Mark Scheme (Results) Summer Pearson Edexcel International A Level in Statistics S2 (WST02/01) [Type here]

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1 Mark Scheme (Results) Summer 07 Pearson Edexcel International A Level in Statistics S (WST0/0) [Type here]

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: June 07 Publications Code WST0_0_706_MS All the material in this publication is copyright Pearson Education Ltd 07

3 General Marking Guidance All candidates must receive the same treatment. Examiners 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. Examiners 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. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners 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 UNLESS the candidate has replaced it with an alternative response.

4 PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75. The Edexcel 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.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. 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) d or dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given or d The second mark is dependent on gaining the first mark. All A marks are correct answer only (cao.), unless shown, for example, as A 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.

5 5. 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 NOT 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. 5

6 June 07 WST0 STATISTICS Mark Scheme Question Scheme Marks.(a) X ~ Po( ) B - P(X = 0) = e = awrt 0.77 B () (b) [(P(X )) ]= ( ) = awrt M A () (c) Y ~ B(7, ) P(Y = 5) = 7C5 (0.7788) 5 ( ) = 0.86 awrt 0. Bft M A () (d) H0: 8 or 0. 5 H: 8 or 0. 5 B () (e) W ~ Po(8) P(W ) = 0.0 (< 0.05) P(W ) = 0.06 (> 0.05) Largest possible value of f is (a) st B for writing or using Po( ). May be implied by a correct answer or by awrt 0.78 B M A () Total (b) M for 0.77 or ( their (a) ) (c) Bft for writing or using B(7, their a ). May be implied by M scored. 7 M for a correct binomial expression for P(Y = 5) (ft their value of p). Allow etc or May be implied by the correct answer but if p 0.77 or better we must see expression ALT (d) (e) They may use W ~ B(7, their (a) ) for Bft then P(W = ) for the M B for both hypotheses correct. Must use or for either 8 or 0.5 [ Use of is B0] If (d) is blank but correct hypotheses are seen in (e) can award retrospectively BUT if hypotheses are given in (d) and (e) award this mark for answer in (d). B for writing Po(8) can be awarded if seen in (d) (may be implied e.g. by scoring M) M for using Po(8) to find a lower-tail critical region Need to see one of the given probability statements or implied by Po(8) and f = seen. A for [f ]= but allow f Correct answer only scores / 6

7 Question Scheme Marks. (a)(i) (ii) X~B(6, 0.5) Prizes are randomly placed in packets. Each packet has a 5% chance of containing a prize B B Each packet contains a prize independently of others () (b) 6 P(X = ) = (0.5)( 0.5) 5 [= ] or [=0.55] P(only box contains exactly prize) = P(X = ) ( P(X = )) = answer in the range 0.58~0.5 (inc) M M A () (c) P(X ) = P(X ) = 0.5 = 0.66 awrt 0.66 M A () (d) Y ~ B(80, 0.66 ) N(awrt 7., awrt.) [Calc : 7.85.,.078 ] Bft P(Y 0.5 '7.' 0) PZ '.' P Z.5= 0.57 = 0.06 (calc: ) awrt 0.06 M dmaft A (5) Total (a)(i) B for a completely specified distribution. Condone B(6,5%) must be in (a)(i) (ii) B for a contextualised reason involving randomness, independence or constant probability Must mention prize and packet and for constant prob 0.5 in correct statement. (b) st M for a correct expression for P(X = ) may use P(X ) P(X = 0) from tables with X ~ B(6, 0.5) (May be implied by sight of awrt 0.56 or answer in range ) nd M for writing or using P(X = ) ( P(X = )) NB M0MA0 is possible Allow just P(X = ) ( P(X = )) or a numerical expression with any p = P(X = ) except p = 0.5 provided 0 <p < (c) M for writing or using P(X ) A for awrt 0.66 (calc: ) (d) st Bft for mean = np and variance = np( p) where p = their (c) ft their Any ft values must be correct to at least sf st.5 or 0 or 0.5 their mean M their sd nd M dependent on st M for using a continuity correction st Aft for (+) correct standardized expression ( ft their and ) or z = awrt +.5 nd A awrt 0.06 [Use of p = 0.5 giving N(0, 5) can score B0MMAA0 i.e. max /5] NB Use of binomial (leads to or ) but scores 0 marks. 7

