Y542/01 Mark Scheme June 2022 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. Y542/01 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. 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. • 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. Y542/01 Mark Scheme June 2022 5 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. Y542/01 Mark Scheme June 2022 6 Question Answer Marks AO Guidance 1 (a) (i) Geo(0.01) B1 1.1 Allow if 0.01 not stated here but used in (ii) (a) (ii) qm – qn where m = 50, 49, 51, n = 150, 149, 151 M1 2.1 e.g. 0.378, 0.386, 0.387, 0.381, 0.390, 0.392. Their q Or pq50 + … + pq149 = pq50(1 – q100)/(1 – q) q50 – q150 = 0.384 (0.38355..) A1 1.1 In range [0.383, 0.384], cao  1 term at either end for M1 M1 needs method for summing GP (b) E(R) = 1/p and correct Var(R) = (1 – p)/p2 M1 1.1 Art least one correct general formulae stated = (1/p)2 – (1/p) = [E(R)]2 – E(R) A1 [2] 2.1 Correctly show required relationship, either way round, don’t need conclusion, not numerical 2 (a) Independent (and no others) B1 2.5 Fully correct only (b) 0.713 BC B2 [2] 1.1 1.1 SC: If B0, give B1 for any 2 of 16.02, 14.24, 10.77, or any 2 of 192.25, 170.9, 129.25 seen e.g. 769/4, 2051/12, 517/4 (c) (i) “More computer failures when temperatures are higher”, rather than “number of failures is related to higher or lower temperatures” B1 [1] 2.4 Oe Allow “Looking for positive correlation”, etc (c) (ii) H0:  = 0, H1:  > 0 where  is the population PMCC between temperature and number of failures B2 1.1 2.5 Treat association and correlation as equivalent One error, e.g. 2-tail, or  not defined: B1 (allow H0:   0). Allow omission of “population” (or “true”) or of context, but not of both See Appendix for exemplars Verbal, e.g. H0: no correlation between temperature and number of failures, H1: positive correlation: max B1, one error B0 CV = 0.6581 B1 1.1 Correct CV, or p = 0.00462 0.713 > 0.6581 B1ft 1.1 Explicit comparison, their r, allow “pmcc > CV” Their CV must be from tables for r or rs Reject H0. Significant evidence of (positive) correlation between temperature and number of failures M1ft A1ft [6] 1.1 2.2b FT on their r, needs 0.6581, 0.6851, 0.7079, 0.4973 or 0.5760; contextualised, not too definite If inconsistent with comparison, M0A0 Not “insufficient evidence of no correlation between …” (d) 0.713 B1ft [1] 1.2 Their (b). Allow “unchanged”, etc. 3 DR E(V) = q + 0.24 + 0.6 B1 3.1a E(V 2) = q + 0.48 + 1.8 B1 1.1 Or: E(V – E(V))2: M1 q + 0.84 = q + 2.28 – (q + 0.84)2 M1 2.1 Use E(V 2) – [E(V)]2 then eliminate p: M1 (q + 0.84)2 = 1.44 A1 1.1 Correct simplified quad, eg q2 + 1.68q – 0.7344 = 0 Or 625q2 + 1050q – 459 = 0 so q = –0.84  1.2 M1 1.1 Solve quadratic, valid method seen or implied Can be implied by one correct root Reject q = –2.04 as negative B1 2.3 Explicitly reject invalid solution, with reason Or reason to take + root of 1.44 q = 0.36 A1 2.2a 0.36 or exact equivalent seen, e.g. 9/25 p = 1 – 0.32 – q = 0.32 B1ft [8] 1.1 p = 0.68 – their q. Withhold if extra solutions seen SC: E(V 2) – E(V): (B1B1) q = 0.6 B1, p = 0.08 B1 E.g. 8/25 Use Poisson: 0/8 Y542/01 Mark Scheme June 2022 7 Question Answer Marks AO Guidance 4 (a) Punctures must occur at constant average rate throughout the period of 24 hours B1 [1] 3.3 Constant average rate (or uniform rate), with context (not events must occur …), no extras Not constant rate, not constant probability, not average constant rate, not singly (b) 2.7 = 1.64(3…) B1 1.1 Any equivalent answer, e.g. 