Y422/01 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 Y422/01 Mark Scheme June 2023 8 some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. e. The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. Y422/01 Mark Scheme June 2023 9 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. 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. Y422/01 Mark Scheme June 2023 10 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. Y422/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 1 (a) P(4 sixes) = ( 1 6) 4 = 1 1296 B1 1.1 AG [1] 1 (b) Random or independent Fixed probability of success Fixed number of trials The random variable is the number of successes oe There are only 2 outcomes B1 B1 2.4 2.4 For two correct statements For at least 4 correct comments Because n (= 10 000) is large and p (= 1 1296) is small a Poisson distribution is also appropriate B1 2.4 For explanation of Poisson. Must mention Poisson. . [3] 1 (c) Poisson( 10000 1296 ) or Poisson(7.716…) or Poisson( 625 81 ) M1 3.3 P(X = 10) = 0.0919 A1 1.1 BC 0.091864… P(X > 10) = 0.157 A1 1.1 BC 0.156963… [3] 1 (d) P(A person does not throw 4 sixes in 20 tries) = ( 1295 1296) 20 = [0.98468 … ] M1 3.1b B(50, 1 – 0.98468…) [=B(50, 0.015319…)] M1 3.3 Or Poisson approx. mean 0.765975 (from 50×0.015319) P(No more than 2) = 0.9586 A1 1.1 BC 0.958647… Poisson approx. leads to 0.9573… Allow awrt 0.96 NB Using B(50, 20/1296 or 5/324) gets answer 0.9579 but scores zero [3] NB: Throughout the paper allow probabilities given to 2sf if the distribution being used is stated. Y422/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 2 (a) Prediction for 5 C is 960 or 962 B1 1.1 Only allow 1 mark max if either given to more than 4sf Prediction for −4 C is 1050 or 1054 B1 1.1 [2] 2 (b) Although prediction for 5 °C lies within the data (interpolation), the points do not lie too close to the line, so it is not too reliable. B1 2.2a Allow first B1 for any correct comment about 5°C Condone ‘Near the centre of the data’ and the value of r2 is not too close to 1 so the estimate is only moderately reliable. B1 3.5b Allow second B1 for all 3 correct comments about 5°C The prediction for −4 °C is even less reliable since it is an extrapolation. B1 3.5b [3] Y422/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 3 (a) P(none in 3 serves) = 0.453 = 0.0911 B1 1.1 (0.091125…) [1] 3 (b) B(20, 0.55) M1 3.3 s.o.i. P(Ten or more) = 1 – 0.2493 = 0.7507 A1 1.1 BC [2] 3 (c) P(fifth serve ) = 0.454 × 0.55 M1 3.3 = 0.0226 A1 1.1 (0.022553…) [2] 3 (d) 4 successes in first nine so B(9, 0.55) then × 0.55 M1 3.1b P(fifth on tenth serve) = 0.2128 × 0.55 = 0.117 A1 1.1 BC (0.117016…) [2] 3 (e) 𝑝(1 −𝑝) = 0.2496 M1 3.1a 𝑝2 −𝑝+ 0.2496 = 0 M1 1.1 For expressing in the form 𝑎𝑥2 + 𝑏𝑥+ 𝑐= 0 𝑝= 0.48 or 𝑝= 0.52, so answer 𝑝= 0.48 A1 2.2a [3] Y422/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 4 (a) 10 sheets thickness ⁓ N(10 × 3.125, 10 × 0.032) M1 3.3 For Normal and mean i.e. N(31.25, 0.009) A1 1.1 For correct variance P(thickness < 31) = 0.0042 B1 3.4 BC (Exact answer 0.00420…) [3] 4 (b) Thickness ~ N(10 × 3.125, 102 × 0.032) i.e. N(31.25, 0.09) B1 3.3 For correct distribution P(thickness < 31) = 0.2023 B1 1.1 BC (Exact answer 0.202328……) Alternative method P(10 sheets < 31) = P(1 sheet <3.1) B1 P(thickness < 31) = 0.2023 B1 BC [2] 4 (c) One of each ⁓ N(3.125+3.117+3.109, 3 × 0.032) M1 3.3 For method i.e. N(9.351, 0.0027) A1 1.1 P(Total ≥ 9.4) = 0.1728 B1 1.1 BC (Exact answer 0.172839…) [3] 4 (d) Distribution of difference of 10 sheets of each has oe mean = 10 × 3.125 − 10 × 3.117 M1 3.3 Method for