 1 sin2 ϕ = – 4 3 is impossible B1 2.3 oe, eg sin ϕ ≠ − 4 3 Not with incorrect reason, eg sin2 ϕ = 16 9 >1 ϕ = sin–1 (±1) M1 1.1 solve for ϕ Allow ϕ = sin–1 (1), may be implied ϕ = 12 2 π π , 3 No extras within range Allow "correct" extras outside range A1 2.2a Both. dep sin2 ϕ = – 4 3 and sin2 ϕ = 1 (or sin ϕ = 1) seen [5] SC ϕ = 12 2 π π , 3 with no working: B2 5 (a) n2 – 1 or n2 + 1 is even OR n2 is odd or n2 = 2k + 1 (k integer) OR 2 1 2 n − > 0 or 2 1 2 n − > 1 oe eg n2 > 3 B1 2.4 B1 for any of these. Numerical examples insufficient Ignore extra, eg 2 1 2 n + > 0 ⇒ n2 > –1 or n > −1 or n ≠ –1 Allow > 0 for this mark Assuming n is a positive integer: Not assuming n is a positive integer: n is odd oe eg n = 2k + 1 (k integer) B1 2.2a n >1 (or n<–1) or |n|>1 or n > 3 Not n > 0 B1 2.2a n = odd integers > 1 or n = 3 , 5 etc oe B2 indep NOT n > ±1 but ignore this if followed by correct, eg |n| > 1 2nd and 3rd B1 marks are independent & can be gained without explanation [3] 5 (b) n2 + ( 2 1 2 n − )2 M1 3.1a ( 2 1 2 n + )2 – ( 2 1 2 n − )2 correct expression = n2 + 4 2 2 1 4 n n − + = 4 2 2 1 4 n n + + = 4 2 2 1 4 n n + + – 4 2 2 1 4 n n − + = 2 4 n4 = 2 2 1 2   n +     A1 1.1 = n2 Correctly obtained [2]H240/02 Mark Scheme November 2020 9 Question Answer Mark AO Guidance 5 (b) ctd n2 + ( 2 1 2 n − )2 = 2 1 2 2   n +     M1 n2 + 4 2 2 1 4 n n − + = 4 2 2 1 4 n n + + 2 4 2 4 2 1 4 n n n + − + = 4 2 2 1 4 n n + + A1 6 LHS ≡ 2 (cos 2θ cos 45o – sin 2θ sin 45o) M1 3.1a correct use of cos(A+ B) formula ≡ 2 × 1 2 (cos 2θ–sin 2θ) or (cos 2θ–sin 2θ) B1 1.1 cos 45o or sin 45o = 1 2 seen or implied ≡ cos2 θ – 2sin θ cos θ – sin2 θ A1 2.2a correctly obtained – use of double angle formulae clear Alternative method RHS ≡ cos 2θ – sin 2θ M1 ≡ 1 1 2 2 2( cos 2 sin 2 ) θ θ − or Rcos α = 1, Rsin α = 1, R2 = 2, tan α = 1, α = 45o ≡ 2 (cos 2θ cos 45o – sin 2θ sin 45o) M1 ≡ 2 cos(2θ + 45o) A1 [3] 7 all Allow a and b without "squiggles" beneath 7 (a) Length of AB oe B1 1.2 Magnitude of AB  or distance from A to B Allow Magnitude of AB Not magnitude of |a – b| or magnitude of a – b Not distance from a to b Not distance from position vector A to position vector B [1] 7 (b) Midpoint of AB oe B1 1.2 or Halfway between A and B Allow Midpoint of AB  Must refer to A and B, not a and b Not Midpoint of the vectors [1] 7 (c) (i) 12 ( ) a b + B1 2.2a [1] 7 (c) (ii) 12 | | a b − oe B1 2.2a [1]H240/02 Mark Scheme November 2020 10 Question Answer Mark AO Guidance 7 (d) Centre is (3, 2) B1 1.1 Allow this mark for (3, 2) or 3 2       or 1 2 6 4       oe seen May be implied by answer r 2 = 10 or r = √10 or 3.16 (3 sf) B1 1.1 May be implied by answer. Must imply radius M1 1.1 M1 for (x – a)2+(y – b)2 = r2 for any non-zero numerical a, b and r (x – 3)2 + (y – 2)2 = 10 A1 1.1 A1 for all correct. ISW [4] 8 (a) Summary scheme Attempt separate variables using (100 – P) M1 3.1a Correct integral, but allowing |100 – P| or (P – 100) or (100 – P) A1 1.1 Allow without + c Attempt t = 0, P = 2000 to find c or A or e±c M1 3.4 dep M1 c = –ln 1900 or A = 1900 or e±c = 1900 OR Allow c = ln1900 or –ln(–1900) or A or e±c = –1900 or 1 − 1900 A1 3.4 dep M1M1 Attempt make P the subject M1 3.4 dep M1M1A1 Correct use of mod & change to P – 100 M1 2.1 P = 1900e–t + 100 A1 1.1 dep M1A1M1A1M1M1 ie dep all correct working seen Examples of correct methods d 100 P −P = dt M1 –ln|100 – P| = t + c or |100 – P| = Ae–t A1 Substitute t = 0, P = 2000 M1 ⇒ c = –ln 1900 or A = 1900 A1 ln |100 | 1900 −P = –t or |100 – P| = 1900e–t 100 1900 P− = e–t P = 1900e–t + 100 M1 A1 A1H240/02 Mark Scheme November 2020 11 Question Answer Mark AO Guidance 8 (a) ctd P−d100 P = –dt M1 M1 ln(P – 100) = –t + c or P – 100 = Ae–t A1 Substitute t = 0, P = 2000 M1 ⇒ c = ln 1900 correct or A = 1900 A1 ln(P – 100)= –t+ln 1900 or P – 100=1900e–t 100 1900 P− = e–t P = 1900e–t + 100 M1 A1 Example of incorrect methods ( d 100 P −P = dt M1 –ln(100 – P) = t + c or 100 – P = Ae–t A1 100 – P = e–t – c Substitute t = 0, P = 2000 M1 ⇒ e–c = –1900 or A = –1900 A1 100 – P = –1900e–t oe No change to P – 100 M0 P = 1900e–t + 100 M1 A0 Correct answer but incorrectly obtained, not using modulus d 100 P −P = dt M1 ln(100 – P) = t + c A0 Substitute t = 0, P = 2000 M1 ln(–1900) = c A1 ln(100–P) = t + ln (–1900) 100 – P = –1900et No change to P – 100 M0 P = 100 + 1900et M1 A0H240/02 Mark Scheme November 2020 12 Question Answer Mark AO Guidance 8 (b) (Starts at 2000) Decreases B1f 3.4 B1 for correct process or ft (a) dep (a) includes exponential Approaches 100 B1f 3.4 B1 for correct limit or ft (a) dep (a) includes exponential [2] 9 (a) (40000 × 0.002 =) 80 B1 1.1 [1] 9 (b) Frequency per £ or No. of cars per £ B1 1.1 Allow cars / £ Allow as fraction NOT cars / price NOT cars / money [1] 9 (c) Show fds for four separate classes within £50000 - £90000 B1 2.3 Make price intervals smaller Show the median (or similar) Include more bars Give a title Explain y-axis better or give units of f.d. Give the frequencies [1] 9 (d) 20 B1f 1.1 or (a) ÷ 4 [1]H240/02 Mark Scheme November 2020 13 Question Answer Mark AO Guidance 10 Allow rounded or truncated to 2 sf throughout H0: p = 0.9, where p = P(a random customer is satisfied) B1 1.1 or p is proportion satisfied Allow other letters OR p = 90%, where p is % of customers satisfied H1: p < 0.9 B1 2.5 Subtract B1 for each error eg: 2-tail B1B0 Use of < with definition B1B0 undefined p B1B0 Not include value 0.9 B0B0 not in terms of parameter B1B0 H0 = 0.9 etc: B0B0 X~Bin(15, 0.9) and X < 10 or 11 or 12 (condone < or = or > or >) M1 3.3 Stated or implied eg by 0.0556 or 0.184 or 0.0127 or 0.944 or 0.816 or 0.987 or 0.0428 or 0.129 P(X<11) oe = 0.0556 P(X>11) oe = 0.944 A1 3.4 BC cao Comp 0.05 Comp 0.95 A1 1.1 Dep 0.0556 or 0.184 or 0.0127 or 0.944 or 0.816 or 0.987 Must be correct comparison, eg not 0.944 comp with 0.05 Alternative method for middle two A-marks P(X < 10) = 0.0127 P(X < 11) = 0.0556 A1 Both needed Hence rejection region is X < 10 (or X < 11) or critical value is X = 10 A1 Dep on M1 Do not reject H0 Condone Accept H0 M1 1.1 Dep 0.0556 or 0.184 or 0.0127 (2 sf) or P(X < 10 or 11 or 12)seen or 0.944 or 0.816 or 0.987 P(X > 10 or 11 or 12) seen And dep correct comparison, eg, not 0.944 comp with 0.05 There is insufficient evidence that Pierre is overconfident (or that < 90% are satisfied) A1f 2.2b In context. Not definite. Full statement oe, eg There is insufficient evidence that Yvette's suspicion is correct Not: There