H240/02 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. H240/02 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. H240/02 Mark Scheme June 2022 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 June 2022 6 Sig figs: “0.348 (3 sf)” means “answer that rounds to 0.348”, ISW. eg 0.347652 = 0.35 scores A1, 0.348 = 0.35 scores A1, but 0.35 alone scores A0 Other forms for probabilities Allow eg 20% or 1 in 5, but not odds eg 1:4 Question Answer Mark AO Guidance 1 (a) DR ( 2) ( 1)( 1) ( 1)( 2) x x x x x x + − − + + + or 2 2 2 2 1 3 2 x x x x x + − + + + oe (= 0) M1 M1 1.1 1.1 M1 for x(x + 2) − (x + 1)(x − 1) oe Multiply out brackets. Allow one error Ignore denominator even if “= 0” x = 1 2 − A1 1.1 NB correct with no working: SC B1 Alternative method x(x + 2) = (x + 1)(x − 1) M1 M1 for attempt “cross-multiply”. x2 + 2x = x2 − 1 or 2x = −1 oe M1 Multiply out brackets. Allow one error x = 1 2 − A1 [3] 1 (b) DR Solve quadratic in 1 3 x or x3 or u (= x3 or 1 3 x ) M1 3.1a or cubic in x Condone quadratic in x with x = 1 3 x or x = x3 using any correct method. Must see attempt at correct method for this mark Allow arithmetical errors 3 1 x (or u) = 1 & 1 8 − or x3 (or u) = 1 & –8 B1 1.1 Can be scored without M1 Condone x = 1, 1 8 − or x = 1, −8 or correct factorisation of quadratic Ignore x3 = 0, if seen, for this mark x = 1& x = –2 with no extras B1f 1.1 ft their x3 or 3 1 x If also x = 0, B0 NB correct with no working: M0B0B1 [3] 1 (c) DR Condone incorrect or omitted brackets eg (x2 – 7)ln3 = ln 1 243 or x2 − 7 = log3( ) 1 243 M1 3.1a Any correct step after log(both sides) or 2 7 3x − = 3−5 or x2 – 7 = –5 or 2 3x = 32 or ANY correct step using indices x = ± 2 or ±1.41 (3 sf) A1 1.1 NB correct with no working or T & I: SC B1 H240/02 Mark Scheme June 2022 7 Question Answer Mark AO Guidance [2] 2 (a) (4i + 2j −5k) − (3i + 2j) (= i − 5k) M1 1.1 b − a or a − b attempted, using i, j, k or column vectors or (3i + 2j) − (4i + 2j −5k) (= 5k − i) May be implied by calculation seen AB = 26 or 5.10 (3 sf) or 5.1 A1 1.1 www. Correct answer, no working: M1A1 [2] Mark(s) cannot be gained retrospectively in (b) 2 (b) '26' = (p – 3)2 + 4 + 9 +(p – 4)2 +4 + 4 M1 3.1a Attempt AB2 = BP2 + PA2 (involving p) ft their AB Alternative methods for M1 Attempt |PC|2 = (their radius)2 M1 or ( 7 2 – p)2 + 4 + 1 4 = 13 2 Attempt . PA PB = 0 M1 or ((3–p)i + 2j + 3k).((4–p)i + 2j – 2k) = 0 p2 – 7p + 10 = 0 oe or (p − 7 2 )2 = 9 4 A1f 1.1 Correct simplified equation, ft their (a), ie: or p2 – 7p + 2 46 their 2 a − = 0 or (p − 7 2 )2 = 2 their 17 4 a − p = 2 or 5 A1f 1.1 ft only their (a) [3] 3 (a) (No because) they differ only by a constant B1 1.2 oe, eg They may have different constants of integration or eg c2 = c1 + 1 3 , or 1 3 is part of Ben’s c Only the “c”s are different If definite integral found, answers are same Not “Both are correct” or “just different correct methods” If differentiate, answers same [1] 3 (b) (i) 1 (1 ) 1 1 x a − + −         or 1 1 2 u a +   −   or 2 ( 1) 1 4 a u   + −     oe M1 1.1 Attempt integral, must be of form k(1 + x)-1 or ku−1 or ku−0.5 (if from substitution u = (1 + x)2) Ignore limits = 1 (1 ) 1 1 2 a − + − + oe M1 1.1 Attempt substitute appropriate limits into their integral (= 1 1 2 1 a + − ) = 1 2( 1) a a − + A1 1.1 cao oe single fraction [3] H240/02 Mark Scheme June 2022 8 Question Answer Mark AO Guidance 3 (b) (ii) 1 2( 1) a a − + = 1 3 or their (b)(i) (limits subst’d) = 1 3 M1 1.1 or their