8 Question Scheme Marks. (a) Using area of triangle: Solve: Using integration: P(X > ) ( 6 ) f () 6 (6 ) P(X > ) = x x = ( x) dx x x 6 8 M = [so median of X is ] = [so median of X is ] = [ so median of X is ] Acso () (b) Area of triangle from < x < = 0.5 ( ax b) dx 0.5 ( ) (a b) 0.5 a b 6a a M b b 0.5 5a 6b M f() = 0 a b 0 f() = 0 a b 0 dm Solving simultaneously: Solving simultaneously: a a 5a 6a (c) a x b a b ( t ) dt or ( x ) dx with + c and F() = 0 or F() = 0.5 x ( t) dt or ( x) dx with + c and F(6) = or F() = x 8 x x 8 x F( x) 7 x 8 x x 6 x 6 A A (5) M M B A A (5) Total (a) M for a correct expr area above using triangle or correct integral & attempt to integrate Acso for ½ or x = with no errors. NB f()=.5 = 0.5 or.5 0.5x = 0.5 is M0A0 Allow use of correct F(x) = 0.5 provided F() = 0.5 not used to establish F(x) (b) st M for a correct equation in a and b for an area (need not be simplified) nd M for a correct equation in a and b using f() = 0 rd dm dep on at least M; for solving two linear equations in a and b by eliminating one variable (allow one slip). If one equation is incorrect we must see explicit method to solve. st A for a and nd A for b (allow exact equivalents) (c) st M for attempt to integrate ax + b with correct limits or + c and F() = 0 or F() = 0.5 nd M for attempt to integrate x with correct limits & or +c and F(6) = or F() = 0.5 For the nd M allow (their ax b ) d x or F() based on their cdf for [, ) instead of B for st and th line correct st A correct nd line with limits NB nd A correct rd line with limits NB ( x ) 8 ( x 6) 8 Allow < or anywhere for the last marks 8

9 Question Scheme Marks. (a)(i) mean = np =.5 B (ii) standard deviation = ( ) awrt.866 M A (b)(i) (ii) p H: p 00 Po(.5) H0: 00 X ~ B ( 500, 00) P(X 5) = P(X ) = 0.8 = [0.088 > 0.05] therefore do not reject H0, not significant, 5 does not lie in CR The doctor s claim is not supported. or Past records are not out of date/reliable or Number/ Proportion / Probability of people/ with allergy is not higher. (c) Y ~ B(n, 0.0) n P(Y = 0) = ( 0.70) n >.85 n = 5 P(Y w) < P(Y 8) = 0.88 or P(Y ) = 0.05 [>0.005] P(Y ) = 0.6 or P(Y 0) = 0.007[<0.005] w = 0 () B B M A M A cso (6) M Acao () Total (a)(ii) M for a correct expression for standard deviation including root NB Assuming Poisson will get.5 =.8708 but scores M0A0 [ Ans only /] (b)(i) B H0 and H correct with p or π ( may be seen in (ii)) [Use of or =.5 for (i) is B0] (ii) B for writing or using Po(.5) st M for P(X ) and use with bin or Poisson or for CR P(X 6) = 0.0 [<0.05] st A for awrt 0.0 or for CR X 6 nd M for a non-contradictory statement which follows from their probability/cr nd Acso correct contextual statement and fully correct solution with all other marks scored in (b) (ii) NB Use of Binomial leading to 0.08 in (b) can score BB0MA0MA0 Use of Normal [ may get 0.0 ( Po) or 0.0 (bin)] in (b) can score BB0M0A0MA0 M A (c) ft st M for a correct expression for P(Y = 0) and comparison with [allow inequality or equation]. Use of tables alone is M0 st A for n = 5 cao [ Answer only is M0A0 unless we see the 0.7 n compared to 0.005] nd M for using B( 5, 0.0) to try and find an upper tail, need sight of one of given probs If st M scored can ft their n but need sight of probability expression and probability one of which must be a correct ft and must be just above or just below (o.e.) nd A for w = 0 [but w 0 is A0] Allow Y 0 or Y > [ Correct answer only scores M0A0MA]

