Cambridge A Level Mathematics 9709 — 2007 Oct/Nov Paper 7 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level MATHEMATICS 9709/07 Paper 7 Probability & Statistics 2 (S2) October/November 2007 1 hour 15 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF9) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 50. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. This document consists of 3 printed pages and 1 blank page. ©UCLES 2007 [Turn over *8304470861*

2 1 Isaac claims that 30% of cars in his town are red. His friend Hardip thinks that the proportion is less than 30%. The boys decided to test Isaac’s claim at the 5% significance level and found that 2 cars out of a random sample of 18 were red. Carry out the hypothesis test and state your conclusion. [5] 2 In summer the growth rate of grass in a lawn has a normal distribution with mean 3.2 cm per week and standard deviation 1.4 cm per week. A new type of grass is introduced which the manufacturer claims has a slower growth rate. A hypothesis test of this claim at the 5% significance level was carried out using a random sample of 10 lawns that had the new grass. It may be assumed that the growth rate of the new grass has a normal distribution with standard deviation 1.4 cm per week. (i) Find the rejection region for the test. [4] (ii) The probability of making a Type II error when the actual value of the mean growth rate of the new grass is m cm per week is less than 0.5. Use your answer to part (i) to write down an inequality for m. [1] 3 (i) Explain what is meant by the term ‘random sample’. [1] In a random sample of 350 food shops it was found that 130 of them had Special Offers. (ii) Calculate an approximate 95% confidence interval for the proportion of all food shops with Special Offers. [4] (iii) Estimate the size of a random sample required for an approximate 95% confidence interval for this proportion to have a width of 0.04. [3] 4 The cost of electricity for a month in a certain town under scheme A consists of a fixed charge of 600 cents together with a charge of 5.52 cents per unit of electricity used. Stella uses scheme A. The number of units she uses in a month is normally distributed with mean 500 and variance 50.41. (i) Find the mean and variance of the total cost of Stella’s electricity in a randomly chosen month. [5] Under scheme B there is no fixed charge and the cost in cents for a month is normally distributed with mean 6600 and variance 421. Derek uses scheme B. (ii) Find the probability that, in a randomly chosen month, Derek spends more than twice as much as Stella spends. [5] © UCLES 2007 9709/07/O/N/07

3 5 The length, X cm, of a piece of wooden planking is a random variable with probability density function given by f(x) = ⎧⎪⎨ ⎪⎩ 1 b 0 ≤x ≤b, 0 otherwise, where b is a positive constant. (i) Find the mean and variance of X in terms of b. [3] The lengths of a random sample of 100 pieces were measured and it was found that Σ x = 950. (ii) Show that the value of b estimated from this information is 19. [2] Using this value of b, (iii) find the probability that the length of a randomly chosen piece is greater than 11 cm, [1] (iv) find the probability that the mean length of a random sample of 336 pieces is less than 9 cm. [4] 6 The random variable X denotes the number of worms on a one metre length of a country path after heavy rain. It is given that X has a Poisson distribution. (i) For one particular path, the probability that X = 2 is three times the probability that X = 4. Find the probability that there are more than 3 worms on a 3.5 metre length of this path. [5] (ii) For another path the mean of X is 1.3. (a) On this path the probability that there is at least 1 worm on a length of k metres is 0.96. Find k. [4] (b) Find the probability that there are more than 1250 worms on a one kilometre length of this path. [3] © UCLES 2007 9709/07/O/N/07

4 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. 9709/07/O/N/07

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2007 question paper 9709 MATHEMATICS 9709/07 Paper 7, maximum raw mark 50 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9709 07 © UCLES 2007 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. • When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

Page 3 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9709 07 © UCLES 2007 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) CWO Correct Working Only - often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR -1 A penalty of MR -1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures - this is regarded as an error in accuracy. An MR-2 penalty may be applied in particular cases if agreed at the coordination meeting. PA -1 This is deducted from A or B marks in the case of premature approximation. The PA -1 penalty is usually discussed at the meeting.