new attempt at 2 1 (1 ) 1 d a x x +  = 1 3 a = 5 A1 1.1 cao [2] 3 (c) DR Allow incorrect use of brackets throughout Allow ln(…) instead of ln|…| 1 2   1 12 0 ln | sin 2 2 | x  + M1 1.1 Allow kln(sin 2x + 2), k any constant. Ignore limits = 1 2 [ln( 1 6 sin  + 2) – ln(0 + 2)] M1 1.1 Attempt substitute both correct limits into their log integral. Allow numerical errors = ( ) 1 2 ln( ) ln 2 5 2 − A1 1.1 Allow × any k, otherwise any correct form without trig. = 1 2 5 4 ln oe, eg 5 2 ln A1 2.1 Correct one-term exact result. ISW, eg ignore decimal [4] NB No working, no marks. Alternative methods u = sin 2x + 2, or u = sin 2x 2 5/2 1 2 1 du u  or 0 1/2 1 2 1 d 2 u u +  1 2 5 / 2 ln 2 u     or 1 2   1/ 2 ln( 2) 0 u + M1 Attempt substitute and integrate and obtain klnu or kln(u + 2), k any constant; ignore limits (= 1 2   1 12 0 ln(sin 2 2)  + x ) (May not see this step) M1 Attempt substitute their limits into their log integral. but not limits for wrong variable, eg not π 12 ln − ln 0 Allow numerical errors 1 2 5 2 (ln( ) ln2) − A1 Correct exact result, any form without trig. Allow × any k H240/02 Mark Scheme June 2022 9 Question Answer Mark AO Guidance = 1 2 5 4 ln oe, eg 5 2 ln A1 cao, ISW, eg ignore decimal answer 4 20 + 20×r + 20×r2 + .... or 20 × 1 1 n r r − − M1 3.1b Sum of a GP implied. Allow any r, eg r = 0.05 20 × 1 0.95 1 0.95 n − − = 205 A1 1.1 Correct equation 0.95n = 195 400 or 39 80 or 0.4875 A1 1.1 Allow 0.487 or 0.488 n = ln0.4875 ln0.95 oe or n = log0.95( 39 80 ) oe M1 2.1 or 0.9514 = 0.4875 or 0.487 or 0.488 seen. Can be implied by their answer ft their equation of form an = b (dep M1 gained and b > 0) (Number of steps =) 14 A1 1.1 cao. Allow n = 14. Allow 14.0. Allow ≈ 14 [5] Alternative method Sum of GP implied M1 20 + 20×r + 20×r2 + .... M1 Attempt add > 10 terms. Allow any value of r for this mark 20 + 20×0.95 + 20×0.952 + .... + 20×0.9513 A1 Correct 14 terms added = 205 (3 sf) A1 Number of steps = 14 A1 NB Unsupported correct answer: SC B3 [5] Alternative (incorrect) methods using r = 1.05, or 1 0.95 or 20 19 (For info’ only: r = 1 0.95 or 20 19 comes from misinterpreting “lowest” to mean “shortest”) 20 + 20×r + 20×r2 + .... or 20 × 1 1 n r r − − M1 Allow any value of r for this mark 20 × ( ) 1 0.95 1 0.95 1 1 n − − = 205 or 20 × ( ) 20 19 20 19 1 1 n − − = 205 A1 oe using 1.05. Correct equation ( ) 20 19 n = 117 76 or 1.05n = 1.51 or 1.54 A1 H240/02 Mark Scheme June 2022 10 Question Answer Mark AO Guidance n = 117 76 20 19 ln ln or 1.053 ln 1.539 or 1.05 ln 1.51 M1 oe, eg ln1.539 ln1.053 or ln1.51 ln1.05 ft their “ 117 76 ” Can be implied by their answer Number of steps = 8 or 9 A0 5 (a) DR d d y x = 3x2 – 6x + 4 = 0 M1 3.1a Differentiate & equate to 0. May be implied by calc of D b2–4ac =–12 or D=−12 or 3(x−1)2 + 1 = 0 oe or x = 6 36 48 6  − or x = 6 i 12 6  oe No (real) roots Must see justification as line above, no errors, & statement or no value of x, or can’t negative A1 1.1 Other correct forms of the quadratic equation and or gradient always +ve. justification may be seen. [2] (b) DR 2 2 d d y x = 6x – 6 = 0 M1 1.1 Differentiate their d d y x and = 0. Can be implied by x = 1 x = 1 gives a point of inflection A1 2.2a Statement “x = 1 gives a point of inflection” is enough. or x = 1 & show that, either side of this point, gradient does not change sign or This equation has one root. (so curve has one inflection) or second derivative does change sign Not just “x = 1” [2] Ignore y-coordinate 