Mark Scheme (Results) Summer 017 Pearson dexcel GC Further Mathematics Statistics S3 (6691)
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General Marking Guidance All candidates must receive the same treatment. xaminers must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. xaminers should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. xaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. xaminers should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLSS the candidate has replaced it with an alternative response.
DXCL GC MATHMATICS General Instructions for Marking 1. The total number of marks for the paper is 7.. The dexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epn. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as A1 ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.
. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.
Question Number 1(a) Scheme Notes Marks -(accurate) estimates for each strata / job -more representative of the population -reflects population structure Any 1 oe (1) (b) Total staff=70 May be implied by calculations 7 For one correct calculation, follow Managers 40 4 70 through their 70. 108 Drivers 40 6 70 180 Administrators 40 10 70 360 4, 6, 10, 0 only. Must identify which Warehouse 40 0 70 job the values relate to. A1 (c) Label all managers 1 7 o.e. Using random numbers in range 1-7 or 0-71 select 4 (managers). Idea of sampling frame or list of managers. Need not give the specific term. Use of random numbers to select required number of managers. Must mention use of random numbers or some random selection process. If they are describing systematic sampling score B0. (3) () Total 6
Question Number.(a) (b) Scheme Notes Marks 4 a 100 1. 360 For one correct calculation 90 b 100 360 13 c 100 37. 360 1.,, 37. only A1 H 0 : Continuous uniform distribution is a suitable model H: 1 Continuous uniform distribution is not a suitable model ( ) 18 1..4.9 16 1. 0.98 0.48 18 1.96 1.96 9 37. 1.97.43 19 1. 3.38 8.88 100 100 10.67 110.67 ( ) 10.67 or 100 10.67 Both Accept in terms of Linda s claim. Accept U[0,360). ( ) for attempting with at least 3 correct expressions or values. or A1 for all values correct to dp or as fractions. Can be implied by correct answer if all not listed. awrt 10.7 A1 A1 () 1 4, (%) 9.488 4 (10.7>9.488) Reject H 0 Linda s claim is not supported. 4 can be implied by awrt 9.488 seen or p awrt 0.03 Correct for their test stat and cv or thier p to %. Can be implied by correct conclusion from their test stat and cv. A correct comment suggesting that continuous uniform model is not suitable or Linda s claim is not correct. Linda s claim must be described in full if Linda is not mentioned. Follow through from their test stat and cv, but hypotheses must be correct.,ft A1ft (8) Total 10
Question Number 3. (a) (b) Scheme Notes Marks Senior Junior d d for an attempt to rank Judge Judge judges lists ( at least 4 correct A 1 3-4 for each judge) B 1 1 1 A1 for correct rankings for C 4-1 1 A1 both (may be reversed). Can D 3 1 1 6 6 0 0 be implied by correct d or r s F 4 1 1 8 N.B. Table could be ordered in terms of Senior Judge. d 8 or 6 8 or 6 or correct d row A1 for use of the correct formula, follow through their d (Dependent on 1st ) r s 68 7 1 0.771 63 3 If answer is not correct, a correct expression is required. A1 exact fraction or awrt ( )0.771 Both hypotheses in terms of H 0 : 0 or. s H 1 : 0 Hypotheses just in words e.g. no correlation score B0 Critical value 0.886 Accept 0.887 if -tailed H 1. Follow through their r s and (0.771<0.886) so insufficient evidence to reject H 0 There is insufficient evidence to suggest a positive correlation between the judges. (c) (For positive correlation c.v.is 0.886>0.771) Training of junior judge was ineffective. their c.v. if cv <1 and rs 1 A correct contextualised comment that includes judges. o.e. Follow through from their cv and r s da1 A1 ft ft () (4) (1) Total 10
Question Number 4(a) 70 4, 10 Scheme Notes Marks 780 6 10 Row Total Column Total For one correct ; can Grand Total be implied by correct answers. (b) (c) 764 7864 30.7, 33.8 10 10 H 0 :Perceived (body) weight is independent of gender (no association) H:Perceived 1 (body) weight is not independent of gender (association) ( ) 0 17.8 0.48148 3.1481 4 0.166667 0.16667 30 30.7 0.01687 9.9688 16 18.7 0.3914 13.671 8 6 0.13846 30.138 34 33.8 0.0177 34.738 10 10 1.17637 11.1763 ( ) or 10 1.18 4, 6, 30.7, 33.8 only. A1 Both hypotheses required. Must mention Perceived, weight and gender at least once. Use of relationship or correlation or connection or link award B0. for at least correct terms (as in 3rd or 4th