H630/02 Mark Scheme June 2023 7 12. 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. 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. H630/02 Mark Scheme June 2023 8 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. H630/02 Mark Scheme June 2023 9 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. 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 H630/02 Mark Scheme June 2023 10 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. H630/02 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 1 (a) The given data (on the graph) is continuous, not discrete oe E1 2.4 Need a comment on the suitability of the histogram for the given data – we need a comment on the type(s) of data. The following score -The given data was discrete, but this diagram represents continuous data -The diagram is for continuous data, but the original data was discrete -‘It is continuous data’ as the question asked about the histogram - The data is discrete so there should be gaps between the bars -‘The histogram uses ranges e.g. 6- 7 but the data is discrete and cannot have values between 6-7’ etc Comments such as ‘there are no gaps between the bars’ or ‘it should be a bar chart’ or FD should be used or ‘it is discrete data’ (ambiguous) score 0 [1] 1 (b) (i) 3.183… to 3.184 or 3.18 BC Mark at most accurate B1 1.1 Exact value is 3.183673469 or 3 9 49 or 156 49 o.e. Allow if seen in (b)(ii) if not seen in (b)(i) SC: 3.2 allow B1 [1] 1 (b) (ii) 1.61… to 1.62 BC OR 1.59… to 1.60 B1 1.1 Exact value is 1.615980989 (sample) Exact value is 1.599406387 (population) [1] H630/02 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 2 (a) (𝑥−3)2 seen M1 1.1 eg in (𝑥−3)2 −9 + 1 (𝑥−3)2 −8 A1 1.1 [2] 2 (b) (3, ‒8) B1FT 1.1 FT their completed square [1] 3 6n ‒ 1 evaluated for any positive integer M1 1.1 eg 6×1 ‒ 1 = 5 n 1 2 3 4 5 6 7 8 9 6n - 1 5 11 17 23 29 35 41 47 53 n 10 11 12 13 14 15 16 17 18 6n – 1 59 65 71 77 83 89 95 101 107 n 19 20 21 22 23 24 25 26 27 6n – 1 113 119 125 131 137 143 149 155 161 eg 6×6 ‒ 1 = 35 = 5×7 which is not prime A1 1.1 may see eg n = 11, 17, 23, 29- sight of any value in the table would imply the M1 Must show it’s factorisation so show it’s not prime and give a concluding comment e.g ‘not prime’ If they say ’35 is divisible by 5’ so not prime etc then A1 BUT ’35 isn’t prime’ is A0 [2] H630/02 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 4 (a) 4 B1 1.1 [1] 4 (b) B1 Dep B1 1.1 1.1 Correct shape in both quadrants; Must not cut either axis or bend away excessively from either axis Reasonably Symmetrical about y-axis Condone slight feathering, and slight asymmetry along the y - axis [2] 5 √(11 −5)2 + (−1 −2)2 o.e. OR √(5 −11)2 + (2 −(−1))2 M1 2.1 allow one sign error √45 A1 1.1 3√5 A1 2.2a Condone √45 for A1A1 and √45 or 3√5 only implies full marks [3] 6 (a) Population because all the available data are used E1 2.4 This is an ‘explain’ question, so we do need ‘population’ and a correct justification. Accept ’population as it is data from every single day the phone was used’ scores ‘Population as it is every day the phone is used’ scores [1] 6 (b) Negative skew B1 1.2 ‘Negative’ is B0 [1] H630/02 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 6 (c) Q3 = 58 or Q1 = 42 identified B1 1.1 IQR = 16 B1 1.1 ‘IQR = 16’ implies B1B1 [2] 6 (d) 42 −1.5 × 16 = 18 M1 FT (c) 