6 (a) One- one Many- one Own inverse Not a function 1 √ 2 √ 3 √ √ 4 √ 5 √ B4 [4] 1.2 1.2 1.1 2.2a B4 for all 5 rows correct B3 for 3 or 4 rows correct B2 for 2 rows correct B1 for 1 row correct 6 (b) > 1 2 B1 1.2 > 1 2 soi, no top limit (except ). Allow > 1 2 Allow f(x) or 1 x or any letter or none for 1st B1 H240/02 Mark Scheme June 2022 11 Question Answer Mark AO Guidance {y: y > 1 2 }, {y: 1 2 < y < }, {y: 1 2 < y < } B1 2.5 Correct range in set notation. Any letter (not x) or 1 x or f(x) or [ 1 2 ,) or [ 1 2 ,] [2] 7 (a) NB Other correct methods may be seen (3m + 0)2 = 9m2 B1 3.1a 9m2 alone, not as part of longer expression (3m + 1)2 (3m + 2)2 = 9m2 + 6m + 1 = 9m2 + 12m + 4 M1 1.1 At least one of these expansions attempted using r = 1 or 2. Must include three (or four) terms, Allow one error = 3(3m2 + 2m) + 1 = 3(3m2 + 4m +1) + 1 or 3(3m2 + 4m) + 4 A1 2.1 At least one of these seen explicitly None of these is of the form 3n + 2 A1 3.2a Must see the statement oe. Allow “≠ 3n + 2” Can be seen once at end or with each separate case Dep complete method, with all three cases seen [4] Alternative method 1 (3m + r)2 (= 9m2 + 6mr + r2 ) M1 Attempted. Must include 3 (or 4) terms, Allow one error = 3(3m2 + 2mr) + r2 A1 Explicit = 3n + r2 But r2 = 0, 1 or 4 B1 Hence not in the form 3n + 2 for any r A1 Must see the statement oe Dep complete method Alternative method 2 Let (3m + r)2 = 3n + 2 M1 3(3m2 + 2mr − n) = 2 − r2 A1 Hence 2 − r2 is divisible by 3 B1 But 2 − 02 = 2, 2 − 12 = 1, 2 − 22 = −2 None of these is divisible by 3 A1 H240/02 Mark Scheme June 2022 12 Question Answer Mark AO Guidance 7 (a) ctd Alternative method 3 (3m)2 = (9m2 − 2) + 2 B1 (3m + 1 )2 = (9m2 + 6m − 1) + 2 M1 Allow one arithmetical error (3m + 2 )2 = (9m2 + 12m + 2) + 2 A1 Both correct (9m2 − 2) = 3(3m2) − 2 or 3(3m2 − 2 3 ) + 2 (9m2 + 6m − 1) = 3(3m2 + 2m) − 1 or 3(3m2 + 2m − 1 3 ) + 2 (9m2 + 12m + 2) = 3(3m2 + 4m) + 2 or 3(3m2 + 4m + 2 3 ) + 2 Hence none is divisible by 3 A1 None of the brackets is an integer 7 (b) Either imply three digits all of the same type M1* 3.1a Could be numerical or algebraic or in words or imply three digits all of different types If listed, must be clear which ones are selected P(0, 0, 0) or P(1, 1, 1) or P(2, 2, 2): ( ) 3 1 3 M1 dep 1.1 M1 for ( ) 3 1 3 associated with at least one of these P( 0, 1, 2): ( ) 3 1 3 × 6 M1 dep 2.1 M1 for ( ) 3 1 3 ×k where k = 4, 5 or 6, associated with (0, 1, 2) or 1 2 1 3 3   oe Alternative method for 2nd & 3rd M1M1 No. of cases = 33 = 27 M1 No. divisible by 3 = (3 + 6 =) 9 M1 Allow 7 or 8 9 27 or 1 3 or 0.333 (3 sf) A1 1.1 Correct answer with no working: M0M0M0A0 H240/02 Mark Scheme June 2022 13 Question Answer Mark AO Guidance [4] 8 Summary method: Express V in terms of h B1 3.3 Correct substitution Differentiate V with respect to h M1 3.4 NOT if h = 50 or r = 50tan30 used Attempt chain rule, M1 Resulting equation must involve exactly 2 variables Attempt separate variables M1 Their equation must involve exactly 2 variables Correct integrals A1 Ignore limits Substitute correct limits M1 Integrals must be of correct forms (see examples below) Answer A1 Note 1 Candidates who substitute numerical values for h or V or r may be able to score the 2nd and/or 3rd M1 marks, but probably nothing else. See the example of this below. Note 2. There is a special case for candidates who use r =h sin30 (answer 625 4 π or 491). These can score all 4 M-marks and the final A1 Note 3. The