Y432/01 Mark Scheme June 2022 3 2. Subject-specific Marking Instructions for AS Level Mathematics B (MEI) 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. Y432/01 Mark Scheme June 2022 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. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. 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. Y432/01 Mark Scheme June 2022 5 f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) 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.  When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. 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. E marks are lost unless, by chance, the given results are established by equivalent working. 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” and “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. Y432/01 Mark Scheme June 2022 6 Question Answer Marks AOs Guidance 1 (a) 0.64 = 0.1296 B1 [1] 1.1 Given answer 1 (b) E(X) = 1.3(056…) B1 1.1 Var(X) = 1.96(3…) B1 [2] 1.1 Accept 2.0 1 (c) E(Y) will be less since the value of r = 1 for X becomes r = 0 for Y and r = 4 becomes r = 5 but P(X = 1) > P(X = 4). E1 E1 3.1a 2.1 Need to refer to change from X to Y Var(Y) will greater since the distribution is now more spread out. E1 [3] 2.2a Allow other suitable answers Y432/01 Mark Scheme June 2022 7 Question Answer Marks AOs Guidance 2 (a) Both (variables are random) B1 [1] 1.1 2 (b) Sdh = 866.63 – 1 15 × 84.9 × 124.7 (= 160.828) Sdd = 624.55 − 1 15 × 84.92 (= 144.016) M1 1.1a For Sdh or Sdd Could be embedded 𝑎 = 160.828 144.016 (= 1.1167 … ) A1 1.1 For a May have a and 𝑏 notation reversed 𝑏 = 124.7 15 −𝑡ℎ𝑒𝑖𝑟1.1167 … × 84.9 15 (= 1.9926…) M1 3.3 For method for b h = 1.12d + 1.99 A1 [4] 1.1 Allow final mark only if equation explicitly stated with values to 2 dp y = 1.12x + 1.99 scores M1 A0 2 (c) Prediction for d = 7.5 is h = 10.4 B1FT 1.1 Allow only 1 mark if either prediction given to more than 1 dp. Use of d on h regression line scores Prediction for d = 20.0 is h = 24.3 B1FT [2] 1.1 Allow 24.4 from equation to 2dp B0 2 (d) The prediction for d = 7.5 is likely to be reliable since the points lie close to the regression line, and this is interpolation. The prediction for d = 20.0 is less likely to be reliable since this is extrapolation, well beyond any data values. B1 E1 B1 [3] 3.5a 3.5a 3.5a B1 for interpolation E1 for noting strong linear correlation B1 for extrapolation Use of ‘accurate’ for ‘reliable’ scores B0 2 (e) e.g. The predicted height for a tree of diameter 60 cm is 69(.0) metres. E1 1.1 69(.2)m from equation to 2dp (d > 42.9cm gives h > 50m) FT their regression line E1 for “The regression line will predict heights (cm) greater than diameters (m)” The regression line is no longer valid (for mature trees). E1 [2] 2.4 Relationship non-linear for mature trees Use of ‘accurate’ for ‘valid’ scores B0 2 (f) Coordinates are (5.66, 8.31) B1 [1] 1.1 Accept ( 283 50 , 1247 150 ) Y432/01 Mark Scheme June 2022 8 Question Answer Marks AOs Guidance 3 (a) 1 × 2 + 2 × 5 + 3 × 5 + 4 × 12 + 5 × 10 + 6 × 10 + 7 × 11 + 8 × 1 + 9 × 4 60 M1 1.1 May see extra 0 terms in numerator Accept e.g. 2 + 10 + ⋯+ 36 306 60 only scores M0 = 5.1 A1 [2] 1.1 Given answer Needs to be correctly obtained 3 (b) C3 = 0.1348 B1 3.4 4 decimal places required for all values D5 = 10.5177 B1 2.2a E8 = (5−8.6414)2 8.6414 = 1.5344 M1 A1 [4] 1.1a 1.1 3 (c) Because if they were not then (some of) the expected frequencies would be too low (<5) (and so the test would not be valid) E1 [1] 2.4 To ensure the expected frequencies large enough 3 (d) H0: Poisson model is a good fit H1: Poisson model is not a good fit B1 1.2 Reference to ‘mean 5.1’ in hypotheses Scores B0 Allow omission of context at this stage X2 = 6.23 B1FT FT their value of E8 Refer to X25 M1 1.1 Critical value at 5% level = 11.07 A1 3.4 or 𝜒5 2(6.231) = 0.7156 6.23 < 11.07 (Accept H0) M1 1.1 0.7156 < 0.95 Comparing their test and critical values leading to a conclusion. There is insufficient evidence to suggest that the Poisson model is not a good fit for (the number of) wasps (entering the nests). A1 [6] 2.2b Correct test and critical values required Conclusion must follow correct hypotheses, not be too assertive and refer to context. 