Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 6 · Variant 2

*8124672967* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/62 Paper 6 Probability & Statistics 1 (S1) October/November 2014 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. 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. JC14 11_9709_62/FP © UCLES 2014 [Turn over

2 1 The 50 members of a club include both the club president and the club treasurer. All 50 members want to go on a coach tour, but the coach only has room for 45 people. In how many ways can 45 members be chosen if both the club president and the club treasurer must be included? [3] 2 Find the number of different ways that 6 boys and 4 girls can stand in a line if (i) all 6 boys stand next to each other, [3] (ii) no girl stands next to another girl. [3] 3 (i) Four fair six-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown. Find the probability that the numbers shown on the four dice add up to 5. [3] (ii) Four fair six-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown on 7 occasions. Find the probability that the numbers shown on the four dice add up to 5 on exactly 1 or 2 of the 7 occasions. [4] 4 Sharik attempts a multiple choice revision question on-line. There are 3 suggested answers, one of which is correct. When Sharik chooses an answer the computer indicates whether the answer is right or wrong. Sharik first chooses one of the three suggested answers at random. If this answer is wrong he has a second try, choosing an answer at random from the remaining 2. If this answer is also wrong Sharik then chooses the remaining answer, which must be correct. (i) Draw a fully labelled tree diagram to illustrate the various choices that Sharik can make until the computer indicates that he has answered the question correctly. [4] (ii) The random variable X is the number of attempts that Sharik makes up to and including the one that the computer indicates is correct. Draw up the probability distribution table for X and find EX. [4] 5 (a) The time, X hours, for which people sleep in one night has a normal distribution with mean 7.15 hours and standard deviation 0.88 hours. (i) Find the probability that a randomly chosen person sleeps for less than 8 hours in a night. [2] (ii) Find the value of q such that PX < q = 0.75. [3] (b) The random variable Y has the distribution N-, 32, where 23 = 3- and - ≠0. Find PY > 4-. [3] © UCLES 2014 9709/62/O/N/14

3 6 On a certain day in spring, the heights of 200 daffodils are measured, correct to the nearest centimetre. The frequency distribution is given below. Height (cm) 4 −10 11 −15 16 −20 21 −25 26 −30 Frequency 22 32 78 40 28 (i) Draw a cumulative frequency graph to illustrate the data. [4] (ii) 28% of these daffodils are of height h cm or more. Estimate h. [2] (iii) You are given that the estimate of the mean height of these daffodils, calculated from the table, is 18.39 cm. Calculate an estimate of the standard deviation of the heights of these daffodils. [3] 7 In Marumbo, three quarters of the adults own a cell phone. (i) A random sample of 8 adults from Marumbo is taken. Find the probability that the number of adults who own a cell phone is between 4 and 6 inclusive. [3] (ii) A random sample of 160 adults from Marumbo is taken. Use an approximation to find the probability that more than 114 of them own a cell phone. [5] (iii) Justify the use of your approximation in part (ii). [1] © UCLES 2014 9709/62/O/N/14

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. 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. © UCLES 2014 9709/62/O/N/14

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2014 series 9709 MATHEMATICS 9709/62 Paper 6, 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 2014 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9709 62 © Cambridge International Examinations 2014 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. The following abbreviations may be used in a mark scheme or used on the scripts:

Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9709 62 © Cambridge International Examinations 2014 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 Cambridge International AS/A Level – October/November 2014 9709 62 © Cambridge International Examinations 2014 1 48C43 = 1712304 (1710000) B1 B1 B1 3 48 seen in a single term combination oe 43 or 5 seen in a single term combination oe Both can be mult by integer k [ 1 Correct final answer 2 (i) 6! ×5! = 86400 B1 B1 B1 3 6! oe seen multiplied by integer k [ 1 5! oe seen multiplied by integer k [ 1 Correct final answer (ii) 6! × 7 × 6 × 5 × 4 = 604800 B1 B1 B1 3 6! seen mult by integer k [ 1 Mult by 7P4 oe Correct final answer 3 (i) 1 1 1 2 or 1 1 2 1 or 1 2 1 1 or 2 1 1 1 Prob = 4 6 1 6 1 6 1 6 1 × × × × = 324 1 (0.00309) M1 M1 A1 3 One of 1 1 1 2 seen Mult a prob by 4 or ( 6 1 )4 × integer k [ 1 seen Correct answer (ii) P(1,2) = 7C1 × (1/324) (323/324)6 + 7C2(1/324)2(323/324)5 = 0.0214 M1 M1 M1 A1 4 Bin term 7Cxpx(q)7–x , 0.99 Y p + q Y 1 Using their p from (i) in a bin term Correct unsimplified answer Correct answer 4 (i) W = wrong, C = correct C 1 W 2 1 W 3 1 2 1 C C 3 1 2 1 W 1 W 2 1 3 1 C C C OR 1 2 1 W W 3 2 2 1 C 3 1 C M1 M1 B1 A1 4 M1 M1 B1 A1 3 branches first qn and 2 by 2 for second qn only One branch twice for third qn or two branches twice with 0 and 1 seen on branches Any two of 3 1 , 2 1 and 1 seen as probs Probs all correct and sensible labels NB SR for 4 outcomes instead of 3, M1 B1 only 2 branches first qn and 1 by 2 for second qn only One branch once for third qn or two branches with 0 and 1 seen on branches Any two of 3 1 or 3 2 , 2 1 and 1 seen as probs Probs all correct and sensible labels