chain rule may be used to find d d V t or d d h t or d d V h or d d V r or other derivatives. Two of the example methods below illustrate use of d d V t and d d V r , but use of other derivatives can also lead to correct methods. H240/02 Mark Scheme June 2022 14 Question Answer Mark AO Guidance 8 ctd Example method 1 V = o 2 3 ( tan30 ) π h h or V = 2 3 3 π h h       oe B1 3.3 or V = 3 9 π h oe 2 d d 3 V π h h = M1 3.4 Attempt differentiate their V in terms of h only NOT if h = 50 or r = 50tan30 used. 2 d d d 3 d V π h t t h = oe or 2 d 3 d d d π h V t t h = M1 2.1 Attempt use chain rule for d d V t or d d h t in terms of t & h only (" 2 d 3 d π h t h " = –2h oe or d 6 d π h t h − = ) (Set their d d V t = –2h) 0 50 0 d 6d t π h h t = −   oe M1 1.1 Attempt separate variables in their equation in terms of h and t only (not V or r). Integral signs not essential 2 2 0 50 πh       =   6 0 t t − oe A1 2.1 Correct integrals, any limits or none –π× 2 50 2 = –6t M1 1.1 Substitute correct limits into integrals of forms ah2 & bt OR substitute t = 0 & h = 50 to find c and substitute h = 0 Time = 625 3 π secs or 654 secs (3 sf) oe A1 3.4 Allow without secs or 10.9 mins or 10 mins 54 secs or: SC. Use of r =h sin30 (answer 625 4 π or 491) can score all 4 M-marks and final A1 [7] H240/02 Mark Scheme June 2022 15 Question Answer Mark AO Guidance 8 ctd Example method 2 V = o 2 3 tan30 π r r or V = 3 3 π r oe B1 Subst h = o tan30 r into correct formula for V 2 d d 3 V r πr = M1 2 d d d d 3 V r t t πr = oe M1 Attempt use chain rule to find d d V t or d d r t in terms of t and r ( 2 d d " 3 " r t πr =– 2 3 r oe) (Set their d d V t = – 2 3 r oe) 50 3 0 0 d 2d t π r r t = −   oe M1 Attempt separate variables in their equation in terms of r and t only (not V or h). Integral signs not essential 2 50 2 3 0 πr       =   2 0 t t − oe A1 Correct integrals, any limits or none – 2 50 6 π = –2t M1 Substitute correct limits into integrals of the form ar2 & bt OR substitute t = 0 & r = 50 3 to find c and substitute r = 0 Time = 625 3 π secs or 654 secs (3 sf) oe A1 3.4 Allow without secs or 10.9 mins or 10 mins 54 secs SC. Use of r = hsin30 (answer 491) can score M4A1 H240/02 Mark Scheme June 2022 16 Question Answer Mark AO Guidance 8 ctd Example method 3 (NOT using chain rule) This method is different from the summary method above V = o 2 3 ( tan30 ) π h h or V = 2 3 3 π h h       oe B1 3.3 or V = 3 9 π h oe h = 9 3 π V M1 Allow h = kV1/3 d d V t = −2 × 9 3 π V M1 d d V t = –2×(their h in terms of V) 3 50 9 0 1/3 π 3 d 9 V V  −  = −20 t t M1 Attempt separate variables in their equation in terms of V and t only (not h or r). Integral signs not essential π 3 3 9 2  3 0 2/3 50 9 V        = −2t A1 Correct integrals, any limits or none 3 2/3 π 3 π50 3 9 2 9   −      = −2t M1 Substitute correct limits into integrals of forms aV2/3 & bt OR substitute t=0 & V = 3 50 9  to find c and substitute V = 0 Time = 625 3 π secs or 654 secs (3 sf) oe A1 Allow without secs or 10.9 mins or 10 mins 54 secs or: SC. Use of r = hsin30 (answer 625 4 π or 491) can score all 4 M-marks and final A1 H240/02 Mark Scheme June 2022 17 Question Answer Mark AO Guidance 8 ctd Example incorrect method r = 50/ 3 V = π 2500 3 3 h   B0 d d V h = 2500π 9 M0 d d V h = d d V t × d d t h 2500π 9 = −2h d d t h M1 dh h = − 18 2500π dt M1 9 (a) Area of 20-30 block ÷ total area M1 1.2 attempted, using any units, eg small squares or cm2 = 110 750 or 22 150 or 4.4 30 = 0.147 (3 