Page 4 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9709 07 © UCLES 2007 1 H0 p = 0.3 H1 3.0 < p P(0, 1, 2 ) = 0.718+0.3× 0.717× 18C1 + 0.32× 0.716× 18C2 = 0.001628+0.01256+0.04576 = 0.0599 This is > 0.05 Accept Isaac’s claim. OR Using N(0.3,0.0116) H0 p=0.3 H1 p<0.3 z=0.111 +1/36 – 0.3 = -1.49159 √0.0116 -1.49159>-1.645 Accept Isaac’s claim OR Using N(5.4,3.78) H0 µ=5.4 H1 µ<5.4 z=2.5 –5.4= -1.49159 √3.78 -1.49159>-1.645 Accept Isaac’s claim B1 M1 A1 M1 A1ft B1 M1 A1 M1 A1ft B1 M1 A1 M1 A1ft 5 Both hypotheses correct For finding P(0, 1, 2) at least two terms of this sum needed Correct answer accept 0.06(0) Comparing with 0.05 must be 0.05 Correct conclusion ft their test statistic – no contradictions Both hypotheses correct For attempt at z with or without cc For correct z For comparison Correct conclusion ft their test statistic Both hypotheses correct For attempt at z with or without cc For correct z For comparison Correct conclusion ft their test statistic 2 (i) -1.645 = 10 / 4.1 2.3 − c c = 2.47 rejection region is < x 2.47 (ii) m < 2.47 M1 B1 A1 A1ft 4 B1ft 1 For standardising, must have sq rt. and z value For ± 1.645 used For 2.47 For inequality correct way round (ft their 2.47 but must be <3.2) ft on their (i) 3 (i) a sample where every element has an equal chance of being chosen OR a random sample of size n is a sample chosen in such a way that each possible group of size n has the same chance of being picked. (ii) 130/350 (0.371) 0.371 ± 1.96× 350 ) 629 .0 )( 371 .0 ( = 0.371± 0.050609 = (0.321, 0.422) (iii) 1.96 n ) 629 .0 )( 371 .0 ( = 0.02 n = 2241or 2242 or 2243 or 2240 B1 1 B1 M1 B1 A1 4 M1* M1*dep A1 3 For proportion used Correct shape n zs x / ± Correct z value 1.96 used Correct limits (written as interval) Seeing an equation involving 0.02 or 0.04, n in denom and a sq rt and proportions used For equation of correct form Correct whole number answer

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9709 07 © UCLES 2007 4 (i) E (cost to Stella) = 600 + 5.52× 500 = 3360 Var (cost to Stella) = 5.522× 7.12 = 1540 (1536) (ii) P (D > 2S) = P(D – 2S >0) D – 2S ~ N(-120, 421 + 4× 1536) ~ N(-120, 6565) P(D-2S > 0) = P( 6565 120 > z ) = P(z > 1.481) = 0.0693 M1 A1 M1 M1 A1 5 M1 B1 A1ft M1 A1 5 For multiplying by 5.52 and adding 600 Correct mean For mult 7.1/7.12 /50.412 by 5.522 For 5.52(2) x 7.1(2) or 50.412 with no addition/subtraction For correct answer For attempt (D-2S) (or equiv) either < or > 0 For correct mean (seen or implied) For correct unsimplified variance For standardising attempt For correct answer, accept 0.069 5 (i) E(X) = ∫ b dx b x 0 = b b x 0 2 2       = 2 b Var (X) = ∫ − b b x 0 2 4 2 b = 12 2 b (ii) 9.5 = b/2 b = 19 AG (iii) 8/19 or 0.421 (iv) ) 336 / 08 . 30 ,5.9 ( ~ N X or using totals N(3192,10106.88) P(       − < = < 336 / 08 . 30 5.9 9 ) 9 z P X or equiv = P(z < -1.671) = 1 – 0.9526 = 0.0474 B1 M1 A1 3 M1 A1 2 B1 1 M1 A1ft M1 A1 4 Correct answer (accept unsimplified) For (substituted) attempt at ∫ − ] ) ( [ ) ( 2 2 X E dx x f x ie )] ( [ 2 X E − must be seen even if ignored in next line Correct answer. Accept unsimplified – but must be a single fraction. Equating their mean to their 9.5 Correct answer Correct answer Dividing their b2/12 by 336 Correct mean and variance Standardising (must involve 336) and area < 0.5 or consistent with their figures Correct answer

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9709 07 © UCLES 2007 6 (i) !4 3 !2 4 2 λ λ λ λ − − = e e 2 = λ new λ = 7 P(X > 3) =       + + + − − !3 7 !2 7 7 1 1 3 2 7 e = 0.918 (ii) (a) λ = 1.3k P(X > 0) = 1 – e-1.3k = 0.96 0.04 = e-1.3k k = 2.48 (b) X~ N(1300, 1300) P(X > 1250) =       − > 1300 1300 5. 1250 z P = P(z > -1.373) = 0.915 M1 A1 B1ft M1 A1 5 B1 M1 A1 A1 4 B1 M1 A1 3 Poisson equation involving λ Correct mean New mean ft 3.5 × previous one Poisson probs with their mean (at least 3 probs) and 1- Correct answer Correct new mean Equation with k or λ in involving 1 – P(0) = 0.96 correct equation correct answer correct mean and variance standardising must have sq rt with or without cc correct answer