mean variance = 10 × 0.032 + 10 × 0.032 M1 1.1 Method for variance [so distribution is N(±0.08, 0.018)] P(difference < 0) = 0.2755 A1 3.4 BC Allow 0.275 or 0.276 (exact answer 0.27549…) [3] Y422/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 5 (a) Est of pop variance = 15065−7652 40 39 M1 1.1 Accept denominator of 40 rather than 39 for M1 leading to est of var = 10.859 or sd =3.295 = 11.138 =[ 3475 312 ] A1 1.1 Allow 11.1 or sd = 3.34 (3.33736…) Confidence interval is 19.125 B1 1.1 Or 765 40 seen anywhere ± 1.96 M1 3.3 Accept t-value of 2.02 to 2.03 × √11.138 40 M1 1.1 For √𝑡ℎ𝑒𝑖𝑟 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 40 = 19.125 ± 1.034 or (18.09, 20.16) A1 [6] 3.4 Accept based on t-distribution from correct working t = 2.02 leads to (18.06, 20.19) t = 2.0227 leads to (18.044, 20.206) t = 2.03 leads to (18.0422, 20.1961) 5 (b) It does to some extent because the confidence interval does contain 20 E1 3.5a Must be unassertive FT their interval But there is a lot of variation so estimates are not very accurate. E1 [2] 2.2b Allow other suitable comments. Allow valid comment on variation relating to the claim EG Might not be representative of the population. Lower confidence level could result in 20 being outside. Near the end of interval so not very reliable. The interval is wide. 5 (c) If the population from which the sample was drawn was Normally distributed then you could have formed it using the t distribution. E1 2.2a Allow ‘don’t know if population is Normally distributed so cannot use the t distribution. If not then you could not have formed it, due to the small sample size E1 [2] 2.4 Max 1 mark if t distribution not mentioned. Marks are independent so can get mark for second comment only 5 (d) 19.25 < µ < 20.11 B1 1.1 [1] 5 (e) It suggests that he may be correct because the interval again contains 20 E1 2.2b Must be unassertive. Do not allow ‘Amari is correct’ And the variance is much lower this time E1 3.5a Allow ‘The centre of the interval is nearer to 20 this time’ Allow ‘This interval is narrower. [2] Y422/01 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 6 (a) Scatter diagram appears to be roughly elliptical E1 3.5a so the distribution may be bivariate Normal E1 2.4 Condone ‘The data is bivariate normal’ or ‘The data comes from a bivariate normal distribution’?? [2] 6 (b) DR 1 20 4713.62 342.10 273.65 xy S = −   [= 32.837] M1 1.1a For xy S 2 1 20 5989.53 342.10 xx S = −  [= 137.91] M1 1.1 For either xx S or yy S 2 1 20 3919.53 273.65 yy S = −  [= 175.31] 32.837 137.91 175.31 xy xx yy S r S S = =  M1 3.3 For general form including sq. root = 0.211 A1 1.1 Allow awrt 0.21 without incorrect working [4] 6 (c) H0:  = 0, H1:  ≠ 0 B1 3.3 For both hypotheses Allow any symbol in place of ρ if defined as population pmcc where  is the population pmcc between x and y B1 2.5 For defining  NB Hypotheses in words only get B1 unless population mentioned ‘between x and y’ may be seen in the hypotheses alongside the correct hypotheses in symbols, rather than in the definition of  For n = 20, the 5% critical value is 0.4438 B1 3.4 For correct critical value Since 0.211 < 0.4438 the result is not significant, so there is insufficient evidence to reject H0 M1 1.1 For comparison and conclusion There is insufficient evidence at the 5% level to suggest that there is correlation between download and upload speed A1 2.2b Must be in context FT their pmcc and cv (provided between -1 and +1) for M1 but not for A1 [5] Y422/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 6 (d) Because download speed would not have a probability