sf) (AG) A1 1.1 Correct calculation seen and answer 0.147 seen Not any method starting with 0.147, eg 0.147 × 150 = 22.05 [2] 9 (b) (i) Roughly bell-shaped B1 2.2b or Roughly symmetrical and peaks in middle or has one peak and tails off at each end, or drops off either side All 3 of these must be seen (except “Bell-shaped” scores B1) [1] Not “Shape is like normal curve” Ignore all else 9 (b) (ii) Roughly symmetrical about x = 40, or calculate mean and obtain 5915 150 or 39.4 or area to left of 40 ≈ area to right of 40 B1 2.4 Allow 40 has the highest frequency or frequency density or the peak is at 40 or 40 is in the middle Ignore all else 70 – 40 ≈ 3σ, hence σ ≈ 10 or calculate sd and obtain 10.3 or most values within 20 of mean, so 20 ≈ 2σ Most data is within 6σ H240/02 Mark Scheme June 2022 18 Question Answer Mark AO Guidance or (Area within 40±10)/total eg 510/750 or 102/150 = 0.68 or ≈ 2 3 B1 3.3 Must see correct fraction and ≈2 3 , or 68% or 0.68 [2] 9 (c) 0.136 (3 sf) B1 1.1 BC [1] 9 (d) m = 39.4 or 5915 150 or 1183 30 , B1 3.1a Allow 39.1 < m < 39.7 Ignore method BC s = 10.3 (3 sf) or s2 =106 (3 sf) B1 1.1 Allow 105.5< s2< 108.5 or 10.27< s < 10.42 Ignore method (Use of denominator n or (n – 1) is OK for full marks) OR if neither mark scored, M1 for attempting find frequencies or areas (NOT heights) or at least five of these seen: 4, 22, 23, 28, 29, 22, 20, 2 or 20, 110, 115, 140, 145, 110, 100, 10 0.150 or 0.151 or 0.152 (3 sf) Allow 0.15 B2 3.4 cao Correct with unclear or no working; B1B1B1B1 B1 1.1 or B1 for 0.145 to 0.158 NB No retrospective marks if 0.151 seen in table for (e)(i) [4] 9 (e) (i) x < 20 20 to 30 30 to 35 35 to 40 40 to 45 45 to 50 50 to 60 > 60 No FT Histogram 0.027 0.147 0.153 0.187 0.193 0.147 0.133 0.013 N(40, 100) 0.023 0.136 0.150 0.191 0.191 0.150 0.136 0.023 B1 1.1 B1 for middle row correct ±0.001 NB N(m, s2) 0.030 0.151 0.153 0.189 0.183 0.142 0.130 0.023 B1 3.4 B1 for bottom row correct ±0.003 NB [2] 9 (e) (ii) Nina's model better fit for lower values of X Allow “more accurate” or “less accurate” or similar Nina's model better fit for any ranges < 40 B1 3.5a Nina’s model less good fit for 40-45 (or >60) Sam's model better fit for higher values Sam's model better fit for any ranges > 40 B1 3.5a BUT SC: “Both less good fit for >60” alone: B1 only Sam’s model less good fit for 20-30 (or >60) NOT “Both are fairly good fit” B0B0 H240/02 Mark Scheme June 2022 19 Question Answer Mark AO Guidance Ignore all else NB No ft [2] 10 all Allow “percentage” or “value” or “number” or “rate” etc for proportion in all parts of qu 10 10 (a) (i) High(er) or increased proportion 18–24 B1 2.2b eg “many 18-24” Ignore any LA mentioned Ignore extras only if they don’t contradict High 18-24 only [1] 10 (a) (ii) High(er) or increased proportion either/both B1 2.2b or high proportion of younger. Ignore any LA mentioned [1] Ignore extras 10 (a) (iii) Low(er) or decreased proportion either/both B1 2.2b or low proportion of younger. Ignore any LA mentioned eg “LA F because low % in younger ages” B1 [1] Ignore extras 10 (b) (i) G, H, K, M B1 2.2b No extras or omissions [1] 10 (b) (ii) F, N, R B1 2.2b No extras or omissions [1] 10 (c) Imply need to consider other age range(s) B1 2.3 Examples: Low 0-17 & 18-24 not  attractive to older May be a large % of 25-64 (or 65+) High % of young people does not necessarily imply low % Some LAs have low 0-17 and 18-24 and 65+ of older people Low 0-17 & 18-24 does not