Cambridge A Level Mathematics (US) 9280 — 2015 May/June Paper 6 · Variant 1

*8260927905* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS (US) 9280/61 Paper 6 Probability & Statistics 1 (S1) May/June 2015 1 hour 15 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulas (MF9) (US) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Center number, candidate number, and name on 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. JC15 06_9280_61/FP © UCLES 2015 [Turn over

2 1 The lengths, in meters, of cars in a city are normally distributed with mean - and standard deviation 0.714. The probability that a randomly chosen car has a length more than 3.2 meters and less than - meters is 0.475. Find -. [4] 2 The table summarises the lengths in centimeters of 104 dragonflies. Length (cm) 2.0 −3.5 3.5 −4.5 4.5 −5.5 5.5 −7.0 7.0 −9.0 Frequency 8 25 28 31 12 (i) State which class contains the upper quartile. [1] (ii) Draw a histogram, on graph paper, to represent the data. [4] 3 Jason throws two fair dice, each with faces numbered 1 to 6. Event A is ‘one of the numbers obtained is divisible by 3 and the other number is not divisible by 3’. Event B is ‘the product of the two numbers obtained is even’. (i) Determine whether events A and B are independent, showing your working. [5] (ii) Are events A and B mutually exclusive? Justify your answer. [1] 4 x Take fewer than 100 photos Take at least 100 photos 0.76 0.90 View fewer than 3 times View at least 3 times View fewer than 3 times View at least 3 times A survey is undertaken to investigate how many photos people take on a one-week holiday and also how many times they view past photos. For a randomly chosen person, the probability of taking fewer than 100 photos is x. The probability that these people view past photos at least 3 times is 0.76. For those who take at least 100 photos, the probability that they view past photos fewer than 3 times is 0.90. This information is shown in the tree diagram. The probability that a randomly chosen person views past photos fewer than 3 times is 0.801. (i) Find x. [3] (ii) Given that a person views past photos at least 3 times, find the probability that this person takes at least 100 photos. [4] © UCLES 2015 9280/61/M/J/15

3 5 The table shows the mean and standard deviation of the weights of some turkeys and geese. Number of birds Mean (kg) Standard deviation kg Turkeys 9 7.1 1.45 Geese 18 5.2 0.96 (i) Find the mean weight of the 27 birds. [2] (ii) The weights of individual turkeys are denoted by xt kg and the weights of individual geese by xg kg. By first finding Σx2 t and Σx2 g, find the standard deviation of the weights of all 27 birds. [5] 6 (i) In a certain country, 68% of households have a printer. Find the probability that, in a random sample of 8 households, 5, 6, or 7 households have a printer. [4] (ii) Use an approximation to find the probability that, in a random sample of 500 households, more than 337 households have a printer. [5] (iii) Justify your use of the approximation in part (ii). [1] 7 (a) Find how many different numbers can be made by arranging all nine digits of the number 223 677 888 if (i) there are no restrictions, [2] (ii) the number made is an even number. [4] (b) Sandra wishes to buy some applications (apps) for her smartphone but she only has enough money for 5 apps in total. There are 3 train apps, 6 social network apps, and 14 games apps available. Sandra wants to have at least 1 of each type of app. Find the number of different possible selections of 5 apps that Sandra can choose. [5] © UCLES 2015 9280/61/M/J/15

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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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 2015 9280/61/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the May/June 2015 series 9280 MATHEMATICS (US) 9280/61 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 May/June 2015 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 – May/June 2015 9280 61 © Cambridge International Examinations 2015 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 Cambridge International AS/A Level – May/June 2015 9280 61 © Cambridge International Examinations 2015 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 Cambridge International AS/A Level – May/June 2015 9280 61 © Cambridge International Examinations 2015 1 P(x < 3.273) = 0.5 – 0.475 = 0.025 z = –1.96 96 .1 714 .0 2.3 − = −µ µ = 4.60s M1 A1 M1 A1 [4] Attempt to find z-value using tables in reverse ±1.96 seen Solving their standardised equation z-value not nec Correct ans accept 4.6 2 (i) UQ 5.5 – 7.0 cm B1 [1] (ii) fd 5.33, 25, 28, 20.7, 6, fd 30 25 20 15 10 5 0 2 4 6 8 10 length in cm M1 A1 B1 B1 [4] Attempt at fd or scaled freq [fr/cw] Correct heights seen on graph Correct bar widths no gaps Labels (fd and length/cm) and correct bar ends 3 (i) P(A) = 1 3 × 2 3 + 2 3 × 1 3 = 4 9 P(B) = 27 36 = 3 4 P(A∩B) = 12 36 = 1 3 P(A)×P(B) = 4 9 × 3 4 = 1 3 Independent as P(A∩B) = P(A)×P(B) M1 M1 B1 M1 A1 [5] Sensible attempt at P(A) Sensible attempt at P(B) correct P(A∩B) Cf P(A∩B) with P(A)×P(B) need at least 1 correct Correct conclusion following all correct working (ii) Not mutually exclusive because P(A∩B) ≠0 Or give counter example e.g. 1 and 6 B1 [1] ft their P(A∩B) 4 (i) (1 – x)0.9 + x × 0.24 = 0.801 x = 0.15 M1 A1 A1 [3] Eqn with sum of two 2-factor probs = 0.801 Correct equation Correct answer

