Mark Scheme (Results) January Pearson Edexcel International A Level in Statistics 2 (WST02/01)

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1 Mark (Results) January 05 Pearson Edexcel International A Level in Statistics (WST0/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere 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: January 05 Publications Code IA04068 All the material in this publication is copyright Pearson Education Ltd 05

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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 EDEXCEL IAL 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) 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) 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 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. 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. Ignore wrong working or incorrect statements following a correct answer.

5 January 05 WST0 Statistics S Mark Question (a) X~Po(.) B. P(X = ) = e.! M = 0.6 awrt 0. A () (b) Y~Po(.6) B P(Y ) = P(Y = 0) M.6 = e = awrt A () (c) X~Po(0.8) e 0.8 ( e 0.8) P ( X = ) P( X = )! = M M.6 4 P( Y = 4) e.6 M A 4! = = 0.5 A (5) (d) A~Po(7) approximated by N(7,7) B = M P(A 84) = P Z 7 M M = P(Z.55 ) = awrt 0.087/0.088 A (5) Notes (a) B for writing or using Po(.) (b) M e λ λ! B for writing or using Po(.6) λ e M P(Y = 0) or (c) st M using Po(0.8) with X= or X= (may be implied by or ) nd λ M ( e λ ) marks λ e λ (consistent lambda) awrt 0.08 implies st M! rd M correct use of conditional probability with denominator = st A fully correct expression nd A 0.5 (allow awrt 0.50) e.6 4!.6 4 (d) B Writing or using N(7,7) st M for exact fraction or awrt 8. (may be implied by 84) (Note: Use of N(40,40) can score B and st M) nd M Using 84 +/ 0.5 rd M standardising using 8.5, 8, 8. (awrt 8.), 8.5, 8.8, 84 or 84.5, their mean and their sd

6 Question (a) P(X > 4) = F(4) M = 5 = 5 oe A (b) B (c) df( x) M f( x ) = = d x 5 x 6 A f( x) = 5 0 otherwise (d) E(X) =.5 B (6 ) 6 (e) Variance = or d 5 x x (.5) M 5 A = awrt.08 (f) E(X ) = Var(X) + [E(X)] 5.5 = + or = 4 or 6 d 5 x x 6 ( x + ) d x M 5 E(X + ) = E(X ) + x x = dm = 44 = 44 Acao () Notes (a) M writing or using F(4) oe (c) M for differentiating to get /5 A both lines correct with ranges (e) (6 ) 6 M or d 5 x x their.5 (f) st M their Var(X) + [ their E(X) ] (which must follow from the st method in (e)) n+ 6 or d 5 x x n x and integrating x 6 n + (may be seen in (e)) or writing ( x + ) d x 5 (May be implied by 4 seen) nd M (dependent on previous M) using their E(X ) + n+ 6 n x or ( x + ) d x and integrating x 5 n + 6 () () () () ()

7 Question (a) (b) (c) (A random variable) that is a function of a (random) sample involving no unknown quantities/parameters or A quantity calculated solely from a random sample If all possible samples are chosen from a population; then the values of a statistic and the associated probabilities is a sampling distribution or a probability distribution of a statistic 4 Mean = = 7 awrt 4 B = + Variance = awrt 450 (to sf) (d) (00,00,00) (00,00,00) (00,00,00) (00,00,00) or x (00,00,00) B (00,00,00) (00,00,00) (00,00,00) or x (00,00,00) (00,00,00) (e) (00,00,00) (00,00,00) (00,00,00) (00,00,00) 4 7 = = = = awrt 0.87 awrt B B M A awrt 0.40 (allow 0.4) M awrt 0.5 B both A () () () () m awrt 44 4 or awrt 0.40 (allow 0.4) P(M = m) 64 4 or awrt 0.87 awrt or awrt or awrt A (4)

8 Question (a) Notes B for a definition which includes each of the following aspects A function of a (random) sample involving no unknown quantities/parameters. function/quantity/calculation/value/random variable. sample/observations/data. no unknown parameters/no unknown values/solely (from a sample) B requires all underlined words: All values of a statistic with their associated probabilities (b) or probability distribution of a statistic 4 (c) M ( their mean ) 7 7 (d) B any of (00,00,00), (00,00,00) any order, (00,00,00) any order, (00,00,00) B all correct, allow (00,00,00) and (00,00,00) and (00,00,00) and (00,00,00) Note: Allow other notation for 00 and 00 e.g. Small and Large (e) B Both probabilities for (00,00,00) and (00,00,00) correct M p ( p) A either correct A all means correct and all probabilities correct (table not required but means must be associated with correct probabilities)