mean high 65+ Older people may want live near young relatives Need to consider other factors or anomalies Eg May be reasons for low % younger people eg no schools [1] 10 (d) NB. No ft for either mark State all 3 LAs are > 1.5×IQR above UQ B1 1.2 Or 16.76 + 1.5×(16.76 − 14.56) (= 20.06) Ignore attempt at lower limit Confirms F, N, R (implied) despite (c) B1 2.2a Independent mark. But must mention (c) [2] H240/02 Mark Scheme June 2022 20 Question Answer Mark AO Guidance 10 (e) Mean > UQ B1* 1.1 or mean is in 4th quartile Median better B1dep 2.2b Ignore all else Not Mean skewed by F, N, R so median better Not Median not skewed by F, N, R so better Not Mean because need take account of outliers (or F,N,R) [2] 11 See the exemplars at the end of the MS NB. Use of a “continuity correction” loses 1st A1 only Hypotheses: Allow other letter (including X) only if clearly defined H0: μ = 3300 B1 1.1 Subtract B1 for each error eg: H1: μ > 3300 2-tail B1B0 Undefined μ B1B0 where μ = (population) mean mass B1 2.5 not in terms of parameter B1B0 Not include 3300 B0B0 X stated or implied B0B0 H0 = 3300 etc: B0B0 µ = sample mean implied B0 & (B1or B0) Calculation and comparison X ~ N(3300, 2 450 200 ) or N(3300, 1012.5) oe and X > 3360 M1* 3.3 Correct distribution and X (allow 3359.5, 3360.5, 3659) stated or implied eg by 0.0297 or 0.0307 or 0.0286 even if within incorrect statement eg P(X = 3360) = 0.0297 Allow 2 450 200  or 2 2 450 200  Not 0.0297 from µ = 3360, P( X < 3300) P( X > 3360) = 0.0297 (NB 3 sf) A1 3.4 BC cao H240/02 Mark Scheme June 2022 21 Question Answer Mark AO Guidance 0.0297 > 0.025 A1 1.1 Explicit comparison Allow compare (any value < 0.35) with 0.025 11 ctd Alternative method 1 for M1A1A1: X ~ N(3300, 2 450 200 ) and X = 3360 M1* Correct distribution and X (allow X = 3359.5 or 3360.5) stated or implied eg by 0.970 or 0.969 or 0.971 even if within incorrect statement eg P(X = 3360) = 0.970 Allow 2 450 200  or 2 2 450 200  P( X < 3360) = 0.970 (NB 3 sf) A1 BC cao 0.970 < 0.975 A1 Explicit comparison Allow compare (any value > 0.65) with 0.975 Alternative method 2 for M1A1A1: 3300 450 200 a−  = 1.96 M1* May be implied, eg by 3362 Allow 2 450 200  or 2 2 450 200  CR, a = 3362 (NB 4 sf) A1 3360 < 3362 or 3360 is in acceptance region A1 Explicit comparison of their a with 3360 Alternative method 3 for M1A1A1: 3360 3300 450 200 −  M1* Allow 3359.5 or 3360.5 May be implied, eg by 1.89 Allow 2 450 200  or 2 2 450 200  CV of z = 1.886 or 1.89 (NB 3 sf) A1 cao 1.89 (or 1.90 or 1.9) < 1.96 A1 Explicit comparison H240/02 Mark Scheme June 2022 22 Question Answer Mark AO Guidance 11 ctd Conclusion Not reject H0 M1 1.1 Allow Reject H1 or Accept Ho dep Dep M1A1A1 or M1A0A1 or M1A1A0 Dep also on comparing like with like, eg not 0.970 > 0.025 May be implied by their conclusion, if M1 criterion is met ft with opposite conclusion if, eg P( X > 3360) = 0.024 There is insufficient evidence that A1f 2.2b Dep M1A1A1 or M1A0A1 or M1A1A0 (mean) mass (or mean) is > 3300 (g) or has In context, not definite; eg not “Mean mass not > 3300 g” increased Not “There is evidence that mean mass hasn’t increased” Not “This suggests that mean mass hasn’t increased” Not “Unlikely (or unsure) that mean has increased” ft with opposite conclusion if, eg P( X > 3360) = 0.024 Alternative scheme for incorrect method using 2-tails: Hypotheses B1B0 Calculation: as above M1A1 Compare 0.0125 oe A1 Conclusion M0A0 [7] 12 (a) X ~ B(600, 0.02) M1 3.3 