3 (e) e−𝜆(𝜆0 0!) = 0.0053 M1 3.1a λ = 5.2(4004…) A1 [2] 1.1 Y432/01 Mark Scheme June 2022 9 Question Answer Marks AOs Guidance 4 (a) 0.72×0.3 = 0.147 M1 A1 [2] 3.3 1.1 Use of Geo(0.3) 4 (b) Use B(9, 0.3) to get 0.2668... 0.2668... × 0.3 = 0.080(048…) B1 M1 A1 [3] 2.2a 3.1a 1.1 Use of correct binomial distribution Require P(3 in first 9 then also 10th) BC Value may be implied 0.08 scores A0 (only 1sf) Y432/01 Mark Scheme June 2022 10 Question Answer Marks AOs Guidance 5 (a) Because the scatter diagram does not appear to be elliptical (but more of a funnel shape) so the distribution is probably not bivariate Normal. E1 E1 [2] 3.5a 2.4 For not elliptical For full answer (dependent on first mark) “data is not bivariate Normal” is E0 Normal bivariate is E0 5 (b) Rank A 1 2 3 4 5 6 Rank P 8 12 6 10 11 7 Rank A 7 8 9 10 11 12 Rank P 5 9 4 3 1 2 M1 M1 1.1 1.1 For ranking Age For ranking Protein consistent with ranking for age Ranks may be reversed Spearman’s rank coefficient = − 0.78(32) (= − 112 143) A1 [3] 1.1 BC 5 (c) H0: There is no association between age and protein (level) in the population B1 3.3 H1: There is some association between age and protein (level) in the population B1 1.2 Need to see context and population in at least one of the hypotheses Critical value is (±)0.5874 B1 3.4 n = 12, 2-tailed 5% | − 0.7832 | > 0.5874 (so reject H0.) M1 1.1 For comparison of their rs and sensible critical value provided |rs| < 1 There is sufficient evidence to suggest that there is association between age and protein level (in the population) A1FT [5] 2.2b FT their rs and sensible critical value Hypotheses need to have been stated the right way round Conclusion must not be too assertive and refer to context 5 (d) (Because a random sample) enables (proper) inference about the population to be undertaken B2 [2] 2.4 2.4 B2 for correct explanation, as shown SC B1 for partially correct explanation, eg a random sample is less likely to be biased 5 (e) Because as the sample size increases, the random variation in the sample tends to decrease. E1 2.2b Allow E1 for ‘the influence of outliers is reduced’ or for ‘gives a more The sample Spearman’s rank correlation coefficient tends to get closer to the population correlation coefficient. E1 [2] 2.2b reliable result’ oe if there is no further explanation. Y432/01 Mark Scheme June 2022 11 Question Answer Marks AOs Guidance 6 (a) X = 3W – 2 where W has a uniform distribution over the values {1, 2, …, n} soi B1 3.1a Var(X) = 32 × Var(W) M1 1.2 Var(X) = 3 4 (𝑛2 −1) oe A1 [3] 1.1 Var(X) = 9 × 1 12 (𝑛2 −1) Mark final answer 6 (b) E(X) = 149.5 SD(X) = √3 4 (1002 −1) (= 86.5 … ) B1FT 3.1a For both FT on their Var(X) So require 62.9 … ≤𝑥≤236.0 … B1FT 1.1 Evaluate bounds for x (accept as integers) FT on their Var(X) and their E(X) ⇒21.6 … ≤𝑤≤79.3 … M1 3.1a Convert to bounds for w (accept as integers) Probability = 58 100 oe A1 1.1 From correct working and values throughout Alternative for candidates who realise that this can be done with a uniform distribution on {1, 2, …, n} For general method stating use W in place of X B1 E(W) = 50.5 SD(W) = √1 12 (1002 −1) (= 28.8 … ) B1 For both ⇒21.6 … ≤𝑤≤79.3 … M1 Convert to bounds for w (accept as integers) Probability = 58 100 oe A1 From correct working and values throughout [4] Need to get in touch? 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