Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 7 · Variant 2

*9670120708* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level MATHEMATICS 9709/72 Paper 7 Probability & Statistics 2 (S2) October/November 2012 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. JC12 11_9709_72/FP © UCLES 2012 [Turn over

2 1 x f( )x 0 –1 1 2 3 2 1 The diagram shows the graph of the probability density function, f, of a random variable X. Find the median of X. [3] 2 The heights of a certain type of plant have a normal distribution. When the plants are grown without fertilizer, the population mean and standard deviation are 24.0 cm and 4.8 cm respectively. A gardener wishes to test, at the 2% significance level, whether Hiergro fertilizer will increase the mean height. He treats 150 randomly chosen plants with Hiergro and finds that their mean height is 25.0 cm. Assuming that the standard deviation of the heights of plants treated with Hiergro is still 4.8 cm, carry out the test. [5] 3 The cost of hiring a bicycle consists of a fixed charge of 500 cents together with a charge of 3 cents per minute. The number of minutes for which people hire a bicycle has mean 142 and standard deviation 35. (i) Find the mean and standard deviation of the amount people pay when hiring a bicycle. [3] (ii) 6 people hire bicycles independently. Find the mean and standard deviation of the total amount paid by all 6 people. [3] 4 A cereal manufacturer claims that 25% of cereal packets contain a free gift. Lola suspects that the true proportion is less than 25%. In order to test the manufacturer’s claim at the 5% significance level, she checks a random sample of 20 packets. (i) Find the critical region for the test. [5] (ii) Hence find the probability of a Type I error. [1] Lola finds that 2 packets in her sample contain a free gift. (iii) State, with a reason, the conclusion she should draw. [2] © UCLES 2012 9709/72/O/N/12

3 5 A random variable X has probability density function given by f(x) =   k x −1 3 ≤x ≤5, 0 otherwise, where k is a constant. (i) Show that k = 1 ln 2. [4] (ii) Find a such that P(X < a) = 0.75. [4] 6 In order to obtain a random sample of people who live in her town, Jane chooses people at random from the telephone directory for her town. (i) Give a reason why Jane’s method will not give a random sample of people who live in the town. [1] Jane now uses a valid method to choose a random sample of 200 people from her town and finds that 38 live in apartments. (ii) Calculate an approximate 99% confidence interval for the proportion of all people in Jane’s town who live in apartments. [4] (iii) Jane uses the same sample to give a confidence interval of width 0.1 for this proportion. This interval is an x% confidence interval. Find the value of x. [4] 7 A random variable X has the distribution Po(1.6). (i) The random variable R is the sum of three independent values of X. Find P(R < 4). [3] (ii) The random variable S is the sum of n independent values of X. It is given that P(S = 4) = 16 3 × P(S = 2). Find n. [4] (iii) The random variable T is the sum of 40 independent values of X. Find P(T > 75). [4] © UCLES 2012 9709/72/O/N/12

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. 9709/72/O/N/12

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2012 series 9709 MATHEMATICS 9709/72 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, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9709 72 © Cambridge International Examinations 2012 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 LEVEL – October/November 2012 9709 72 © Cambridge International Examinations 2012 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 LEVEL – October/November 2012 9709 72 © Cambridge International Examinations 2012 1 2 2 ) (m 2 2 ) (m = 2 1 m = √2 or 1.41 (3 sfs) M1 M1 A1 [3] y = x 2 1 (attempt at linear equ with c = 0) ∫ x x m d ) ( 0 2 1 = 2 1 (Note: ±√2 as final answer scores A0) 2 H0: Pop mean = 24.0 H1: Pop mean > 24.0 150 8.4 24 25− = 2.55(2) Comp z = 2.054 or 2.055 Evidence that Hiergro has incr hts B1 M1 A1 M1 A1ft [5] Allow ‘µ’ but not just ‘mean’ Standardise, with √ 150. Ignore cc. Accept sd/var mixes. OR find xcrit For correct z or area or xcrit Valid comparison (z values/areas/x values ) Correct conclusion No contradictions (Note 2 tail test can score B0 M1 A1 M1 (z = 2.326) A1ft) 3 (i) Mean = 500 + 3 × 142 = 926 (cents) SD = 3 × 35 = 105 (cents) B1 M1 A1 [3] Or 9 × 352 seen Accept √11025 (ii) Mean = 6 × ‘926’ = 5556 (cents) 6 × ‘105’2 (= 66150) (SD = √66150) = 257 (cents) (3 sf) B1ft M1 A1 [3] or SD = √6 × ‘105’. ft their (i) Accept √66150 4 (i) P(X Y 1) = (0.75)20 + 20(0.75)19(0.25) = 0.0243 P(X Y 2) = (0.75)20 + 20(0.75)19(0.25) + 20C2(0.75)18(0.25)2 = 0.0913 or 0.0912 Critical region is 0 or 1 pkt contain gift or < 2 pkts contain gift oe M1 A1 M1 A1 A1 [5] Attempt correct expression Attempt correct expression OR Find P(2) = 0.0669 or 0.0670 dep M1M1 & their P(X Y=1) < 0.05 < their P(X Y=2) (S.R. Use of Normal: N(5.3.752) used B1 –1.645=(x + 0.5 – 5)/√3.75 M1 x < 1.31 A1 (3/5)) (ii) P(Type I) = 0.0243 (3 sfs) B1ft [1] ft their P(X Y 1) dep < 0.05 ft Normal (iii) 2 is outside rej reg No evidence to reject claim M1 A1ft [2] or P(X Y 2) > 0.05 No contradictions