soi, eg H0: p = 0.02 and B(600, p). Allow n = 600, p = 0.02 Attempt P(X > n) for 17 < n < 20 M1 2.1 May be implied by 0.0991 or 0.0202 or 0.9798 or 0.9009 or correct values (P(X > 18) = ) 0.0610 or 0.061 (2 sf) A1 3.4 or (P(X < 17) = ) 0.939 or 0.94 (2 sf) (P(X > 19) = ) 0.0359 or 0.036 (2 sf) A1 1.1 or (P(X < 18) = ) 0.964 or 0.96 (2 sf) These two probabilities seen imply M1M1A1A1 Condone errors such as P(X > 18) = 0.0610 P(concludes claim incorrect) = 0.0359 (3 sf) A1 2.2a Ignore hypotheses and/or “Reject H0” or similar H240/02 Mark Scheme June 2022 23 Question Answer Mark AO Guidance Unsupported answers; 0.0359: M1M1A1A0A0 [5] Critical region is X > 19 M1M1A0A0A0 12 (a) ctd Alternative method (normal with no cc) X ~ N(600×0.02, 600×0.02×0.98) or X ~ N(12, 11.76) M1 soi. Can be scored either for N(12, 11.76) or B(600, 0.02) Attempt P(X > n) for 17 < n < 20 M1 P(X > 17) = 0.0724 or 0.072 (2 sf) A1 P(x > a) = 0.05 a = 17.64 only gets M1 if a probability is calculated P(X > 18) = 0.0401 or 0.040 (2 sf) A1 P(concludes claim incorrect) = 0.0401 A0 Alternative method (normal with cc) X ~ N(600×0.02, 600×0.02×0.98) or X ~ N(12, 11.76) M1 soi. Can be scored either for N(12, 11.76) or B(600, 0.02) Attempt P(X > n) for 17 < n < 20 M1 P(X > 18) = P(X > 17.5) = 0.054 (2 sf) A1 P(X > 19) = P(X > 18.5) = 0.0290 (2 sf) A1 P(concludes claim incorrect) = 0.0290 A0 12 (b) (Incorrect because eg:) You have to consider P(X > 18) or 18 is under the significance level or18 is in the acceptance region (for 5% test) B1 2.3 Allow You have to do a proper hypothesis test or critical region is > 19, or CV is 19 No other answers acceptable [1] 13 (a) Single Venn diagram drawn showing 3, 14, x and 3x correctly placed B1 3.1a or showing 3, 14, 2 and 6 correctly placed Allow omission of rectangle, so long as 14 seen outside Allow probabilities in the diagram 3 + 14 + x + 3x = 25 oe or x = 2 M1 1.1a May be implied, eg by 2 seen in correct place in diagram Number who study English = "2" + 3 or 5 M1 1.1 Their x + 3. May be implied by answer H240/02 Mark Scheme June 2022 24 Question Answer Mark AO Guidance P(E) = 5 25 or 1 5 or 0.2 A1 1.1 [4] If x is total English, giving x = 5, use an equivalent scheme. 13 (a) ctd Alternative (incorrect) method for H E Diagram B0 3 + 14 + x + 3x = 25 oe or x = 2 M1 Or implied in diagram, History only = 2, or total History= 5 Number who study English = "6" + 3 or 9 M1 Their 3x + 3. May be implied by answer P(E) = 9 25 A1 If x is total History, giving x = 5, use an equivalent scheme 13 (b) P(exactly one English) = 5 20 25 24  ×2 oe M1 1.1 Allow omit ×2. Allow 5 20 25 25  or 0.16 or 0.32. Allow + …. = 1 3 or 0.333 (3sf) A1 1.1 NB No ft from (a) in (b) P(exactly one E and exactly one H) = P( HEand H E ) + P(EH and E H ) = ( 6 2 25 24  + 3 14 25 24  ) ×2 oe M2 3.1b M1 for one of 6 2 25 24  or 3 14 25 24  oe 2.4 OR 6 2 25 25  + 3 14 25 25  (both terms) (= 9 50 or 0.18 ) OR 6 25 24 a  + 3 25 24 b  or 2 25 24 a  + 14 25 24 b  (a,b integer < 24) Allow any of the above + extras for M1 ( ) ( ) exactly one E and exactly one H exactly one E P P (= 9 50 ÷ 1 3 ) M1 1.1 Divide attempted probs of correct events dep > M1M1 = 27 50 or 0.54 (3 sf) cao A1 1.1 Careful!! SCs for correct answer by incorrect methods: “×2” omitted throughout: 9 100 ÷ 1 6 = 27 50 : M1A0M1M0M1A1 (Total 4) Denominator 25×25 instead of 25×24: H240/02 Mark Scheme June 2022 25 Question Answer Mark AO Guidance 54 27 4 625 25 50  = : M1A0M1M0M1A1 (Total 