1.1a Ignore checking of upper tail For calculating 𝑄1 −1.5 × 𝐼𝑄𝑅 for their values Smallest value is 19 which is not an outlier, so no outliers in lower tail A1 1.1 Comparison of lower bound with 19 and conclusion e.g. ’18 < 19 or 19 > 18 (or equivalent in words) so no’ etc If they calculate 18 then and mention 19 and ‘no’ then A1 ‘..18 so no as all values > 18’ is A1 ’18 so no as all values more than 18’ is A1 ’18 so no etc’ is A0 as they need to compare to smallest value (directly or indirectly) and explain why. Command word ‘determine’- justification needed [2] H630/02 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 7 (a) 23 + 6 × 22 −2 −30 =0 so (𝒙−𝟐) is a factor B1 1.1 Factor theorem must be used, and a concluding statement needed Statement might be at the start e.g 𝑓(2) = 0 ⇒(𝑥−2) a factor e.g. Synthetic Division or Long Division is B0 Must see evidence of the substitution- ‘show that’ so e.g simply 𝑓(2) = 0 is B0 [1] 7 (b) (𝑥−2)(𝑥2 + 8𝑥+ 15) M1 A1 By inspection or from long division, allow sign errors only Fully correct linear × quadratic (𝑥−2)(𝑥+ 5)(𝑥+ 3) A1 1.1 Fully correct and fully factorised [3] Alternatively f(k) evaluated, where k is -3 and -5 M1 1.1a Allow a slip with either but not both (𝑥+ 3) 𝑜𝑟 (𝑥+ 5) identified as a factor A1 1.1 (𝑥−2)(𝑥+ 5)(𝑥+ 3) A1 8 (a) Remove any data where #N/A is in the column, as there is no data available B1 2.4 LDS advantage Comments such as exclude 20/81 year old female are B0 as this is only an extract of the full set of data Must refer to N/A/missing data but accept ‘not available/non- existent’ Remove data without a pulse reading is B0 as we could have missing BMI data too H630/02 Mark Scheme June 2023 16 Question Answer Marks AO Guidance [1] 8 (b) (62.77, 84) ringed and no others B1 1.1 [1] 8 (c) No evidence of a linear relationship oe so unlikely to be reliable B1 2.2b Any comment relating to interpolation or extrapolation is B0 as we want a comment on the appropriateness of the model considering the scatter diagram Condone ‘there appears to be little correlation/weak positive correlation’ etc ‘No/Zero correlation’ is B0 e.g the PMCC could be 0.1 etc [1] H630/02 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 8 (d) None of the pulse rates are that unusual so should not be removed B1 2.2b LDS advantage Need ‘no/keep’ and reason Condone ‘No as they are not outliers’ Accept ‘Higher pulse rates are not uncommon’ [1] 9 (a) 0.1 + 0.3 + 𝑞+ 2𝑞+ 3𝑞= 1 M1 1.2 Setting sum of values equal to 1 𝑞= 0.1 A1 1.1 [2] 9 (b) (0.1 + 0.3 + 0.1 + 0.2 =) 0.7 B1 1.1 Or 1 −𝑝(𝑋= 5) [1] 9 (c) 0.1 × 0.3 seen or 0.03 seen M1 1.1 0.1 × 0.3 + 0.1 × 0.3 = 0.06 A1 1.1 0.06 o.e 6 100 or 3 50 etc [2] 9 (d) 0.098314…correct to 2 or more sf BC B1 1.1 By using e.g 𝑋~𝐵(50,0.3) and finding 𝑃(𝑋= 17) [1] H630/02 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 10 Cos A = 3.52+3.92−4.52 2×3.9×3.5 oe M1 2.1 Or Cos B = 4.52+3.92−3.52 2×3.9×4.5 or Cos C = 3.52+4.52−3.92 2×4.5×3.5 Correct use of cosine rule – might not see the ‘Cos A’ etc till next line, but must be a correct use so no ‘sin’ etc Can be in form 𝑎2 = 𝑏2 + 𝑐2 −2𝑏𝑐Cos𝐴 e.g. 4.52 = 3.52 + 3.92 −2 × 3.5 × 3.9 × Cos 𝐴 OR 3.52 = 4.52 + 3.92 −2 × 4.5 × 3.9 × Cos 𝐵 OR 3.92 = 3.52 + 4.52 −2 × 3.5 × 4.5 × Cos 𝐶 cos A = 0.2641… correct to 2 or more sf soi A1 1.1 cos B = 0.66125… or cos C = 0.5488889 A = 