Cambridge A Level Mathematics 9709 — 2014 May/June Paper 6 · Variant 1

1 The petrol consumption of a certain type of car has a normal distribution with mean 24 kilometres per litre and standard deviation 4.7 kilometres per litre. Find the probability that the petrol consumption of a randomly chosen car of this type is between 21.6 kilometres per litre and 28.7 kilometres per litre. [4]

Mark scheme: 1 P (21.6 < x < 28.7)    216. − 24   287. − 24   M1 Standardising; no cc, no sq rt = P   < z <      7.4   7.4   A1 One rounding to Φ (0.841 or 0.695) = P (–0.5106 < z < 1) = Φ (1) – Φ (–0.5106) M1 Φ1 + Φ2 – 1 = 0.8413 – (1 – 0.6953) = 0.537 (0.5366) A1 4 Correct answer 12 − µ

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2 Lengths of a certain type of white radish are normally distributed with mean - cm and standard deviation 3 cm. 4% of these radishes are longer than 12 cm and 32% are longer than 9 cm. Find - and 3. [5]

Mark scheme: 12 µ 2 1.751 = B1 Rounding to ±1.75 seen σ 9 − µ 0.468 = B1 ±0.468 seen σ M1 An eqn with a z-value, µ and σ no √σ, no σ2 σ = 2.34 M1 Sensible attempt to eliminate µ or σ by substitution or subtraction, need a value µ = 7.91 A1 5 correct answers

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3 (i) State three conditions which must be satisfied for a situation to be modelled by a binomial distribution. [2] George wants to invest some of his monthly salary. He invests a certain amount of this every month for 18 months. For each month there is a probability of 0.25 that he will buy shares in a large company, there is a probability of 0.15 that he will buy shares in a small company and there is a probability of 0.6 that he will invest in a savings account. (ii) Find the probability that George will buy shares in a small company in at least 3 of these 18 months. [3]

Mark scheme: 3 (i) constant / given p, independent trials, B1 Any one correct fixed / given no. of trials, only two outcomes B1 2 Any 3 correct (ii) P (x [ 3) = 1 – P (0, 1, 2) M1 Any binomial expression pr (1 – p) 18–r 18Cr seen = 1 – [(0.85)18 + (0.85)17(0.15) × 18 + (0.85)16(0.15)2 × 18C2] M1 1 – P (0, 1, 2 ), any n,p,q = 0.520 A1 3 Correct answer 6C 2 15 3 M1 6C / 8C 4C lt b 4 f ti

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5 Playground equipment consists of swings (S), roundabouts (R), climbing frames (C) and play-houses (P). The numbers of pieces of equipment in each of 3 playgrounds are as follows. Playground X Playground Y Playground Z 3S, 2R, 4P 6S, 3R, 1C, 2P 8S, 3R, 4C, 1P Each day Nur takes her child to one of the playgrounds. The probability that she chooses playground X is 14. The probability that she chooses playground Y is 4.1 The probability that she chooses playground Z is 2.1 When she arrives at the playground, she chooses one piece of equipment at random. (i) Find the probability that Nur chooses a play-house. [4] (ii) Given that Nur chooses a climbing frame, find the probability that she chose playground Y. [4]

Mark scheme: 1 4 15 (i) P(X and P) = × = M1 Mult a playground prob with a P prob 4 9 9 1 2 1 P(Y and P) = × = A1 One correct prob 4 12 24 1 1 1 × = P(Z and P) = M1 Summing at least two 2-factor probs 2 16 32 53 P(P) = = 0.184 A1 4 Correct answer 288 P (Y ∩ C ) (ii) P(Y | C) = M1 Attempt at P (Y ∩ C ) as numerator of a P (C ) fraction 1 1 × M1 Attempt at P(C) in form of summing two 2- 4 12 factor products, seen anywhere 1 1 1 4 × + × A1 Correct unsimplified P(C) seen anywhere 4 12 2 16 1 48 1 = = 7 7 A1 4 Correct answer 48 !6 B1 6! Seen alone

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6 Find the number of different ways in which all 8 letters of the word TANZANIA can be arranged so that (i) all the letters A are together, [2] (ii) the first letter is a consonant (T, N, Z), the second letter is a vowel (A, I), the third letter is a consonant, the fourth letter is a vowel, and so on alternately. [3] 4 of the 8 letters of the word TANZANIA are selected. How many possible selections contain (iii) exactly 1 N and 1 A, [2] (iv) exactly 1 N? [3]

Mark scheme: !6 B1 6! Seen alone 6 (i) = 360 !2 B1 2 Dividing by 2! only !4 !4 B1 4! seen mult (ii) × !2 !3 B1 Dividing by 2! or 3! (Mult by 4 implied B1B1) = 48 B1 3 Correct answer (iii) 1N and 1A: N A xx in 3C2 M1 3Cx or xC2 seen alone = 3 ways A1 2 Correct answer (iv) 0 A : Nxxx = 1 way M1 Finding ways with 0 or 2 or 3 As 2 As: NAAx in 3C1 = 3 ways M1 Summing 3 or 4 options 3 As: NAAA in 1 way Total = 8 ways A1 3 Correct answer GCE AS/A LEVEL – May/June 2014 9709 61

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