4) [6] Both the above 27 27 2 625 25 50  = M1A0M1M0M1A1 13 (b) ctd Alternative method 1: n(exactly one English) = n(E) × n(E) M1 = 5 × 20 = 100 A1 ( ) exactly one E and exactly one H n = n(EH ) n(E H) + n(EH) n(E H )      = 2×6 + 3×14 (= 54) M1 M1 M1 for one of 2×6 or 3×14 or M1 for 2×a + 3×b (a, b integers, a < 23, b < 22) Attempt ( ) ( ) exactly one E and exactly one H exactly one E n n M1 (= 54 100 ) = 27 50 oe or 0.54 (3 sf) cao A1 Alternative method 2 P(H/E) = 3/5 P(H/E') = 6/20 M1 M1 for three of these fractions seen P(H'/E) = 2/5 P(H'/E') = 14/20 A1 A1 for all four fractions seen 6/20×2/5 + 3/5×14/20 M3 M2 for one of these products 6/20×2/5 or 3/5×14/20 or M1 for a/20×2/5 + 3/5×b/20 or M1 for 6/a×2/5 + 3/5×14/b = 27/50 A1 H240/02 Mark Scheme June 2022 26 Question Answer Mark AO Guidance 13 (b) ctd Alternative (incorrect) method for H E P(exactly one English) = 9 16 25 24  (×2) M1 Allow without ×2. Allow 9 16 25 25  or 0.230 or 0.461. = 6 25 A0 P(exactly one E and exactly one H) = P( HEand H E ) + P(EH and E H ) M1 Same as main scheme M1 (= 9 50 or 0.18 ) ( ) ( ) exactly one E and exactly one H exactly one E P P M1 Attempt divide attempted probabilities of correct events dep at least M1M1 (= 9 50 ÷ 6 25 ) = 3 4 or 0.75 (3 sf) A0 SC answer 3 4 , but omit ×2 and/or denominator of 25, M1A0M1M0M1A1 H240/02 Mark Scheme June 2022 27 Exemplars for Q11 Hypotheses A H0: μ = 3300 H1: μ > 3300 where μ = (pop) mean mass B1B1 B H0: μ = 3300 H1: μ > 3300 B1B0 C H0: The (pop) mean mass is 3300 H1: The (pop) mean mass is greater than 3300 B1B0 See Specimen paper q10 MS "Must be in terms of parameter values" D H0 = 3300 H0 > 3300 B0B0 E H0: μ = 3300 H1: μ ≠ 3300 where μ = (pop) mean mass B1B0 F H0: μ = 3300 H1: μ ≠ 3300 B0B0 Calculation, comparison and conclusion G No statement of distribution P( X = 3360) = 0.0297 M1A1 0.0297 > 0.025 A1 Don't reject H0 M1 There is no evidence that mean mass has increased A1 H P( X = 3360.5) = 0.0286 M1A0 0.0286 > 0.025 A1 Accept H0 M1 There is evidence that mean mass hasn't increased A0 I P( X > 3360.5) = 0.0286 M1A0 A0 Accept H0 M1 There is evidence that mean mass hasn't increased A0 J P( X = 3359.5) = 0.024 M1A0 0.024 < 0.025 A1 Reject H0 M1 There is evidence that mean mass has increased A1ft K P( X < 3360) = 0.970 M1A1 0.970 < 0.975 A1 Reject H1 M1 Insufficient evidence that mean mass has changed A0 L P( X > 3360) = 0.970 M1A1 0.970 > 0.025 A0 Insufficient evidence that mean mass has increased M0A0 H240/02 Mark Scheme June 2022 28 M X ~ N(3300, 1012.5) P( X > 3360) = 0.297 M1A0 0.297 > 0.025 A1 Insufficient evidence that mean mass has increased M1A1 N μ ± 1.96σ = 3237 to 3362 M1A1 3360 lies within this range A1 Can’t reject H0 M1 Mean mass hasn’t increased A0 O CV = 3362 M1A1 3360 < 3362 A1 Reject H0. Evidence that level of poll't has increased. M0A0 P (3360 − 3300) ÷ (450÷ 200 = 1.886 M1A1 1.866 < 1.96 A1 Don't reject H0. Mean mass hasn't increased. M1A0 2-tail Q H0: μ = 3360 H1: μ ≠ 3360 B0B0 0.0297 > 0.0125 A1 Don't reject H0 M0 There is no evidence that mean mass has changed A0 R H0: μ = 3360 H1: μ ≠ 3360 where μ = (pop) mean mass B1B0 0.0297 > 0.025 A0 Don't reject H0 M0 There is no evidence that mean mass has changed A0 S H0: The (pop) mean mass = 3360 H1: The (pop) mean mass ≠ 3360 B0B0 0.97 < 0.9875 A1 Accept H0 M0 There is no evidence that mean mass has changed A0 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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