distribution (so the distributional assumption could not be met). E[1] 2.2a Allow ‘the situation is not sampling from a bivariate Normal population’ or ‘Statistical inference cannot be carried out for non-random data’. Allow other suitable answers. [1] E0 for ‘pmcc can only be found for random variables’ Question Answer Marks AO Guidance 7 (a) The Normal probability plot is roughly straight and the p- value is not too low E1 1.1 No marks if Wilcoxon suggested which are both consistent with the data coming from a Normal distribution E1 2.2b A t test should be carried out (since this test requires the population to be Normally distributed) B1 3.3 Dependent on at least 1 of the previous marks [3] 7 (b) DR No marks for Wilcoxon except for mean and sd if found H0: μ = 1.0 H1: μ > 1.0 B1 1.1a Hypotheses in words only must include “population”. Allow H0: μ = 0.01 H1: μ > 0.01 Where μ is the population mean concentration B1 1.2 For definition in context. Must include population. Sample mean = 1.015 B1 1.1 Sample SD = 0.0981 B1 1.1 Test statistic is 1.015−1.0 0.0981/√12 M1 3.3 FT their mean and/or sd = 0.530 A1 1.1 Use of t11 M1 3.4 No FT if not t11 Can be implied by 1.796 or 2.201 OR p-value = 0.3034 and compare with 5% Critical value (1-tailed) at 5% level is 1.796 A1 1.1 0.530 < 1.796 not significant (do not reject H0) M1 2.2b Or confidence interval method [0.9641, 1.0659] Insufficient evidence to suggest that the mean concentration exceeds 1.0% E1 3.5a Answer must be in context FT their sensibly obtained test statistic and cv (provided from t11) for M1 but not for A1 [10] Y422/01 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 8 (a) 2 × 3 10 × 7 10 M1 1.1 For 3 10 × 7 10 or 0.3 × 0.7 = 42 100 or 21 50 or 0.42 A1 1.1 [2] 8 (b) P(T ≤ 25) = 0.5 B1 1.1 [1] 8 (c) P(𝑇≤25) = 48 100 or 12 25 or 0.48 B1 1.1 P(𝑇> 35) = 7 100 or 0.07 B1 1.1 [2] 8 (d) DR E(X) = 5 B1 3.1a s.o.i. Var(X) = 1 12 (10 −0)2 M1 1.2 = 25 3 A1 1.1 Allow M0A0 SCB1 if 25 3 used below but not explicitly found without full explanation. E(T) = 25 Var(T) = 125 3 M1 1.1 Allow Var(T) = 5× their Var(X) [E(Y) = 25] Var(Y) = 125 300 [= 5 12] M1 1.1 Allow Var(Y) = their Var(T)/100 By CLT distribution is approx N (25, 5 12) M1 2.2a P(Y > 26) = 0.0607 A1 1.1 BC (0.060667…) Do not allow a continuity correction Allow equivalent method M1 for Var(T) = 125 3 , M1 for Var(total of 100 values) = 12500 3 , M1 for N(2500, 12500 3 ), A1 for P(Total > 2600) = 0.0607 [7] Y422/01 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 9 (a) DR H0: no association between where registered and passing distance H1: some association between where registered and passing distance B1 3.3 Allow hypotheses and conclusion in terms of independence, but not relationship or correlation which results in B0 and final E0 Context needed in at least one of the hypotheses Expected Local Non-local Under 1.5m 15.55 7.45 At least 1.5m 153.45 73.55 M1 A1 3.4 1.1 For at least 1 correct expectation For all correct Contribution Local Non-local Under 1.5m 0.8096 1.6893 At least 1.5m 0.0820 0.1712 B1 1.1 At least one correct value must be seen; this mark cannot be implied merely by a correct final value of X 2. Allow 1 error. 