Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 6 · Variant 2

2 On Saturday afternoons Mohit goes shopping with probability 0.25, or goes to the cinema with probability 0.35 or stays at home. If he goes shopping the probability that he spends more than $50 is 0.7. If he goes to the cinema the probability that he spends more than $50 is 0.8. If he stays at home he spends $10 on a pizza. (i) Find the probability that Mohit will go to the cinema and spend less than $50. [1] (ii) Given that he spends less than $50, find the probability that he went to the cinema. [4]

Mark scheme: 2 (i) P(C ∩ < 50) = 0.35 × 0.2 = 0.07 B1 [1] P (C ∩ < 50) (ii) P(C │ < 50) = M1 [4] Summing three 2-factor products seen P ( < 50) anywhere (can omit the 1) .0 35 × 2.0 = A1 0.545 (unsimplified) seen as num or .025 × 3.0 + .035 × 2.0 + 4.0(× )1 denom of a fraction .007 = M1 Attempt at P(C ∩ < 50) as 2-factor prod .0545 only seen as num or denom of a fraction = 0.128 (14/109) A1 Correct answer

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3 The amount of fibre in a packet of a certain brand of cereal is normally distributed with mean 160 grams. 19% of packets of cereal contain more than 190 grams of fibre. (i) Find the standard deviation of the amount of fibre in a packet. [3] (ii) Kate buys 12 packets of cereal. Find the probability that at least 1 of the packets contains more than 190 grams of fibre. [2]

Mark scheme: 3 (i) z = 0.878 B1 ± 0.878, 0.88, rounding to 0.88 seen 190 − 160 (190 – 160)/σ = something = .0878 M1 σ σ = 34.2 A1 [3] Correct answer (ii) P(at least 1) = 1 – P(0) M1 Using 1 – P(0), 1 – P(0, 1), P(1,2 … 12) or P(2, … 12) with p = 0.19 or 0.81, terms must be evaluated to get the M1 = 1 – (0.81)12 = 0.920 A1 [2] Correct answer accept 0.92

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4 The following histogram summarises the times, in minutes, taken by 190 people to complete a race. Frequency density 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 Time in 0 100 200 300 400 minutes (i) Show that 75 people took between 200 and 250 minutes to complete the race. [1] (ii) Calculate estimates of the mean and standard deviation of the times of the 190 people. [6] (iii) Explain why your answers to part (ii) are estimates. [1]

Mark scheme: 4 (i) number = 1.5 × 50 = 75 (AG) B1 [1] Must see 1.5 × 50 (ii) freqs are 10, 25, 50, 75, 30 (15, 15) M1 Attempt at freqs not fd A1 Correct freqs Mean = (10 × 125 + 25 × 162.5 + 50 × 187.5 M1 attempt at mid points not cw or ucb or lcb + 75 × 225 + 30 × 300)/190 = 40562.5/190 = 213 (213.48 …) A1 correct mean sd2 = 10 × 1252 + 25 × 162.52 + 50 × 187.52 M1 subst their Σfx2 in correct variance + 75 × 2252 + 30 × 3002)/190 – (213.48 …)2 formula sd = 46.5 or 46.6 A1 [6] (iii) have used the mid-point of each interval and B1 not the raw data [1] GCE AS/A LEVEL – October/November 2013 9709 62 4 4 8

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5 On trains in the morning rush hour, each person is either a student with probability 0.36, or an office worker with probability 0.22, or a shop assistant with probability 0.29 or none of these. (i) 8 people on a morning rush hour train are chosen at random. Find the probability that between 4 and 6 inclusive are office workers. [3] (ii) 300 people on a morning rush hour train are chosen at random. Find the probability that between 31 and 49 inclusive are neither students nor office workers nor shop assistants. [6]

Mark scheme: 5 (i) P(4, 5, 6) = (0.22)4(0.78)48C4 + M1 Bin term with 8Cr pr (1 – p)8-r (0.22)5(0.78)38C5 + (0.22)6(0.78)28C6 M1 seen r ≠ 0 any p < 1 Summing 2 or 3 bin probs p = 0.22, n = 8 = 0.0763 A1 [3] Correct answer (ii) prob = 0.13 B1 Correct prob can be implied mean = 300 × 0.13 = 39 B1ft Correct unsimplified np and npq ft wrong var = 300 × 0.13 × 0.87 = 33.93 0.13 P(30 < x < 50 )= P M1 Standardising a value need sq rt  305. − 39 495. − 39   < z <   33.93 33.93  M1 Cont correction 30.5 / 31.5 or 48.5/49.5 only = P(-1.4592< z < 1.8026) M1 Correct area Φ1 + Φ2 – 1 oe = Φ(1.8026) + Φ(1.4592) – 1 Rounding to correct answer = 0.9643 + 0.9278 – 1 = 0.892 A1 [6] SC P(31,..49)=300C31(0.13)31(0.87)269 + … +300C49 etc.) B1B1

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6 The 11 letters of the word REMEMBRANCE are arranged in a line. (i) Find the number of different arrangements if there are no restrictions. [1] (ii) Find the number of different arrangements which start and finish with the letter M. [2] (iii) Find the number of different arrangements which do not have all 4 vowels (E, E, A, E) next to each other. [3] 4 letters from the letters of the word REMEMBRANCE are chosen. (iv) Find the number of different selections which contain no Ms and no Rs and at least 2 Es. [3]

Mark scheme: 6 (i) 1663200 B1 [1] (ii) M xxxxxxxxx M M1 9! or 9P9 seen !9 Number of ways = = 30240 A1 [2] Correct answer !2!3 (iii) 4 vowels together = 8! × 4/2!2! M1 8!/2!2! seen mult by something = 40320 M1 4 oe 4!/3! or 4C1 etc. seen mult by something 1663200 – 40320 = 1622880 B1 [3] Correct answer SC 7!/2!2! × 8P4 or 7! × 8P4/3! Or 7!/2!2! × 8P4/3! M1 (iv) Exactly 2 Es 4C2 = 6 M1 Summing 2 options Exactly 3 Es 4C1 = 4 B1 One option correct Total = 10 ways A1 [3] Correct answer OR 5C2 M2 M1 for k5C2 = 10 A1 Correct ans GCE AS/A LEVEL – October/November 2013 9709 62

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