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 61 © Cambridge International Examinations 2015 (ii) P([100 times given =Y=3 views) ( ) ( ) views 3 views 3 times 100 [ [ [ P P ∩ = 801 .0 1 or 76 .0 15 .0 1.0 85 .0 1.0 85 .0 − × + × × = 0.427 B1 M1 A1 A1 [4] 0.85×0.1 seen on its own as num or denom of a fraction Attempt at P([ 3 views) either (0.85×p1 + 0.15×p2) or 1 – 0.801 seen anywhere Correct unsimplified P([ 3 views) as num or denom of a fraction Correct answer 5 (i) new mean = 27 2.5 18 1.7 9 × + × = 5.83 M1 A1 [2] Mult by 9 and 18 and dividing by 27 correct answer (ii) 1.452 = so 9 2 tx ∑ = 472.6125 mm 0.962 = 2 2 18 5.2 gx − ∑ so Σxg 2 = 503.3088 New sd2 117 .2 83 .5 27 3. 503 6. 472 2 2 2 = − + K K K New sd = 1.46 M1 A1 A1 M1 A1 [5] subst in a correct variance formula sq rt or not correct Σxt 2 (rounding to 470) correct Σxg 2 (rounding to 500) using Σxt 2 + Σxg 2, dividing by 27 and subt comb mean2 correct answer 6 (i) P(5, 6, 7) = 8C5(0.68)5(0.32)3 + 8C6(0.68)6(0.32)2 + 8C7(0.68)7(0.32) = 0.722 M1 M1 A1 A1 [4] Binomial term 8Cx px(1–p)8-x seen 0 < p < 1 Summing 3 binomial terms Correct unsimplified answer Correct answer (ii) np = 340, npq = 108.8 P(x > 337) = P       − > 8. 108 340 5. 337 z = P(z > – 0.2396) = 0.595 B1 M1 M1 M1 A1 [5] Correct (unsimplified) mean and var standardising with sq rt must have used 500 cc either 337.5 or 336.5 correct area (> 0.5) must have used 500 correct answer (iii) np (340) > 5 and nq(160) > 5 B1 [1] must have both or at least the smaller, need numerical justification 7 (a) (i) !3!2!2 !9 = 15120 ways B1 2 B1 [2] Dividing by 2!2!3! Correct answer

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 61 © Cambridge International Examinations 2015 (ii) ********3 in !3!2!2 !8 = 1680 ways ********7 in !3!2 !8 = 3360 ways Total even = 15120 – 1680 – 3360 = 10080 ways OR ********2 in 8!/2!3! = 3360 ways ********6 in 8!/2!2!3! = 1680 ways ********8 in 8!/2!2!2! = 5040ways Total = 10080 ways OR “15120” ×6/9 = 10080 B1 B1 M1 A1 [4] B1 B1 M1 A1 M2 A2 Correct ways end in 3 Correct ways end in 7 Finding odd and subt from 15120 or their (i) Correct answer One correct way end in even correct way end in another even Summing 2 or 3 ways Correct answer Mult their (i) by 2/3 oe Correct answer (b) T(3) S(6) G(14) 1 1 3 in 3×6×14C3 = 6552 1 3 1 in 3×6C3×14 = 840 3 1 1 in 1×6×14 = 84 2 2 1 in 3C2×6C2×14 = 630 2 1 2 in 3C2×6×14C2 = 1638 1 2 2 in 3×6C2×14C2 = 4095 Total ways = 13839 (13800) M1 M1 M1 B1 A1 [5] Mult 3 (combinations) together assume 6 = 6C1etc Listing at least 4 different options Summing at least 4 different options At least 3 correct numerical options Correct answer