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9709 62 © Cambridge International Examinations 2014 (ii) x 1 2 3 Prob 3 1 3 1 3 1 P(1) = P(C) say = 3 1 P(2) = P(WC) = 6 1 P(WC) = 6 1 total P (2) = 3 1 P(3) = P(WWC) = 6 1 P(WWC) = 6 1 total P(3) = 3 1 E(X) = 1 × 3 1 + 2 × 3 1 + 3 × 3 1 = 2 B1 B1 B1 B1 4 1, 2, 3 seen only oe 2 correct probs 3 correct probs Correct answer ft their probs provided 0.999 Y Σp Y 1 5 (a) (i) P(x I 8) =P       − < 88 .0 15 .7 8 z = Φ (0.9659) = 0.833 M1 A1 2 Standardising ± , no cc no sq rt no sq Correct answer (ii) z = 0.674 88 .0 15 .7 − q = 0.674 q = 7.74 B1 M1 A1 3 Accept ± 0.674 or 0.675 only Standardised eqn = ± their z-value, allow sq or sq rt if already penalised in (i) Correct answer (b) P(Y > 4µ) = P (       − > ) 2 / 3 ( 4 µ µ µ z ) = P(z K 2) = 1 – 0.9772 = 0.0228 M1 A1 A1 3 Standardising no sq rt, no cc, no sq, one variable z = ±2 seen correct ans SR B1 if made-up values used and 0.0228 obtained

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9709 62 © Cambridge International Examinations 2014 6 (i) ht I10.5 I15.5 I20.5 I25.5 I30.5 CF 22 54 132 172 200 cf 200 144 100 3.5 10.5 20.5 30.5 ht (cm) B1 B1 M1 A1 4 At least 4 CFs correct seen on graph Labels correct, i.e. all of ht, cm, cf Attempt at upper end points either 10 or 10.5 or 11 at least 4 upper end points All correct, i.e. points joined up from (3.5, 0) to (10.5, 22)….to (30.5, 200) Straight lines or curve (ii) 72% less, i.e. 144 less than ht h. h = 22.5 cm M1 A1 2 144 used can be implied single value in range 21 to 23 inclusive (iii) var = (72 × 22 + 132 × 32 + 182 × 78 + 232 × 40 + 282 × 28)/200 – 18.392 = 74870/200 – 18.392 = 374.35 – 18.392 = 36.1579 sd = 6.01 M1 B1 A1 3 Using mid points attempt 7 ± 0.5in correct var formula incl – mean2 At least 4 correct midpoints Correct ans 7 (i) P (4, 5, 6) = (0.75)4 (0.25)4 × 8C4 + (0.75)5 (0.25)3 × 8C5 + (0.75)6 (0.25)2 × 8C6 = 0.606 M1 M1 A1 3 Bin term pr(1 – p)8– r ×8Cr seen any p Correct unsimplified answer Correct ans (ii) np = 160 × 0.75 = 120 npq = 30 P( > 114) = P             − > 30 120 5. 114 z = P (z > –1.004) = Φ(1.004) = 0.842 B1 M1 M1 M1 A1 5 Unsimplified mean and var correct Standardising, need sq rt Cont correction either 114.5 or 113.5 Correct area consistent with their working Correct ans (iii) np and nq both > 5 B1 1 Need both