330/10 4 (c) Po(18.9) M1 1.1 Po(72.7) stated or implied 1 – P( 21) M1 3.4 Probability from their Po(72.7) E.g. 1 – P( 22) = 0.2(0015) M1M1A0 = 0.267 A1 1.1 Awrt 0.267 (d) Po(3.5) M1 1.1 Stated or implied, e.g. by 1 – 0.858 = 0.142 P(> 6) = 0.0653 A1 3.4 Correct calculation, awrt 0.0653 (e) 0.0653 is some way from 0.12 B1ft 3.5b Compare 0.12 with answer to (d), or other appropriate calculation and comparison Or: B(100, 0.0653), P( 12) = 0.03 or E(R) = 6.5, etc so assumptions look doubtful (but possible as 100 days may not be enough evidence) B1ft [2] 3.5a Appropriate assessment, not definite (and correct calculation, if used). FT on their (d) See Appendix for exemplars 5 (a) 80 + 2X M1 3.1b Consider distribution of 80 + 2X OR: make X subject M1 ~ N(246, … M1 1.1 Normal with mean 246 X < 77.5 (or  77) A1 … 1440) A1 1.1 Correct variance P(X < 77.5) from N(83, 360) M1, allow  77 P(< £235) = 0.386 A1 [4] 3.4 Awrt 0.386 P(< £235) = 0.386 A1 (0.376 A0) Or P(2X < 155) where 2X ~ N(166, 1440) (b) Using costs A1, A2, A3, B1, B2: A1 = 80 + 2X1 ~ N(246, 1440) B1 1.1 Means 246, 244.5, can be implied by later work If in effect using 3X, 2Y: Means 246, 244.5 B1 B1 = 120 + 1.5X4 ~ N(244.5, 810) B1 2.2a Variances 1440, 810, ditto Variances 1440, 810 B1 A1 + A2 + A3 ~ N(738, 4320) B1 1.1 Means 738 & 489, allow 498 & 249 Means 738, 489 B1 B1 + B2 ~ N(489, 1620) B1 1.1 Variances 4320 and 1620 Variances 12960, 3240 B0 Difference ~ N(249, 5940) M1 3.1b Use normal for difference, add variances N(249, 16200) M1 P(diff  300) = 0.254 A1 [6] 3.4 Awrt 0.254, allow 0.02 if cc used 0.344 A0 OR Using distances X1, X2, X3, X4, X5: X1 + X2 + X3 ~ N(249, 1080) B1 Means 249, 166, can be implied by later work If in effect using 3X, 2Y: Means 249, 166 B1 X4 + X5 ~ N(166, 720) B1 Variances 1080, 720, ditto Variances 3240, 1440 B0 A’s charge ~ N(738, 4320) B1 Means 738 & 249, allow 498 & 249 Means 738, 489 B1 B’s charge ~ N(489, 1620) B1 Variances 4320 and 1620 Variances 12960, 3240 B1 NB Difference ~ N(249, 5940) M1 Use normal for difference, add variances N(249, 16200) M1 P(diff  300) = 0.254 A1 Awrt 0.254, allow 0.02 if cc used 0.344 A0 OR Diff in costs is 2(X1 +X2 +X3) – 1.5(X4 +X5) M1A1 M1 needs attempt to include constants ~ N(249, 5940) M1A2 M1 N(249, ...), A2 variance correct, A1 for 16200 SC: variance formula seen correct but P(diff  300) = 0.254 A1 Awrt 0.254, allow 0.02 if cc used wrongly calculated: (M1)M1 SCM1 SC1: insufficient working shown 0.254 www gets 6/6; 0.746 www gets 5/6 [A0] 0.344 www gets 4/6 [A0A0]; all else 0/6 SC2: assume all/both variables equal: M1A1 3X > 300 M1 N(249, 3240) or N(51, 3240) A1 Or N(83, 360) and compare 100: A1 (0.185) Y542/01 Mark Scheme June 2022 8 Question Answer Marks AO Guidance 6 (a) Mean 30.7 B1 1.1 Seen or implied Inequalities can be omitted until final line Biased variance 22.225 (SD 4.714) M1 1.1 Calculate variance, allow biased here 22.225  128/127 = 22.4 (SD 4.733) M1 1.2 128/127 seen, or implied used Independent of previous M1. 