74.686° correct to 2 or more sf soi A1 1.1 B = 48.604° or C = 56.709° 1 2 × 3.5 × 3.9 × sin74.686 M1FT 3.1a or 1 2 × 4.5 × 3.9 × sin′48.604′ or 1 2 × 3.5 × 4.5 × sin′56.709′ Must be using their included angle for their two adjacent sides For the Final two marks: They could also find an altitude, h, using one of the angles e.g using angle at C ‘ℎ= 3.5 their sin 𝐶= 2.9256 … ‘ then 1 2 𝑏ℎ= 1 2 (4.5)(2.9256. . ) = 6.58267.. awrt 6.58 or 6.6 A1 3.2a H630/02 Mark Scheme June 2023 19 Question Answer Marks AO Guidance [5] 11 𝑑𝑦 𝑑𝑥= 6𝑥2 + 18𝑥+ 24 B1 2.1 their derivative 𝑑𝑦 𝑑𝑥= 0 M1 1.1 The 𝑑𝑦 𝑑𝑥= 0 may be implied by their concluding statement e.g if they say ‘no real roots’ Use of discriminant implies this mark their 182 −4 × 6 × 24 calculated (may be seen embedded in an attempt to solve their quadratic with QF etc. If no formula quoted then the the solutions must be correct for the method mark. If the formula is quoted, we can allow one error with the substitution of values) M1 3.1a Most common quadratics seen are: 6𝑥2 + 18𝑥+ 24 = 0 or 3𝑥2 + 9𝑥+ 12 = 0 or 2𝑥2 + 6𝑥+ 8 = 0 or 𝑥2 + 3𝑥+ 4 = 0 May have to check for their quadratic NOTE: They could also use a sketch method here: sketch their quadratic and then complete the square to show that the TP is above the x – axis- will need to check their work carefully ‘−252 < 0’ o.e. for their quadratic A1 1.1 -252 < 0 or -63 < 0 or -28 < 0 or -7 < 0 etc May be implied by correct solutions to their quadratic e.g. −3±√7 𝑖 2 Hence 𝑑𝑦 𝑑𝑥= 0 has no solutions and therefore there are no stationary points on the curve A1* 3.2a Must give a concluding statement e.g. ‘therefore no stationary points’. Depends on all previous marks. Condone SPs or ‘turning points’ or TPs for stationary points H630/02 Mark Scheme June 2023 20 Question Answer Marks AO Guidance [5] 12 (a) Opportunity/Convenience sampling B1 1.2 Condone ‘Opportunistic Sampling’ [1] 12 (b) Because every sample (of size n) does not have the same probability of being selected B1 2.4 Accept ‘all adult males registered at the surgery do not have an equal chance of being selected’ Accept ‘everyone registered at the surgery does not have an equal chance of being selected OR ‘For a SRS each element from the SF must have an equal chance of selection’ OR ‘A subset of the population cannot form a complete sampling frame’ OR ‘The sampling frame would be incomplete’ ‘No random method employed in the process’ B0 (need to have the idea that the SF is incomplete) ‘Only collected data from one week’ is B0 [1] H630/02 Mark Scheme June 2023 21 Question Answer Marks AO Guidance 12 (c) w 50- 65- 70- 80- 90- 100- 120 f 6 8 8 11 6 6 B1 1.1 [1] 12 (d) 2 3 × 𝑡ℎ𝑒𝑖𝑟 6 + 1 2 × 𝑡ℎ𝑒𝑖𝑟 6 Or (10 x 0.4) +(10 x 0.3) M1 1.1 IF part (c) is correct then this could be implied by ‘4 +3’ 7 45 or 0.15̇ or 0.15555 to 0.156 Mark at most accurate A1FT 1.1 FT their 6, 6 and 45. May need to check their calculation. May see interpolation methods, which lead to the same calc. [2] 12 (e) The distribution of the weights within each class is unknown E1 2.4 Accept ‘we assume that the values (individual weights) are equally distributed in each class interval’ Accept ‘the individual values (weights) are not known’ Accept ‘the number of people (frequency) in each category of the histogram may not be spread out equally across the category’ Accept ‘we don’t know exactly