A-level Mathematics. MS03 Statistics 3 Final Mark scheme June Version/Stage: v1.0

Transcription

1 A-level Mathematics MS03 Statistics 3 Final Mark scheme 6360 June 017 Version/Stage: v1.0

2 Mark schemes are prepared by the Lead Assessment Writer 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 associates 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 associate understands and applies it in the same crect way. As preparation f standardisation each associate 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, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. 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 017 AQA and its licenss. All rights reserved. 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.

3 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 A,1 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. 3 of 10

4 General Notes f MS03 GN1 GN GN3 GN4 GN5 GN6 There is no allowance f misreads (MR) miscopies (MC) unless specifically stated in a question In general, a crect answer (to accuracy required) without wking sces full marks but an increct answer ( an answer not to required accuracy) sces no marks In general, a crect answer (to accuracy required) without units sces full marks When applying AWFW, a slightly inaccurate numerical answer that is subsequently rounded to fall within the accepted range cannot be awarded full marks Where percentage equivalent answers are permitted in a question, then penalise by one accuracy mark at the first crect answer but only if no indication of percentage (eg %) is shown In questions involving probabilities, do not award accuracy marks f answers given in the fm of a ratio odds such as 13/47 given as 13:47 13:34 GN7 Accept decimal answers, providing that they have at least two leading zeros, in the fm c 10 -n (eg as ) GN8 Where a candidate's response to a part of a question is simply to label the part (eg (d)(i)) with nothing else (ie no attempt at a solution), then this is still treated as a response and marked as 0 rather than NR. Also, deleted wk, if not replaced, should be marked and not treated as NR. 4 of 10

5 1 98% z =.3 to.33 AWFW (.363) zσ = < n Require ( ) ( 00 0.).3 to ( = < ) to.06 n OE; allow no Accept 0.33 and 0. n ( ) ( ).3 to = > 100 OE; f n Accept 0.33 and 0.1 n (= >) 58.6 to CAO 4 Total 4 Sample is selected at random OE CI f p is ˆp = 77/440 7/ CAO; igne notation 99% z =.57 to.58 AWFW (.5758) ( z ) ± -value within list a to ±.3 to OE Expression f a ± (0.18, 0.) 6 CAO/AWRT ( ) AWRT Total 6 5 of 10

6 3 (a) H 0 : (λ µ) = 8 H 1 : (λ µ) < 8 Both; accept λ/p = P(FE 5 λ = 8) = Any one = AWRT < 0.05 m1 Crect comparison (PI) (b) Accept H 0 no evidence, at 5% level, that upgrade has reduced average number of faulty envelopes per pack H 0 : (λ µ) = (8 400) H 1 : (λ µ) < (8 400) Adep1 () 5 Dep on previous 4 marks OE; but must not be definitive Both; iff not sced in (a) 1% (0.01) z =.3 to.33 p-value of z-calculated = < 0.01 AWFW (.363) AWRT ( to ) z z = = = =.57 to ( x λ) λ ( n) ( x( 0.5) ) A crect expression ( ± λ) λ ( n) CAO (.6) AWFW (.575) Note Reject H 0 evidence, at 1% level, that Dep on previous 4 ( 5) marks refurbishment has reduced average number of Adep1 faulty envelopes per pack OE; but must not be definitive 5 1 Use of Poisson and P( 348 λ = 400) = (AWRT) < 0.01 () (0.01) (Po(400)) A (0.0043) dep1 Total 10 6 of 10

7 4 (a) P(C1) = = CAO; (77/00) (1) (b) P(M C1) = (a) (c) P(E C) = ( ) + ( ) + ( ) = AWRT; (7/77) ( ) () Numerat Denominat (d) P(W C3 ) = P ( M C3 ) + P( E C3 ) P( C3 ) = to AWFW; (85/14) ( ) (3) = ( ) + ( ) ( ) + ( ) + ( ) = () () () Numerat Denominat; accept [Num + ( )] [(a) + (Den(c)] 1 Numerat Denominat = 0.89 AWRT; (14/139) (0.8909) (4) Total 10 7 of 10