1 – ((30.7 – 30)/(22.4/128) M1 1.1 Standardise with 128, answer < 0.5 or 50% E.g. 4.6% seen (from no 128/127). Not 0.441 Range is  0.047 ( 4.7%) A1 [5] 2.2a Allow > or  oe. Awrt 0.047 or 4.7%. CWO. Wrong or no range: A0. Allow = 4.71 so  5% 0.0464 is probably B1M1M0M1A0 No 128/127, or wrong z: max 3/5 Insuff working: 4.7%  5/5, 4.6%  3/5 0 with some evidence [(22.4)/128]: 4/5 6 (b) 128 large enough (for CLT to apply) M1 2.3 Reason, e.g. sample is large, or n > 25 SC1: CLT applies so not invalidated: B1 Hence test not invalidated A1 [2] 2.3 Conclusion, allow “test is valid” or just “no” Wrong extras, eg “all dists approach normal”: M0 SC2: CLT applies as n > (any number other than 25): max B1 7 1 0 d 1 n kx x =  ⇒ k = n + 1 M1 A1 2.1 1.1 Equate integral of f(x), upper limit 1, to 1 Correct equation for n and k 0.8816 0 d 0.5 n kx x = ò M1* 1.1 Equate integral of f(x), upper limit 0.8816, to 0.5 k(0.8816)n+1/(n + 1) = 0.5 and use k = n + 1 depM1 3.1a Set up equation in n or k only 0.8816 and 0.5 reversed: (M1A1M1) 0.8816n+1 = 0.5 A1 1.1 Correct equation [or 0.8816k = 0.5] M1A0A0 (M1A1) n = 4.5 or k = 5.5 A1 3.1a Correct n or k, awrt 4.5(0) or 5.5(0) P(X < 0.8) = 0.85.5 (= 0.293) M1 3.4 Use their f(x), upper limit 0.8, with their n E(number < 0.8) = 2.93 A1ft [8] 2.2a 10(their p) NB: can get 2.93 from wrong integration Not just 3, but ISW if 2.93 rounded to 3 8 (a)  = 264 B1 3.1a Allow even if no CC used ¼n(n + 1) = 264 M1 1.1 Use formula for mean Or: 113 + 415 = ½n(n + 1): M2 n = 32 or –33, but n > 0 so 32 only AG A1 [3] 2.2a Solve n2 + n – 1056 = 0 to obtain 32 Can be implied by both 32 and –33 Just verification: SC B1 (b) Variance = 1 24 ( 1)(2 1) n n n + + = 2860 B1 3.3 Variance 2860 stated or implied (113.5 – 264)/2860 (= 0.002445) M1* 3.4 Standardise, their parameters (or use 414.5) No or wrong cc: (0.503% or 0.518%):  r = 20.002445  100% depM1 3.1b Double their p-value (p < 0.5), oe B1M1M1A0 = 0.489 (%) A1 [4] 2.2a 0.49 or better (0.48 is probably from no cc) Allow 0.50% only if correct CC seen 9 (a) H0: data consistent with belief, H1: not B1 1.1 OE, e.g. “follows the given ratio” Expected frequencies 30, 20, 20, 20, 30 B1 1.1 Stated or implied X 2 values 0.1333, 0.8, 0.8, 0.8, 1.2 M1 3.4 At least two correct, or correct subs in formula Can be implied by answer Total 3.73(333…) (= 56 15 ) A1 1.1 Fully correct SC: Some cells merged: < 9.488 A1 1.1 Compare with 9.488, or p = 0.443 > 0.05 can get B2M1A0A0 M1A1 Do not reject H0. Insufficient evidence that data inconsistent with head’s belief M1ft A1ft [7] 1.1 2.2b Consistent first conclusion, FT on their X 2 Contextualised, not over-definite Sufficient evidence that belief is correct: A0 Needs CV 9.488 or 11.14 or 7.815 Withhold A1 if hypotheses reversed Insuff evidence to reject H0, belief correct M1A1 Y542/01 Mark Scheme June 2022 9 Question Answer Marks AO Guidance (b) “Determine”, so needs proper working 3.73(333…) × n/120 > 9.488 M1 3.1b Inequality for n, using TS and CV 3.73 gives 305.2, 3.733… gives 304.97 ( 7n/225 > 9.488)  n  305 (3 sf) A1 1.1 n > 305 or n  305 stated or implied, allow 306 e.g. answer of 305 or 306 Must be multiple of 30, so least n is 330 B1ft [3] 3.2a Next multiple of 30 above their 305 Needs reason if rounded up by < 10 or to multiple of 300 OR ( ) ( ) ( ) 2 2 2 7 3 4 2 9 3 30 12 30 12 30 12 3 2 3 12 12 12 3 n n n n n n n n n − − − +  + = 1 1 1 900 150 100 3 9.488 n n n +  +  M1 Inequality for n, using 120, TS and CV, needs n in both O and E terms () Or T& I: n = 300 → 9.333 Any n in range 300 to 360, with x 2 in [9.33, 11.2]: M1 n = 330 (9.489) A2 (n = 305 → 9.489 A1) ( 7n/225 > 9.488)  n  305 (3 sf) A1 n > 305 or n  305 stated or implied, allow 306 e.g. answer of 305 or 306 Must be multiple of 30, so least n is 330 B1ft Next multiple of 30 above their 305, needs reason Or n/120 must be a multiple of 0.25 APPENDIX Exemplar responses for Q2(c)(ii) Hypotheses Verbal statements A H0: no correlation between variables, H1: positive correlation between variables [no