column) or correct expressions. A1 for all correct. Accept sf accuracy. Allow truncation e.g. 1.17 Awrt 1.18-1.19 A1 A1 (31)( 1), (10%) 4.60 can be implied by 4.60 seen ft (Accept H 0 ) Perceived (body) weight is independent of gender (no association) ( ) 36 0 3.9.9 0 0 0 0 64 0 3.9 81.9 10 10 7.84 17.84 A correct comment in context - must mention weight and gender. Condone relationship or connection here but not correlation. Follow through from their test stat and cv, but hypotheses must be correct. for i 0, could be implied. for combining values and for ( ) attempting or with at least correct expressions or values. A1 for all correct, can be implied by correct answer below. A1ft () (7) A1 ( ) or 10 7.84, (.%) 7.378 0.0 or.% A1 Awrt 7.84 A1 () Total 14
Question Number. (a) (b)(i) (ii) (c) Scheme Notes Marks 60 x 4 1 4 cao Use of complete, correct formula 1 and attempt to substitute. s 1946 1 4 11.87... 14 A1 awrt 1 or 83,A1 7 (3) 10or"their s" 10 Accept use of xz, x 1.96 4.06 1,A1 1 A1 all correct. Accept x 083. Can be implied from correct interval ( 1.06,9.06) below. A1 Accept (089.94,0840.06) or expressed using words or as an (0896,084004) inequality. A1 Accept answers to the nearest minute ie (0830,0840). Paul samples times of buses randomly or independently of each Context required. other 0 / 0831 / 8.31(am) is contained in the confidence interval Paul s belief is not supported / 0831 arrival time is reasonable Award if comment about their interval is correct. nly accept above the lower limit of etc if the statement taken as a whole clearly means contained in. Must contain some context A1cao () () Total 10
Question Number Scheme Notes Marks 6. H : 8 (a) 0 new old H : 8 1 new old 83 74 8 z 7 6 0 40 1 z 1.86 0.38... Accept equivalent rearranged equation. Definitions of parameters must be clear e.g. use of 1 and without definitions scores B0. Accept x for new and y for old as defined in the question. Accept equivalent rearranged strict inequality. Definitions of parameters must be clear e.g. use of 1 and without definitions scores B0. Accept x for new and y for old as defined in the question. for attempting standard error. Condone swapping 7 and 6. Accept 6.86 and.8 for 7 and 6. d for 1/ their standard error A1 for awrt 1.86 NB -1.86 is A0. If 8 missing from H 1 then accept 9 z awrt16.7 and 0.38... must be consistent with their H 1. da1 cv z 1.6449 Accept or probability of 0.9686 (1.86>1.6449) so reject H 0 (r 0.9<0.9686) Correct comment in context. vidence to support engineer s claim (that Must mention engineers the new battery will last more than 8 hours claim or battery, old, longer than the old battery). new and 8. A1cso (7) (b) Sample sizes are large CLT guarantees sample means ( X and Y ) are approximately normally distributed. Must mention means and normal. No assumptions are being made so B0 if key to answer. () Total 9
Question Number 7(a) L N(110,0 ) and Scheme Notes Marks 1 3 M W L ( M M M ) ( W) 110 30 0 N(0,10 ) Allow L ( M M M) but not L 3M Can be implied by correct Var( W ). May use W L ( M M3) 1 for. Accept 0 if definition reversed. Accept ( W) 110 30 1 6 Attempt Var( W) Var( L) 3Var( M). Do not condone missing squares, cao. Attempting the correct probability and standardising with their mean and sd dependent on 1st. If values for W is not being used or not their variance score M0. Must use 1. Accept 0 6 P( W 0) PZ 700 P( Z.46769...) 0.0069 0.0071 by calc. awrt 0.007 A1 Var( W) 0 10 10 10 700 1 0 P( W 1) PZ 700,A1 (6) (b) X 3M L Can be implied by correct variance. ( X ) 30 110 0 Accept -0 if reversed. Var( X ) 3 10 0 1300 Attempt Var( W) 3 Var( M) Var( S). Do not condone missing squares, cao. Condone 10 3 0 for A0.,A1 0 P( X 0) P Z Attempting the correct 1300 probability and standardising with their mean and sd. d (c) P( Z 1.3867...) 0.9177 0.917 by calc. awrt 0.917-0.918 A1 P(all bags weigh more than 0 grams) = 1 1 0.031 3 d 0.031 ()
M or i1 10 N(0, ) M i ~ N(600,00) d 0 P( M d) PZ 0.031 or 10 d 600 P( T d) PZ 0.031 00 d 0 1.86(7...) 10 Both mean and variance required in either case. Can be implied below. Standardise using d, 0 and 10 or d, 600 and 00. quate to z value or d 600 1.86(7...) 00 d 8.3 awrt 8.3 A1 () ALT (c) Accept use d as difference to 0 provided 0 added to final answer: P(all bags weigh more than 0 grams) = 1 1 0.031 3 10 M N(0, ) or i1 M i ~ N(0,00) d P( M d) PZ 0.031 or 10 d P( T d) P Z 0.031 00 d 1.86(7...) 10 0.031 Both mean and variance required in either case. Can be implied below. Standardise using d and 10 or d and 00. quate to z value d or 1.86(7...) 00 d 0 8.3 8.3 awrt 8.3 A1 () Total 16
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