10 Question Scheme Marks 5. (a) P(customer waits > minutes) = F() = [0.() 0.00( )] = 7 5 or M A (b) P(customer waits > minutes customer waits > minutes) = P( T ) "0.056" "0.056" "0.056", P( T ) F() [0.() 0.00( )] 0. () M,Aft 7 or awrt 0.0 A 5 () (c) F(.7) = 0.7(68 ) or F(.7) 0.75 = 0.0 F(.8) = 0.75( ) or F(.8) 0.75 = (+) 0.00 F(.7) < 0.75 < F(.8) therefore.7 < upper quartile <.8 M Acso () (d) f(t) = t t 0.0t E( ) t(0. 0.0t ) dt T ,= 0 0 M, A 5 = 8 or.875 A () Total (a) M for writing or using F() [ Just writing F() = 0. alone is M0] (b) P( T ) "(a)" M for a correct ratio expression or or better. Ignore e.g. P(T > T > ) P( T ) P( T ) Allow other letters for T. If only numerical values are used, must have num < denom. st Aft for a correct ratio expression with their (a) on numerator and correct numerical expression or 0. or better on denominator. M (c) ALT M for attempting to find F(.7) and F(.8) Acso for both F(.7) = awrt 0.7 < 0.75 < F(.8) = awrt 0.75 and correct conclusion. May use F(x) 0.75 and look for a change of sign. There must be sight of 0.75 Using calculator: M for 0.t 0.00t = 0.75 leading to t = awrt.7 (.787 ) A for correct conclusion (must reject other roots i.e. 6.8 and.7 if found) Conclusion must clearly state that Q is between.7 and.8. Penalise false statements. (d) st M for differentiating F(t) to find f(t) [ Just terms with at least correct] nd n n M for attempting to integrate ttheir f(t) ignore limits[ at least one t t ] st A for correct integration of correct f(t) & use of limits (must see some correct use of 5) This mark may be implied by a correct answer. nd 5 5 A or.875 (condone.88) [Correct answer only of or.875 scores /] 8 8 Special Case Use of 5 E( T ) 5 F( t)dt can score out of if fully correct. 0 0

11 Question Scheme Marks 6. (a) (,, ) (,, ) [(,, ), (,, )] (,, ) [(,, ), (,, )] (,, ) BB () (b) P(M = ) = ( 0.5) = 0.5 B P(M = 5) = (0.8) or P(M = 5) = or P(M = ) = (0.5) (0.) (0.5)(0.) (0.)(0.8) (0.) (0.8) (0.) (0.) P(M = ) = [(P(M = ) + P(M = 5)] or P(M = 5) = [(P(M = ) + P(M = )] M M m 5 P(M = m) A (c) Mode of S = and Mode of M =5 B () () (a) st B for at least correct samples listed e.g. (,, ) and (,, ) nd B for all 8 correct samples listed (with no extra or incorrect ones given) Total 7 (b) B for P(M = ) = 0.5oe Condone e.g. X = and.5% st M for a correct expression or a correct probability for P(M = 5) or P(M = ) nd M for a correct expression for third probability of, or 5 or if B or st M are scored then award this mark for using the sum of the probs = A for both P(M = ) = 0.87oe and P(M = 5) = 0.88oe (c) B for both correct modes with clear S and M labels

12 Question Scheme Marks ( ab) 7.(a) E( X)= E(X) [= ] M = a b A () (b) P( X b a) (c) b ( b a) ( b a) a, or, b a b a b ( ba) a E( X ) x dx b b ( b ) b ( b( b)) x b b [Var( X ) E( X ( ba) E(X E( X ) b (d) Range = b a = 8 or b b = 8 or b = 8 Var (X) = or 0 = 7 ) 0 ) [E( X )] ( b( b)) or ] E( X ) (a) M for using E(X) where E(X) is a linear function of a and b A for a b or (a + b) a b M, A M dm A M A () () () Total (b) M for a correct fraction expression for ( X b a ) in terms of a and b (need brackets!) A for P (c) ALT st M for a correct integral for E(X ) or E(X ) (ignore limits) nd dm dependent on st M for correct integration and correct use of a = b including in limits. Must be E(X ) st M for a correct expression for E(X ) or E(X ) in terms of a and b from substituting into Var(X) and using E(X) = 0 or ( a b ) nd dm dependent on st M for E( X ) and correct use of a = b (d) M for writing or using (b a) = 8 or b b = 8 or b = A for 7 [Correct answer only is /] Pearson Education Limited. Registered company number 8788 with its registered office at 80 Strand, London, WCR 0RL, United Kingdom

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