Y432/01 Mark Scheme June 2023 5 6. Subject Specific Marking Instructions 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 2023 6 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. Y432/01 Mark Scheme June 2023 7 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. 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. NB for Specification A the rubric specifies 3 s.f. as standard, so this statement reads “3 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. Y432/01 Mark Scheme June 2023 8 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” 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. Y432/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 1 (a) 0.72 × 0.3 M1 3.3 For geometric = 0.147 A1 1.1 Accept 0.15 [2] 1 (b) P(At least 5) = 0.24(01) B1 1.1 0.74 or 1 −0.3(1 + 0.7 + 0.72 + 0.73) oe [1] 1 (c) Mean = 10 3 soi B1 3.1b 1 0.3 Accept 3.3 Variance = 1−0.3 0.32 M1 1.1a Standard deviation = √70 3 A1 1.1 (2.78886…) Evaluate 𝑡ℎ𝑒𝑖𝑟 ( 10 3 ± √70 3 ) and identify their correct integer range M1 3.4 P(0.54 < 𝑋< 6.12) = P(𝑋≤6) (1 −0.76) = 0.88(2351) A1 FT 1.1 FT their 𝐸(𝑋) and 𝑆𝐷(𝑋) provided more than one positive integer included in their range. [5] Y432/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 2 (a) (i) For example, one of: • a sample this small might not give any useful information (about the strength) E1 2.4 E0 if comment only refers to the sample being too small. Allow any suitable answer • a sample this small is unlikely to be representative of the population • the sample statistics from such a small sample are unlikely to be close to the population parameters [1] 2 (a) (ii) For example, one of: • it is wasteful as the cans cannot then be used/it is a test to destruction E1 2.4 E0 if comment only refers to the sample being too large. Allow any suitable answer • a (far) smaller sample size is adequate to give useful information (about the strength) [1] 2 (b) For example, one of: • there may have been a fault earlier in the batch which then corrected itself for the later tins • if there has been a fault throughout the batch it is not sensible to only discover that at the end E1 2.2b Allow any suitable answer • this would not be random so it would not be appropriate to make (statistical) inferences (about the population) [1] 2 (c) The sample should be: • unbiased B1 B1 1.2 Any two from these three • representative of the population • random [2] Y432/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 3 (a) Both cases: 2 zeroes, 1 one and 1 zero 2 ones seen B1 3.1a 3 × (1 4) 3 + 3 × (1 4) 3 M1 1.1 Calculation needs to relate correctly to the two cases. Following B0 M0, SC B1 for use of ( 1 4) 3 but without reference to the two distinct cases = 6 64 = 3 32 A1 1.1 AG [3] 3 (b) DR E(𝑋) = 3 with justification B1 2.4 (0 × 1 64) + 1 × 3 32 + 2 × 13 64 + 3 × 3 8 + 4 × 13 64 + 5 × 3 32 + 6 × 1 64 or ‘By symmetry’ E(𝑋2) = (0 × 1 64) + 1 × 3 32 + 4 × 13 64 + 9 × 3 8 +16 × 13 64 + 25 × 3 32 + 36 × 1 64 M1 1.1a E(𝑋2) = 167 16 = 10.4375 quoted without working is M0 Allow one slip in calculation. Alt: M1 M1 for 1 64 {(0 −3)2 + 6(1 −3)2 + 13(2 −3)2 + 24(3 −3)2 + 13(4 −3)2 + 6(5 −3)2 + (6 −3)2} oe Var(𝑋) = 𝑡ℎ𝑒𝑖𝑟 {167 16 −32} M1 1.1 = 23 16 A1 1.1 1.4375 allow 1.4 Must follow M1 M1 [4] 3 (c) (i) The values must be independent B1 2.2a [1] 3 (c) (ii) Var(𝑌) = 10 × 𝑡ℎ𝑒𝑖𝑟 23 16 M1 1.1a 230 16 SD(𝑌) = 3.79 or 1 4 √230 A1 FT 1.1 Allow 3.8 [2] Y432/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 4 (a) 1.3 oe B1 1.1 0×36+1×33+2×14+3×10+4×4+5×1+8×1+10×1 100 = 130 100 [1] 4 (b) C4 probability = 0.2303 B1 FT 3.4 D5 expected frequency = 14.2888 B1 FT 2.2a Allow 14.2911 for use of calculator probability for x=2 and 4dp values of x=1 and x=0 E3 contribution = (33−35.4291)2 35.4291 M1 1.1a = 0.1665 A1 1.1 [4] 4 (c) DR H0: Poisson model is a good fit H1: Poisson model is not a good fit B1 2.5 Reference to ‘mean 1.3’ in hypotheses scores B0 Allow omission of context at this stage X2 = 7.0(283) B1 FT 1.1 FT (6.8618 + 𝑡ℎ𝑒𝑖𝑟E3) Use of ν = 2 M1 3.4 Critical value at 5% level = 5.99(1) A1 1.1 or 𝜒5 2(7.0283) = 0.9702 their 7.0(283) > their 5.99(1) M1 1.1 0.9702 > 0.95 critical value must be from 𝜒2 table (Reject H0) There is sufficient evidence to suggest that the Poisson model is not a good fit for the number of vehicles (passing Eve’s house each minute) A1 3.5a Correct test statistic and critical value required Conclusion must follow correct hypotheses, not be too assertive and refer to context. [6] 4 (d) We would now not reject H0 (insufficient evidence to suggest that the Poisson model is not a good fit). E1 FT 3.2a Comment on impact on conclusion. Allow ‘accept H0’ It is reasonable remove these two values as they are not representative of the normal situation. (vehicles would not be travelling independently, mean rate would not be constant) E1 2.4 Allow alternative answers such as ‘If horse-riders regularly use the lane, even if not very frequently, then the Poisson model may not be valid.’ [2] Y432/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 5 (a) 𝑚= 0.914𝑤−64.1 B1 B1 3.3 1.1 First B1 for either coefficient Second for both given correct to 3sf and 𝑚 and 𝑤 used in an equation, not just stated separately (so 𝑦= 0.914𝑥−64.1 scores B1 B0) [2] 5 (b) Prediction for 99 is 26(.4) B1 FT 1.1 Prediction for 110 is 36(.4) B1 FT 1.1 Allow only 1 mark if either prediction is given to more than 1 decimal place [2] 5 (c) The prediction for 99 is (moderately) reliable as it is interpolation although the points do not appear to be close to a straight line. B1 3.5a First B1 for a correct conclusion and reference to at least one of interpolation/extrapolation/not close to a straight line Use of ‘accurate’ for ‘reliable’ is incorrect. The prediction for 110 is not (at all) reliable as it is extrapolation and the points do not appear to be close to a straight line. B1 3.5b Second B1 for all 3 [2] 5 (d) It would not be sensible/appropriate B1 2.2a Correct conclusion (may be implied by rest of comment) For example: • Because this is the equation of the w on m regression line, not m on w. B1 2.2a Must be linked with a correct conclusion • Because this is found by minimising the squares of the horizontal (𝑤)residuals. • Since this line should only be used to estimate wing length from the mass of a bird. • Since this line only measures the average value of the wing length for a given value of the mass. [2] Y432/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 6 (a) Because the scatter diagram does not appear to be elliptical (due to the possible outlier) E1 3.5a For not elliptical. Alternatively, ‘the points appear to be funnel-shaped’ so the (underlying) distribution is (probably) not bivariate Normal. E1 2.4 For full answer (dependent on first mark) “data is not bivariate Normal” is E0 “Normal bivariate” is E0 [2] 6 (b) Rank saw 7 6 3 2 1 8 5 4 Rank chop 7 4 6 1 5 8 3 2 𝑑2 0 4 9 1 16 0 4 4 M1 depM1 1.1 1.1 For ranking saw or chop correctly For ranking saw and chop consistently (i.e. same way round) with at most one adjacent pair of ranks transposed. If rankings not seen they may be inferred from correct 𝑑 or 𝑑2 values or ∑𝑑2 = 38 seen Spearman’s rank coefficient 23 42 or 0.5476 A1 1.1 Accept 0.55 or better. SC B1 only if 0.5476 stated with no working. [3] 6 (c) H0: There is no association between sawing and chopping (times) in the population B1 B1 3.3 For first B1 need to see one correct hypothesis with context (need not have population at this point). H1: There is positive association between sawing and chopping (times) in the population 1.2 For B1B1 need to see two correct hypotheses with each of context and population in at least one of the hypotheses. Critical value 0.6429 B1 3.4 n = 8, 5%, 1-tailed 0.5476 < 0.6429 (so do not reject H0/accept H0) M1 1.1 For comparison provided 0 < 𝑟𝑠< 1 There is insufficient evidence to suggest that there is positive association between sawing and chopping times (in the population) A1FT 2.2b FT their rs and sensible critical value from correct table A0 for “(Sufficient evidence to suggest) no association”. Needs to include context. [5] Y432/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 7 (a) Number of values of X is 𝑛−100 + 1 or 𝑛−99 soi M1 3.1a (Number < 100 + 𝑛 2 is) 𝑛−100 2 M1 3.1a N.B. may see use of 𝑛= 2𝑘 Probability = 𝑛−100 2(𝑛−99) oe ISW A1 1.1 e.g. 1 2 (1 − 1 𝑛−99) or 1 2𝑛−50 𝑛−99 [3] 7 (b) Var(𝑋) = 1 12 ((𝑛−99)2 −1) M1 3.1a Accept Var(𝑋) = 1 12 ((𝑛−100)2 −1) Var of sum of 50 values = 50 × 1 12 ((𝑛−𝑡ℎ𝑒𝑖𝑟 99)2 −1) M1 1.1 Their 99 must be a positive integer = 25 6 (𝑛2 −198𝑛+ 9800) A1 2.1 𝑎= 25 6 or exact equivalent [3] Need to get in touch? 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