is evidence that Pierre is not overconfident oe [7] N~Bin: μ = 13.5, μ < 13.5 B1B0 dep defined μ. If undefined: B0B0 N(13.5, 1.35) & X =11.5 or 11 M1 soi p = 0.0426 A1 compare 0.05 A1 dep 0.0426 or 0.0157 Conclusion M0A0H240/02 Mark Scheme November 2020 14 Question Answer Mark AO Guidance 10 ctd 2-tail: H0: p = 0.9 defined p B1 H1: p ≠ 0.9 B0 X~Bin(15, 0.9) and X < 11 or 12 (condone < or = or > or >) M1 Stated or implied eg by 0.0556 or 0.184 or 0.944 or 0.816 or 0.0428 or 0.129 P(X<11) oe = 0.0556 A1 Comp 0.025 A1 Dep 0.0556 or 0.184 Conclusion M0 A0 11 (a) (mean =) 201 (3 sf) B1 1.1 (sd =) 60.7 (3 sf) B1 1.1 Allow 60.8 [2] 11 (b) 0.364 (3 sf) B1 3.4 [1] 11 (c) P(X < 160) = 0.252(49) B1 3.4 soi, eg by P(X > 160) = 0.748 or 0.747 or by 0.147 or 0.148 x1 = Φ–1(0.6 + '0.25249') M1 1.1 = 262.83 (5 sf) ISW A1 1.1 T&I: correct answer scores B1M1A1, otherwise max B1 SC Answer 263 with correct working: B1M1A0 SC Answer 263 with inadequate working: B1 only [3] 11 (d) 112 and 288 are within 2 sd from mean (no working needed) P(X < 112) = 0.0708, which is > 0.025 or > 0.0013 or > 0 B1 3.5a or μ + 2σ = 320 (μ + 3σ = 380) which is > than 288 or P(112< M < 288) = 0.858 which is < than 0.95 (or 0.99) or p = 0.858, but model suggests p = 1 NOT 0.858 alone B0 [1] 11 (e) Reduce σ B1 3.5c May be implied by value of σ 288 – 200 = 2σ or 288 – 200 = 3σ or 288 – 112 = 4σ or 288 – 112 = 6σ Allow more precise correct methods σ = 44 σ = 29.3 or about 30 B1 3.3 Allow σ between 25 and 50. No working needed B1B1 or σ2 between 625 and 2500 [2]H240/02 Mark Scheme November 2020 15 Question Answer Mark AO Guidance 12 (a) H0: μ = 45.7, where μ = mean of all new journey times B1 1.1 Allow "where μ = mean journey time" Allow different letters H1: μ < 45.7 B1 2.5 Subtract B1 for each error eg: use of "p" unless defined B0B0 2-tail B1B0 μ= sample mean implied B1B0 undefined μ B1B0 Not include value 45.7 B0B0 not in terms of parameter B1B0 H0 = 45.7 etc: B0B0 [2] 12 (b) N(45.7, 5.62 30 ) and probability = 0.025 soi M1* 3.3 or N(45.7, 392 375 ) or N(45.7, 1.045) and probability = 0.025 soi P( X < a) = 0.025 or a = Φ-1(0.025) M1 1.1 soi Dep M1 a = 43.7 (3 sf) (43.696....) A1f 1.1 Rejection region is ( X ) < 43.7 (3 sf) A1 1.1 Allow <. Answer ( X ) < 34.7 SC B1 (from not ÷ by 30 ) or ( X ) < 43.6 with explanation Correct answer with inadequate or no working: SC B2 If (a) μ > 45.7, allow ( X ) > 47.7, M1M1A1A0 [4] 13 (a) (i) P(AA or BAA) = 0.42 + 0.6×0.42 oe M1 3.1b allow M1 for either 0.42 (×...) or 0.6×0.42 (×...) = 0.256 or 32 125 A1 1.1 [2] 13 (a) (ii) ABA or BAB M1 3.1b both seen or implied P(ABA or BAB) = 0.42 × 0.6 + 0.62 × 0.4 M1 1.1 M1 for either 0.42×0.6 or 0.62 ×0.4 0.24 A1 1.1 Alternative method 1 – ("0.256" + 0.62 + 0.4×0.62) M1 M1 for 1 – P(A wins or B wins) attempted M1 M1 for 1 – ("0.256" (+ ......)) or 1 – ((..... )+ 0.62 + 0.4×0.62) = 0.24 A1 [3] NB 0.4 × 0.6 = 0.24: M0M0A0 13 (b) '0.256' + '0.24'×'0.256' + '0.242'×0.256 +.... M1 3.1b ft (a)(i)&(ii) = 0.256 1 0.24 − M1 2.1 ft (a)(i)&(ii) ie 1 ( )(ii) −( )(i) aa = 32 95 or 0.337 (3 sf) A1 1.1 cao S5 = 0.337 SC B1, but with added comment M1M1A1 [3]H240/02 Mark Scheme November 2020 16 Question Answer Mark AO Guidance 14 (a) (i) The actual number of