Final. Mark Scheme. Mathematics/Statistics MS/SS1B. (Specification 6360/6380) Statistics 1B. General Certificate of Education (A-level) June 2013



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Version 1.0 General Certificate of Education (A-level) June 2013 Mathematics/Statistics MS/SS1B (Specification 6360/6380) Statistics 1B Final Mark Scheme

Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all examiners participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every examiner understands and applies it in the same crect way. As preparation f standardisation each examiner analyses a number of students scripts: alternative answers not already covered by the mark scheme are discussed and legislated f. If, after the standardisation process, examiners encounter unusual answers which have not been raised they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a wking document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this Mark Scheme are available from: aqa.g.uk Copyright 2013 AQA and its licenss. All rights reserved. Copyright AQA retains the copyright on all its publications. However, registered schools/colleges f AQA are permitted to copy material from this booklet f their own internal use, with the following imptant exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even f internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester 5 6EX.

Key to mark scheme abbreviations M mark is f method m dm mark is dependent on one me M marks and is f method A mark is dependent on M m marks and is f accuracy B mark is independent of M m marks and is f method and accuracy E mark is f explanation ft F follow through from previous increct result CAO crect answer only CSO crect solution only AWFW anything which falls within AWRT anything which rounds to ACF any crect fm AG answer given SC special case OE equivalent A2,1 2 1 ( 0) accuracy marks x EE deduct x marks f each err NMS no method shown PI possibly implied SCA substantially crect approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method f any marks to be awarded. Where the answer can be reasonably obtained without showing wking and it is very unlikely that the crect answer can be obtained by using an increct method, we must award full marks. However, the obvious penalty to candidates showing no wking is that increct answers, however close, earn no marks. Where a question asks the candidate to state write down a result, no method need be shown f full marks. Where the permitted calculat has functions which reasonably allow the solution of the question directly, the crect answer without wking earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a crect method f any marks to be awarded.

1(a)(i) Mean = 62.2 to 62.3 B1 AWFW (62.25) SD = 17.4 to 17.6 16.7 to 16.9 B1 2 AWFW (17.519 16.774) (ii) Mean = 16.77 to 16.84 BF1 AWFW (16.806) F on (a)(i) only providing 45 < mean < 65 SD = 9.66 to 9.78 9.27 to 9.39 BF2 3 AWFW (9.733 9.319) F on (a)(i) only providing 10 < SD < 20 (b) r xy = 0.997 B1 CAO Award on value only; igne any explanation wking r xy = r uv with no value stated B0 r not affected by change(s) in/different units Accept Fmula It f r and reference to linear is not necessary r not affected by linear scaling B1 2 Accept Fmula It f r but reference to linear is necessary Scaling/coding/transfmation/change/ conversion to u and v is linear Total 7 OE; but reference to linear is necessary 4

Accept percentage equivalents in (a) Weight, X ~ N(421, 2.5 2 ) 2(a)(i) P(X = 421) = 0 zero nought 0% B1 CAO; accept nothing else but igne additional wds providing that they are not contradicty (eg impossible so = 0) (ii) P(X < 425) = 425 421 P Z < 2.5 Standardising 425 with 421 and 2.5 but allow (421 425) = P(Z < 1.6) = 0.945 to 0.946 A1 AWRT (0.94520) (iii) P(418 < X < 424) = P( a < Z < a) = P(Z < a) (1 P(Z < a)) 2 P(Z < a) 1 = 0.885 (1 0.885) = 0.885 0.115 = 2 0.885 1 A1 OE; a = 1.2 crect standardising are not required May be implied by 0.885 (AWRT) seen anywhere by a crect answer AWRT (0.88493/0.11507) Implied by a crect answer = 0.769 to 0.77 A1 6 AWFW (0.76986) (b) 0.98 z = 2.05 to 2.06 B1 AWFW (2.0537) x 421 2.5 = 2(.0) to 2.4 x = 426 to 426.3 A1 3 Standardising x with 421 and 2.5 but allow (421 x); and equating to a z-value (igne sign) AWFW (426.13) Must be consistent signs throughout (c) 0.01 z = 2.33 to 2.32 B1 AWFW; (igne sign) ( 2.3263) z = 410 µ 3.0 2.5 Standardising 410 with µ and (3.0 2.5) but allow (µ 410) 410 µ Equating to a z-value (igne sign) 3.0 = 2.6 to 2.3 A1 µ = 417 Adep1 4 Total 13 AWRT (416.98) Dependent on previous A1 Must be consistent signs throughout 5

Accept percentage equivalents except f 27 3(a)(i) O ~ B(40, p) P(NS 10) = 0.97 B1 1 AWRT (0.9701) (ii) P(LPE 25) = 1 (0.9231 0.9597) Requires 1 Accept 3 dp rounding Can be implied by (0.0769 to 0.077) but not by (0.04 to 0.0403) = 0.077 A1 2 AWRT (0.0769) (iii) 40 0.175 0.825 2 2 38 P(UPE = 2) = ( ) ( ) Crect expression; may be implied by a crect answer Igne extra terms = 0.016 A1 2 AWRT (0.0160) (iv) p = 0.85 0.50 = 0.35 B1 CAO; award on value only May be implied by any of four probabilities below by a crect answer P(10 < X < 15) = 0.5721 0.6946 (p 1 ) MINUS 0.1215 0.0644 (p 2 ) Accept 3 dp rounding Accept 3 dp rounding = 0.45 to 0.451 A1 4 AWFW (0.4506) (b) p = 0.85 0.175 = 0.675 p = 0.325 B1 CAO; may be implied by 27 Each can be found in several ways CAO; may be implied by 13 27 Number = 40 0.675 = 27 B1 2 CAO; can be found in several ways Total 11 6