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9709 72 © Cambridge International Examinations 2012 5 (i) x x k d 5 3 1 ∫ − = 1 [kln(x – 1)]3 5 = 1 k(ln4 – ln2) = 1 kln2 = 1 (k = 2 ln 1 AG) M1 A1 M1 A1 [4] Attempt integ f(x) & ‘= 1’ ignore limits Correctly integrated; ignore limits Subst of limits 3, 5 No errors seen. No decimals seen (ii) x x x d 3 1 1 2 ln 1 ∫ − = 0.75 2 ln 1 [ln(x – 1)] 3 x = 0.75 2 ln 1 (ln(x – 1) – ln2) = 0.75 ln (x – 1) = (0.75 × ln2 + ln2) ln (x – 1) = 1.75 × ln2 x – 1 = 21.75 or x – 1 = 3.36 x = 4.36 (3 sfs) M1* A1 M1 dep* A1 [4] Attempt integ f(x), unknown limit, & ‘= 0.75’or ‘= 0.25’ oe. Fully correct equn after subst limits oe. Correct manipulation of logs to find x 6 (i) Excludes children Excludes people without phones More than one person in some houses Some ex-directory B1 [1] or other implying directory excludes some people (ii) Var(p) = 200 ) 200 38 1( 200 38 − ( = 0.0007695) z = 2.576 200 ) 200 38 1( 200 38 200 38 − ± z 0.119 to 0.261 (3 sfs) M1 B1 M1 A1 [4] Seen For correct form of CI Accept 0.262 Must be an interval (iii) z × ‘√0.0007695’ = 0.05 z = 1.802 Φ(‘1.802’) (= 0.9642) (‘0.9642’ – (1 – ‘0.9642’) = 0.9284) x = 93 (2 sfs) M1 A1 M1 A1 [4] z × (their sd of p) = 0.05. Allow = 0.1 Attempt Φ(their z) and find 2Φ –1 7 (i) λ = 4.8 E–4.8(1 + 4.8 + ! 2 8.4 2 + !3 8.4 3 ) = 0.294 (3 sfs) B1 M1 A1 [3] P(R = 0, 1, 2 or 3), their λ allow one end error

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9709 72 © Cambridge International Examinations 2012 (ii) e–λ × ! 4 4 λ = 3 16 e–λ × !2 2 λ or without e–λ 12 2 λ = 3 16 or better (λ = 8) λ = 1.6n seen or implied n = ‘8’ ÷ 1.6 = 5 M1 A1 B1 A1 [4] λ = 1.6n seen or implied B1 e–1.6n × ! 4 ) 6.1( 4 n = 3 16 e1.6n × ! 2 ) 6.1( 2 n M1 12 ) 6.1( 2 n = 3 16 or better A1 (1.6n = 8) n = 5 A1 (iii) T~N(64, 64) 64 64 5. 75 − (= 1.4375) 1 – Φ(‘1.4375’) (= 1 – 0.9247) = 0.0753 to 0.0754 B1 M1 M1 A1 [4] May be implied Allow with wrong or no cc. No sd/var mixes Finding correct area consistent with their working