context] B0 B H0: no correlation between number of failures and temperature, H1: correlation between number of failures and temperature [not one-tailed: needs “positive” in H1] B0 C H0: no evidence of correlation between runners’ positions, H1: evidence of positive correlation between number of failures and temperature [“evidence” does not belong in hypotheses; without “evidence of”2 this would be B1] B0 Statements using parameter symbol [mark statements using r in place of  identically] D H0:  = 0, H1:   0 [two errors: H1 wrong, no interpretation of ] B0 E H0:  = 0, H1:   0 [two errors: H1 wrong, no interpretation of ] B0 F H0: r = 0, H1: r > 0 [one error: no interpretation of . Allow use of r] B1 G H0:  = 0, there is no correlation between number of failures and temperature; H1:  > 0, there is positive correlation [two different answers, either of which would score 1 but neither gains 2 and we don’t give 1+1 here] B1 H H0:  = 0, H1:  > 0, where  is the pmcc [neither “population” nor context] B1 I H0:   0, H1:  > 0, where  is the population pmcc OR H0:  = 0, H1:  > 0, where  is the pmcc between temperature and number of failures B2 Conclusions (In general, allow “Accept H0” as a synonym for “Do not reject H0”, etc.) Y542/01 Mark Scheme June 2022 10 A Reject H0. There are more computer failures when temperatures are higher [too definite] M1A0 B Reject H0. There is significant evidence of correlation [no context] M1A0 C Sufficient evidence to reject H0. There are more computer failures when temperatures are higher [“Evidence” used, albeit in wrong sentence, but BOD] M1A1 D Wrong but validly obtained TS leading consistently to Do not reject H0. There is insufficient evidence that there are more computer failures when temperatures are higher [standard FT] M1A1 E Wrong but validly obtained TS leading consistently to Do not reject H0. There is evidence that the number of computer failures is not correlated with temperature. [Non-rejection doesn’t give positive evidence that H0 is correct. When H0 is not rejected, candidates should aim for a “double negative”] M1A0 F (from correct TS and CV) Do not reject H0. (+ anything) [inconsistent] M0A0 G If H0, H1 the wrong way round, but calculations right: Reject H0. (+ anything) M1A0 Exemplar responses for Q4(e) Response Mark A 12 is very high so it does cast doubt [relevant comparison needs to be seen] B0B1 B The probability is so low that 12 is worrying so it does cast doubt [unclear comparison] B0B1 C 0.0653 is not equal to 0.12 so probably not valid [don’t allow “not equal”] B1 D 0.0653 is not close to 0.12 B1 E 12 is much higher than 6.53 so it is not valid [too definite] B1B0 F P(= 12) from B(100, 0.0653) = 0.0166 which is very small so it does cast doubt B0B1 G P(> 12) from B(100, 0.0653) = 0.00784 which is very small so it does cast doubt B0B1 H P( 12) from B(100, 0.0653) = 0.00784 which is very small so it does cast doubt B0B1 I P( 12) from B(100, 0.0653) = 0.0299 which is small so it casts doubt [0.035 from N(6.53, 6.1036)] B1B1 J Expected number is 1000.0653 = 6.53 which is very different from 12 so it does cast doubt B1B1 K Wrong p, e.g.: 0.142 not far from 0.12 so it does not cast doubt (0.142 from Po( 6), 0.0267 from Po( 7) B1B1ft L Likewise: Expected number is 14.2 which is not very close to 12 so it does cast doubt B1B1ft M Likewise: P( 12) from B(100, 0.142) = 0.319 which is not too low so it does not cast doubt B1B1ft Need to get in touch? 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