8 5(a) (i) E() = (0 0.15) + (1 0.4) + ( 0.3) + (3 0.15) = = 1.45 CAO; (9/50) E( ) = (1 0.4) + ( 0.3) + (3 0.15) = =.95 59/0 PI Var() = m1 Use of {E( ) (E()) } iff > 0 (ii) = to AWFW; (339/400) (0.8475) 4 Var(Y) = = 1. to 1.3 AWFW; (491/400) (1.75) Cov(, Y) = FT only on E() from (a)(i) (iii) C(, Y) = = to 0.33 AWFW; ( 133/400) ( 0.335) FT from (a)(i) & (ii) = 0.36 AWRT ( ) (b) (i) E(T) = =.3 CAO Var(T) = PI; use of σ σ + Y ± Y Cov(, ) σ + σy ± σ σy C( Y, ) ; here in (b)(ii) (ii) = 1.41 AWRT (3) E(D) = = 0.6 CAO Var(D) = =.74 AWRT () 5 Total 14 8 of 10

9 6 n x n x E = x p 1 p = Used; igne limits until x (a) ( ) ( ) x= 0 Note (b) ( n 1! ) x 1 n x ( 1 ) = ( x 1! ) ( n x)! np p p x= 1 Using u = x 1 and m = n 1 gives m m! u m u np p ( 1 p) = np 1 = np u! m u! u= 0 ( ) 1 Other valid derivations are possible and acceptable y e E( Y( Y 1) ) y( y 1) λ λ = y! y= 0 Using v = x gives y e λ λ λ = λ y! ( ) y= 3 Fact of np plus n! to (n 1)! and x! to (x 1)! Fully complete & crect derivation but allow min slips in limits AG = Used; igne limits until = v = e v! λ v λ λ 1 = λ Fact of λ plus x! to (x )! and fully complete & crect derivation but allow min slips in limits Var(Y) = E(Y(Y 1)) + E(Y) (E(Y)) (E(Y)) = λ + λ Fully crect deduction AG so Var(Y) = = λ + λ λ = λ 3 Note 1 Other valid derivations are possible and acceptable (c)(i) A ~ B(50, 0.005) 50 P(FS = 1) = ( )( ) 49 1 PI = to AWFW ( ) SC 1 Use of 0.5e 0.5 = (ii) B ~ B(50, 0.005) Po(1.5) PI 1.5 P(FS < ) = e ( 1 1.5) + PI; a crect Poisson expression (iii) C ~ B(50000, 0.005) N(50, 48.75) = to AWFW ( ) 3 PI: nmal and 50 CAO CAO P(FS > 40) = P Z > = Allow and/ 50 Fully crect expression SC P(Z > ) = 0.75 to 0.77 Adep1 Dependent on (0.7653) 5 1 (Po(50)) ( (AWRT)) (0.74 (AWRT)) max of 3 marks Total 16 9 of 10

10 7 (a) H 0 : µ µ Y = 1.5 H 1 : µ µ Y > 1.5 Award f µ µ Y = 0 Allow any valid notation 5% (0.05) z = 1.64 to 1.65 p-value of z-calculated = 0.08 > 0.05 AWFW (1.6449) AWRT (0.0810) x = 4.53 and y =.88 CAO: both s = and s Y = 0.10 s = and s Y = AWRT; both ( and 0.100) AWRT; both ( and ) z = ( ) = = 1.40 Numerat; allow ( ) Denominat; OE AWRT ( ) Accept H 0 no evidence, at 5% level, that Dep on previous 8 marks difference in mean lengths is me than 1.5 Adep1 metres OE; but must not be definitive 9 Notes 1 Invalid pooling of variances z =1.31 M0 A0 Adep 0 (max = 6) Use of σ /σ (0.93/0.541 and 0.116/0.341) M0 A0 Adep 0 (max = 6) unless ( (n 1)) in z 3 Omission of 1.5 throughout z = 15.4 B0 M0 A0 Adep0 (max = 5) (b) CV is given by: ( Y) = 1.64 to 1.65 Fully crect equality (c) ( Y) = ( to 1.65) = (0.175 to 0.177) = 1.68 Power = P(reject H 0 H 0 false) = (( Y ) ( µ µ Y ) ) P > 1.68 = 1.85 = AWRT; AG ( ) P Z > P Z > Crect use of 1.68 (OE) Igne sign of inequality Crect numerical expression = P(Z > 1.58 to 1.63) AWFW (PI); igne sign ( ) = 0.94 to 0.95 AWFW ( ) 4 Total of 10