4(a)(i) 24.15 in r gy = = 0.91 to 0.911 (a)(i) (a)(ii) (c)(i) 0.1196 5880 A1 AWFW (0.91067) (ii) 10.25 r ly = = 0.64 to 0.0436 5880 A1 3 AWFW (0.64017) 0.641 (b) (Very) Strong positive crelation Bdep1 Dependent on 0.9 r gy < 1 (Some) Moderate positive crelation between girth and weight and/ length and weight Bdep1 Dependent on 0.6 r ly 0.7 Bdep0 f any mention of strong B1 3 At least one interpretation in context (c)(i) 5662.97 r xy = 5656.15 5880 = 0.98 to 0.982 B1 AWFW (0.98196) Most strongly crelated with y is x Bdep1 2 CAO; dependent on 0.97 r xy < 1 (ii) x = 69.3 1.25 2 1.15 = 124 to 125 A1 2 AWFW (124.52) (iii) b = 5662.97 5656.15 = 1 to 1.002 A1 116/115.4 (= 1.005) M0 A0 AWFW (1.00121) a = 116 115.4b = 0.3 to 0.6 B1 3 AWFW (0.46085) (iv) r xy /nearly/almost/close to (+)1 very strong/almost exact (positive) crelation (Stating r xy =0.98 to 0.982 Bdep0) b =/ /nearly/almost/close to (+)1 a /nearly/almost/close to 0 (Stating a = 0.4 to 0.6 Bdep0) Bdep1 Bdep1 Bdep1 OE Dependent on 0.97 r xy < 1 OE; strong is not sufficient OE; must reference value of 1 unity Dependent on A1 in (c)(iii) OE; must reference value of 0 igin Dependent on B1 in (c)(iii) Estimate (not 'it' 'this' 'value', etc) is (very/highly/likely to be) accurate/precise/ reliable (almost) exact/crect Bdep1 4 Total 17 OE; dependent on scing at least 2 of the previous 3 marks in (c)(iv) Fairly accurate, good approximation, (quite) likely, (very) close, reasonable, etc Bdep0 7

5(a)(i) P(A = 2) = 0.90 0.95 = 0.85 to 0.86 B1 AWFW (0.855 171/200 OE) (ii) P(A = 1) = (0.90 0.05) + (0.10 0.95) = 1 [0.855 + (0.10 0.05)] Do not igne extra terms = 0.14 A1 3 CAO (7/50 OE) (b)(i) P(A W D W ) = 0.90 0.80 = 0.72 A1 2 CAO (18/25 OE) (ii) P(A B D B ) = (b)(i) 0.95 ( 1) = 0.90 0.80 0.95 ( 1) = (a)(i) 0.80 0.68 to 0.685 A1 2 AWFW (0.684 171/250 OE) (iii) P(A T D T ) = 0.95 0 = 0 B1 1 CAO; award on value only (iv) P(neither) = P([A W D W ] [A T D T ]) (1 0.90) (1 0.15) (1 0.95) (1 0) P(neither) = P(A W A T ) P(D W A W ) P(D T A T ) (1 0.90) (1 0.95) (1 0.15) (1 0) m1 () (m1) Accept 0.085 17/200 OE Award and m1 on value(s) only Accept 0.05 1/20 OE Accept 0.005 1/200 OE Award and m1 on value(s) only Accept 0.85 17/20 OE = 0.085 0.05 0.005 0.85 OE = 0.0042 to 0.0043 A1 3 AWFW (0.00425 17/4000 OE) Total 11 8

6(a)(i) 497.5 x = = 19.9 B1 CAO 25 98% (0.98) z = 2.32 to 2.33 B1 AWFW (2.3263) σ CI f µ is x ± z n Used with z (2.05 to 2.58), x (497.5 19 to 21) and σ (0.4) and n with n > 1 Thus 04. 19. 9 ± 2. 3263 A1 25 z (2.05 to 2.06 2.32 to 2.33 2.57 to 2.58), x (19.9) and σ (0.4) and 25 24 Hence 19.9 ± 0.2 (19.7, 20.1) A1 5 CAO/AWRT (0.186104) AWRT (ii) Clear crect comparison of 20 with CI eg 20 is within CI LCL < 20 < UCL so Agree with claim no reason to doubt claim F on CI providing it contains 20 BF1 Quoting values f CI is not required Bdep1 2 OE; dependent on previous BF1 (iii) Weight of sand in a bag X/x iginal distribution parent population is nmal B1 1 It/mean/data/sample/infmation/sand is nmal B0 Reference only to sample size standard deviation B0 9

Accept percentage equivalent probabilities 6(b)(i) Y ~ N(25.25, 0 35 2 ) V(mean) = 0.35 2 /10 0.0122 to 0.0123 SD (mean) = 0.35/ 10 0.11 to 0.111 25 25.25 P( Y 25) P < = Z < 0.35 10 B1 CAO/AWFW (0.01225) CAO/AWFW (0.11068) Standardising 25 using 25.25 and 0.35/ 10 OE but allow (25.25 25) = P(Z < 2.25877) = 1 P(Z < 2.25877) m1 Crect area change an answer < 0.5 = 1 (0.98809 to 0.98778) = 0.011 to 0.013 A1 4 AWFW (0.01195) (0.987 to 0.989) B1 m0 A0 (ii) 25 25.25 P( Y > 25) = P Z > 0.35 Standardising 25 using 25.25 and 0.35 but allow (25.25 25) = P(Z > 0.71429) = P(Z < 0.71429) = 0.761 to 0.764 A1 AWFW (0.76247) (0.236 to 0.239) A0 P(Y > 25 in each of 10) = p 10 Any p 10 providing 0 < p < 1 = 0.065 to 0.068 A1 4 AWFW (0.06641) Total 8 TOTAL 75 10