how many were less than 60kg and how many were more than 110kg’ (idea of correct frequency at both ends for correct probability calculation) [1] H630/02 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 13 (a) H0 : p = 0.37 and H1 : p < 0.37 B1 1.1 Allow equivalent in words Do not allow percentages [1] 13 (b) p is the probability that an adult selected at random in the United Kingdom never exercises (or plays sport) B1 2.5 Accept ‘proportion’ but not number/amount etc B1B1 in (a)(b) if another symbol instead of p used if correctly defined Underlined words needed [1] 13 (c) P(X) ≤ 35 = 0.058… B1 1.1 their 0.058.. compared with 0.05 M1 1.1 may see 𝑃(𝑋< 35) =0.0386 or 𝑃(𝑋≤36) = 0.0847 using 𝑋~𝐵(118, 0.37) – SC 1 mark Use of 𝑷(𝑿= 𝟑𝟓) = 𝟎. 𝟎𝟏𝟗𝟔𝟏..is M0 There is another approach using Normal approx to Binomial to find 5% CR 𝑌~𝑁(43.66, 27.5058) which gives CV 35.033… Or finding value from ND 𝑃(𝑌≤35.5) = 0.059867.. (must be using a continuity correction so 𝑃(𝑌≤35) = 0.0493465.. is M0) 0.058 > 0.05 or 35.033… > 35 or 0.059867 > 0.05 and ‘so do not reject H0’ A1 2.2b ‘accept H0’ is ok H630/02 Mark Scheme June 2023 23 Question Answer Marks AO Guidance There is no evidence (or insufficient evidence) to suggest/support at the 5% level that the percentage of adults (selected at random in the United Kingdom) who never exercises (or plays sport) is less than 37% A1* 2.4 Fully correct contextual conclusion No assertive statements such as ‘proves that’ or ‘shows that’ ‘concludes that’ etc Accept percentage/proportion or probability with 0.37 Dependent on award of all other marks in (c) [4] 14 𝑦= 16𝑥 1 2 + 8𝑥−1 B1 3.1a May be implied by correct derivative 𝑑𝑦 𝑑𝑥= 8𝑥−1 2 −8𝑥−2 M1 A1 1.1 1.1 At least one term of the form 𝛼𝑥−1 2 or 𝛽𝑥−2 obtained All correct 𝑥= 4, 𝑑𝑦 𝑑𝑥= 7 2 B1FT 1.1 FT their 𝑑𝑦 𝑑𝑥, dep on award of M1 x = 4, y = 34 B1 1.1 y ‒ their 34 = (their 7 2)(x ‒4) oe e.g. sub (4, ‘34’) into their 𝑦= 𝑚𝑥+ 𝑐 to find their ‘c’ M1 FT 1.1 Their 7/2 must come from substituting x = 4 into their derivative H630/02 Mark Scheme June 2023 24 Question Answer Marks AO Guidance 𝑦= 7 2 𝑥+ 20 o.e. A1 3.2a All correct. Depends on all previous marks. We can accept any form of the equation of the line: 7𝑥−2𝑦+ 40 = 0 or 𝑦−34 = 7 2 (𝑥−4) o.e. Once the correct equation is seen in any form we can ISW if they simplify incorrectly etc NOTE: Final answer can be obtained from incorrect working- check their derivative [7] 15 (a) c = 1.14 B1 3.3 [1] 15 (b) 1.20 = 4𝑎+ 2𝑏+ 1.14 oe 1.25 = 16𝑎+ 4𝑏+ 1.14 oe M1 3.3 both equations. FT their c 𝑎= −0.00125, 𝑏= 0.0325 A1 1.1 Fractional equivalents are 𝑎= − 1 800 and 𝑏= 13 400 Equivalents in standard form is acceptable [2] 15 (c) 1.29 = 1.14 + 0.0325𝑡−0.00125𝑡2 M1 3.1b FT their a,b,c (Can be > etc) t = 6 and 20 A1 3.4 6 ≤𝑡≤20 A1 3.5a Set notation such as 𝑡∈[6, 20] is fine but must not be soft brackets 𝑡≥6 and 𝑡≤20 or 𝑡≥6 ⋂ 𝑡≤20 but NOT 𝑡≥6 , 𝑡≤20 [3] H630/02 Mark Scheme June 2023 25 Question Answer Marks AO Guidance 15 (d) It will eventually predict a negative exchange rate oe (will fall below zero etc) B1 3.5a ‘Exchange rate tends to zero’ is B0 Must mention the variable ‘exchange rate’ Underlined words needed [1] Need to get in touch? If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in touch with our customer support centre. 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