2 2.75 X = B1 1.1 Allow 2.76 from expected values given to 2dp Yates’ correction is not expected, but full marks if used correctly Contribution Local Non-local Under 1.5m 0.5975 1.2467 At least 1.5m 0.0605 0.1263 2 2.03 X = Use of 𝜒1 2 B1 2.5 Can be implied by correct critical value of 3.84 or by 2.706 or 5.024. Must be from chi-squared with 1 deg of freedom Critical value at 5% level = 3.84; B1 2.2b OR p-value = 0.09725 and compare with 5% 2.75 < 3.84 so not significant (do not reject H0) M1 2.2b FT their sensibly obtained test statistic and cv for M1 but not for A1 There is insufficient evidence to suggest that there is an association between where registered and passing distance. E1 [9] 3.5a If hypotheses wrong way around do not allow first or last two marks. Must mention ‘insufficient evidence’ in conclusion, not just ‘insufficient evidence to reject H0’ followed by a completely new sentence. 9 (b) Because the cyclist will know whether or not the car is locally registered when she notes the passing distance E1 2.2b Allow other reasonable answers Do not allow ‘She may not be able to estimate accurately’ [1] Y422/01 Mark Scheme June 2023 20 Question Answer Marks AO Guidance 10 (a) F(𝑥) = ∫ 4 15 ( 𝑎 𝑢2 + 3𝑢2 −7 2) 𝑥 1 d𝑢 M1 2.1 For attempt to integrate (Condone ito x). No need for limits. = 4 15 [−𝑎 𝑢+ 𝑢3 −7 2 𝑢] 1 𝑥 M1 1.1a For correct integral (Condone ito x) No need for limits. = 4 15 (− 𝑎 𝑥+ 𝑥3 − 7 2 𝑥) − 4 15 (−𝑎− 5 2) F(𝑥)= 0 for 𝑥< 1 F(𝑥) = 4 15 (− 𝑎 𝑥+ 𝑥3 − 7 2 𝑥+ 𝑎+ 5 2) for 1 ≤𝑥≤2 F(𝑥) =1 for 𝑥> 2 A1 A1 1.1 1.1 A1 for correct answer AEF Fully correct AEF [= 4 15 (𝑎(1 −1 𝑥) + 𝑥3 −7 2 𝑥+ 5 2)] Or using F(2) = 1 leads to 4 15 (− 𝑎 𝑥+ 𝑥3 − 7 2 𝑥+ 11 4 + 1 2 𝑎) [4] 10 (b) F(2) = 1 ⇒4 15 (−𝑎 2 + 8 −7) −4 15 (−𝑎−5 2) = 1 ⇒2 15 𝑎+ 14 15 = 1 M1 3.1a For setting their F(2) = 1.. Their F(x) must be i.t.o. a [⇒F(𝑥) = 4 15 (− 0.5 𝑥+ 𝑥3 − 7 2 𝑥+ 3) = 1] 𝑎= 0.5 A1 1.1 NB Alternative method F(2) −F(1) = 1 leads to same equation if correct but FT their F(𝑥) for M1. Condone a wrong constant term in F(𝑥) which leads to a correct value of a [2] 10 (c) 4 15 (− 𝑎 𝑚+ 𝑚3 − 7 2 𝑚) + 0.8 = 0.5 Or 4 15 (−𝑎 𝑚+ 𝑚3 −7 2 𝑚+ 3) = 0.5 M1 2.1 For setting their F(m) = 0.5 Condone F(x) = 0.5 Or ∫ 4 15 ( 𝑎 𝑢2 + 3𝑢2 − 7 2) 𝑚 1 = 0.5 followed by attempt at integration 8 (−0.5 𝑚+ 𝑚3 −7 2 𝑚) = −9 ⇒8𝑚4 −28𝑚2 + 9𝑚−4 = 0 A1 1.1 AG Must show at least one line of correct working [2] Y422/01 Mark Scheme June 2023 21 Question Answer Marks AO Guidance 10 (d) F(1.735) = 0.4965, F(1.745) = 0.5119 B1 3.4 For either OR evaluation of 8𝑚4 −28𝑚2 + 9𝑚−4 for 1.735 (= −0.1797) and 1.745 (= 0.6217), so change of sign So median is 1.74 to 2 dp E1 1.1 No marks for calculator answer [2] 10 (e) E(𝑋) = ∫ 4 15 𝑥(0.5 𝑥2 + 3𝑥2 −7 2) d𝑥 2 1 M1 1.1 Allow with wrong value of a or ito a. Correct limits required. = 1.69 A1 1.1 BC (1.692419…) or 2 15 ln 2 + 8 5 [2] 10 (f) f ′(𝑥) = 4 15 (−1 𝑥3 + 6𝑥) M1 3.1a Allow with wrong value of a f ′′(𝑥) = 4 15 ( 3 𝑥4 + 6) M1 1.1 Allow with wrong value of a Alternatives for second mark: When 𝑥= √1 6 4 the curve has a minimum; For 1  x  2, 6x is clearly greater than 1 𝑥3 so f ′ is positive hence f is increasing so mode = 2 A1 1.1 Allow third mark if full answer based on the above. Allow SCB1 for mode = 2 without proper justification [3] Y422/01 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 11 (a) E(X) = p B1 1.1 E(𝑋2) = 𝑝 M1 1.1 Var(𝑋) = 𝑝−𝑝2 A1 1.1 [3] 11 (b) 𝑌= 𝑋1 + 𝑋2 + ⋯+ 𝑋50 B1 3.1a where p = 0.2, giving E(𝑋𝑖) = 0.2, Var(𝑋𝑖) = 0.2 × (1 −0.2) = 0.16 B1 1.1 Seen E(𝑌) = E(𝑋1) + E(𝑋2) + ⋯+ E(𝑋50) M1 2.1 = 50 × E(𝑋𝑖) = 50 × 0.2 = 10 A1 1.1 AG SCB1 for use of results of part a and E(Y) = 50×E(X), Var(𝑌) = Var(𝑋1) + Var(𝑋2) + ⋯+ Var(𝑋50) M1 2.1 (Where the Xi are all independent) . Not simply np. = 50 × Var(𝑋𝑖) = 50 × 0.16 = 8 A1 1.1 AG SCB1 for use of results of part a and Var(Y) = 50×Var(X). Not simply npq. [6] 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. Registered company number 3484466. 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