extra pupils determines the number of places needed Shows how many new students there will be Shows trend so LA can provide accordingly Need to know expected number of pupils B1 2.2b The existing numbers are already catered for Increase in provision Not Need to know increase in proportion of pupils [1] 14 (a) (ii) Wigan Increase in number is greatest there B1 B1 2.2b 2.2b Allow “Wigan and Bolton” Ignore mention of % increase. Ignore extras. [2] 14 (a) (iii) E.g. all those in this category stay in the LA Populations continue growing at same rate Populations all growing at same rate The population increases consistently NOT Increase has been steady All LAs have the same teacher/pupil ratio All LAs have same need for teachers in 2011 B1 2.2b No decrease in population Children born in that LA will go to school in that LA Assume no great influx or outflow of children after 2011. The LAs are not currently understaffed Ignore extra eg "between 2001 & 2011" [1] 14 (b) Manchester and Salford B1 Highest % or absolute increase B1dep Manchester and Liverpool B1 The two highest in 2011 B1dep SC Manchester (alone), Highest % or absolute increase: B1B0 B1 B1 2.2b 2.4 Wigan and Bolton B1 Highest numbers in 2011 except Manchester and Liverpool, which are very large B1dep Salford and Trafford B1 They have the largest absolute (or %) increase, but are small (or not huge like Manchester) B1dep [2] 15 (a) DR 2 15 2 64 2! × oe eg 15 64 2 × 4 (= 15 32 AG) B1 1.1 Must see this expression and result [1]H240/02 Mark Scheme November 2020 17 Question Answer Mark AO Guidance 15 (b) DR 2, 2, 5 2, 3, 4 3, 3, 3 M1 3.1a Any two seen, with no more than 2 extra different combinations. eg 0, 4, 5 and 0, 5, 4 count as one extra P(X1 + X2 + X3 = 9) = 15 5 2 3 ( ) × × 32 80 + 6× × × 15 5 5 32 16 32 + ( ) 16 5 3 0.0412 + 0.1373 + 0.0305 3× 225 16384 + 6× 16384 375 + 4096 125 675 16384 + 1125 8192 + 4096 125 (= 0.209045) M1 M1 2.1 2.1 M2: > 1 correct product actually seen & all three products correct M1: 1 correct product seen or all correct except omission of, or incorrect, multiple(s) or all three results or total correct, but without working P(X1+ X2 + X3 =9 and at least 1 X value = 2) = 15 5 2 3 ( ) × × 32 80 + 6× × × 15 5 5 32 16 32 (= 0.178528) M1 1.1 Allow M1 for 1 correct product or omit, or incorrect, multiple(s) or ft their probabilities from their previous calculation '0.178528' '0.209045' M1 2.1 ÷ their attempted probs of correct events = 0.854 (3 sf) or 117 137 A1 2.2a P(X1 + X2 + X3 = 9 and no X value = 2) = 5 3 ( ) 16 (= 0.030518 or 4096 125 ) M1 ft their P(3, 3, 3) 1 – '0.030518' '0.209045' M1 ÷ their attempted probabilities of correct events & subtract from 1 = 0.854 (3 sf) or 117 137 A1 NB 1 – ( ) 16 5 3 alone scores M1 [6]H240/02 Mark Scheme November 2020 18 Question Answer Mark AO Guidance 15 (c) P(two 2's in nine vales of X) or 0.094466 or 9C2 × 15 15 7 2 (1 ) ( ) − × 32 32 M1 3.1a soi eg by 9C2 seen P(two 2's in nine vales of X) × P(X = 2) or 0.094466 × 15 32 or 9C2 × (1 ) ( ) − × 15 15 32 32 7 3 M1 2.1 soi NB (17 15 32 32 )7 3 ×( ) scores 0, unless multiplied by 9C2 0.0443 (3 sf) A1 1.1 [3]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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