9 Question 4(a) X ~ Po(6) M e 6 e 6 M P(5 X < 7) = P(X 6) P(X 4) or + 5! 6! = = 0. awrt 0. A (b) H 0 : λ = 9 H : λ < 9 B () Po(9) therefore X 4 = or CR X B P ( ) Insufficient evidence to reject H 0 or Not Significant or 4 does not lie in the critical region. There is no evidence that the mean number of accidents at the crossroads has reduced/decreased. Notes (a) M writing or using Po(6) λ 5 λ 6 e λ e λ M either P(X 6) P(X 4) or + 5! 6! (b) st B both hypotheses correct ( λ or µ) allow 0.5 instead of 9 nd B either awrt or critical region X dm for a correct comment (dependent on previous B) Contradictory non-contextual statements such as not significant so reject H 0 score M0 (May be implied by a correct contextual statement) A cso requires correct contextual conclusion with underlined words and all previous marks in (b) to be scored. dm Acso (4)

10 Question 5(a) M kx ( + a)dx+ kdx= x k + ax + [ kx] = dm 8 k + a+ + a + 9k 6k = A 6k + ak = 7 k ( x + ax)dx + kx dx = M 4 x ax kx 7 k + + = 4 dm a 7k 7 k 4+ a + 6k = 4 A 45k ak 7 + = 4 5k + 8ak = 7 99k = a=, k = 9 ddm A (b) B Notes (a) st M writing or using kx ( + a)dx+ kdx = ignore limits n+ nd n x dm attempting to integrate at least one x n + and sight of correct limits (dependent on previous M) (8) () st A a correct equation need not be simplified rd M k( x + ax)dx + kx dx ignore limits 4 th 7 dm setting = and attempting to integrate at least one and sight of correct limits (dependent on previous M) x n n+ x n + nd A a correct equation need not be simplified 5 th ddm attempting to solve two simultaneous equations in a and k by eliminating variable (dependent on st and rd Ms) rd A both a and k correct

11 Question (a) P(X = 5) = C 5(0.) (0.7) or M = awrt 0.79 A (b) Mean = 6 B sd = M =.049 awrt.05 A (c) H 0 : p = 0. H : p > 0. B () () B(0,0.) M P(X 8) = 0.77 or P(X > 0) = , so CR X > 0 A Insufficient evidence to reject H 0 or Not Significant or 8 does not lie in the critical region. There is no evidence to support the Director (of Studies ) belief/there is no evidence that the proportion of parents that do not support the new curriculum is greater than 0% (d) B(n, 0.5) B(8, 0.5) P(X 4) = 0.8 B(0, 0.5) P(X 5) = n =0 n = 5 Notes (a) M C5( p) ( p) or using P(X < 5) P(X < 4) (b) M use of (with or without the square root) (c) B both hypotheses correct (p or π) M using B(0,0.) (may be implied by 0.77, 0.77, or 0.) A awrt 0.8 or CR X > 0 dm a correct comment (dependent on previous M) A cso requires correct contextual conclusion with underlined words and all previous marks in (c) to be scored. dm Acso M A A (5) () (d) M for 0.8 or or or 0.99 seen st A B(0, 0.5) selected (may be implied by n = 0 or n = 0 or n = 5) An answer of 5 with no incorrect working seen scores out of Special Case: Use of a normal approximation, n ( n 0.5) M for = z with.8 < z <.9, st A for n=4./4., nd A for n=5 n 8

12 Question 7. n 4n N, 5 5 P(Y 0) =P Z 9.5 n 5 > n 5 B M MA 9.5 n 5 = B n 5 n+ 4 n 47.5 = 0 or 0.04n.44n = 0 n = 0... n = 06.6 or n = 04.7 A dm n = 06 Notes st n 4n B writing or using N, 5 5 st M writing or using 0 +/ 0.5 nd M standardising using 9, 9.5, 0 or 0.5 and their mean and their sd st A fully correct standardisation (allow +/-) nd B for z = +/ or awrt.00 must be compatible with their standardisation rd dm (dependent on nd M) getting quadratic equation and solving leading to a value of n or n nd A awrt 0. or awrt (06 or 07 or 04 or 05) rd A for 06 only (must reject other solutions if stated) 9.5 n Note: 5 = leading to an answer of 06 may score n 5 BMMAB0MAA A cao (8)

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