Cambridge A Level Mathematics 9709 — 2020 May/June Paper 5 · Variant 1

2 (a) Find the number of different arrangements that can be made from the 9 letters of the word JEWELLERY in which the three Es are together and the two Ls are together. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the number of different arrangements that can be made from the 9 letters of the word JEWELLERY in which the two Ls are not next to each other. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) 6! M1 720 A1 2 2(b) Total number: ( ) 9! 30240 3!2! M1 Number with Ls together = ( ) 8! 6720 3! M1 Number with Ls not together = 9! 8! 3!2! 3! − = 30 240 – 6720 M1 23 520 A1 Alternative method for question 2(b) 7! 8 7 3! 2 × × 7! × k in numerator, k integer ≥ 1 M1 8 × 7 × m in numerator or 8C2 × m, m integer ≥ 1 M1 3! in denominator M1 23 520 A1 4

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3 A company produces small boxes of sweets that contain 5 jellies and 3 chocolates. Jemeel chooses 3 sweets at random from a box. (a) Draw up the probability distribution table for the number of jellies that Jemeel chooses. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The company also produces large boxes of sweets. For any large box, the probability that it contains more jellies than chocolates is 0.64. 10 large boxes are chosen at random. (b) Find the probability that no more than 7 of these boxes contain more jellies than chocolates. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) x 0 1 2 3 Probability 1 56 15 56 30 56 10 56 (B1 for probability distribution table with correct outcome values) P(0) = 3 2 1 1 8 7 6 56 × × = P(1) = 5 3 2 15 3 8 7 6 56 × × × = P(2) = 5 4 3 30 3 8 7 6 56 × × × = P(3) = 5 4 3 10 8 7 6 56 × × = (M1 for denominator 8×7×6) M1 Any one probability correct (with correct outcome) A1 All probabilities correct A1 4 3(b) 1 – P(8, 9, 10) = 1 – 10 8 2 10 9 1 10 8 9 C 0.64 0.36 C 0.64 0.36 0.64   + +   M1 1 – (0.164156 + 0.064852 + 0.11529) M1 0.759 A1 3

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4 In a music competition, there are 8 pianists, 4 guitarists and 6 violinists. 7 of these musicians will be selected to go through to the final. How many different selections of 7 finalists can be made if there must be at least 2 pianists, at least 1 guitarist and more violinists than guitarists? [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Scenarios: 2P 3V 2G 8C2 × 4C2 × 6C3 = 28 × 6 × 20 = 3360 2P 4V 1G 8C2 × 4C1 × 6C4 = 28 × 4 × 15 = 1680 3P 3V 1G 8C3 × 4C1 × 6C3 = 56 × 4 × 20 = 4480 4P 2V 1G 8C4 × 4C1 × 6C2 = 70 × 4 × 15 = 4200 (M1 for 8Cr × 4Cr × 6Cr with ∑ݎ = 7) Two unsimplified products correct B1 Summing the number of ways for 3 or 4 correct scenarios M1 Total: 13 720 A1 4

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5 On Mondays, Rani cooks her evening meal. She has a pizza, a burger or a curry with probabilities 0.35, 0.44, 0.21 respectively. When she cooks a pizza, Rani has some fruit with probability 0.3. When she cooks a burger, she has some fruit with probability 0.8. When she cooks a curry, she never has any fruit. (a) Draw a fully labelled tree diagram to represent this information. [2] (b) Find the probability that Rani has some fruit. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that Rani does not have a burger given that she does not have any fruit. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Fully correct labelled tree for method of transport with correct probabilities. B1 Fully correct labelled branches with correct probabilities for lateness with either 1 branch after W or 2 branches with the prob 0 B1 2 5(b) 0.35 × 0.3 + 0.44 × 0.8 (+ 0) M1 0.457 A1 2 Question Answer Marks 5(c) P(not B|not fruit) = ( ) ( ) P B' F' P F' ∩ M1 ( ) 0.35 0.7 0.21 1 1 their × + × − b M1 0.455 0.543 (M1 for 1 – their (b) or summing three appropriate 2-factor probabilities, correct or consistent with their tree diagram as denominator) M1 0.838 or 455 543 A1 4

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6 The lengths of female snakes of a particular species are normally distributed with mean 54 cm and standard deviation 6.1 cm. (a) Find the probability that a randomly chosen female snake of this species has length between 50 cm and 60 cm. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The lengths of male snakes of this species also have a normal distribution. A scientist measures the lengths of a random sample of 200 male snakes of this species. He finds that 32 have lengths less than 45 cm and 17 have lengths more than 56 cm. (b) Find estimates for the mean and standard deviation of the lengths of male snakes of this species. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) ( ) 50 54 60 54 P P 0.6557 0.9836 6.1 6.1 z Z − −   < < = − < <     Both values correct A1 Φ (0.9836) – Φ (–0.6557) = Φ (0.9836) + Φ (0.6557) – 1 = 0.8375 + 0.7441 – 1 (Correct area) M1 0.582 A1 4 Question Answer Marks 6(b) 45 0.994 μ σ − = − B1 56 1.372 μ σ − = B1 One appropriate standardisation equation with ߤ, ߪ, z-value (not probability) and 45 or 56. M1 11 = 2.366 σ (M1 for correct algebraic elimination of µ or σ from their two simultaneous equations to form an equation in one variable) M1 σ = 4.65, μ = 49.6 A1 5

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7 The numbers of chocolate bars sold per day in a cinema over a period of 100 days are summarised in the following table. Number of chocolate bars sold 1 −10 11 −15 16 −30 31 −50 51 −60 Number of days 18 24 30 20 8 (a) Draw a histogram to represent this information. [5] (b) What is the greatest possible value of the interquartile range for the data? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Calculate estimates of the mean and standard deviation of the number of chocolate bars sold. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Class widths: 10, 5, 15, 20, 10 M1 Frequency density = frequency/their class width: 1.8, 4.8, 2, 1, 0.8 M1 All heights correct on diagram (using a linear scale) A1 Correct bar ends B1 Bar ends: 10.5, 15.5, 30.5, 50.5, 60.5 B1 5 7(b) 11 – 15 and 31 – 50 B1 Greatest IQR = 50 – 11 = 39 B1 2 7(c) Mean = 18 5.5 24 13 30 23 20 40.5 8 55.5 2355 23.6 100 100 × + × + × + × + × = = B1 Var = 2 2 2 2 2 2 18 5.5 24 13 30 23 20 40.5 8 55.5 mean 100 × + × + × + × + × − M1 2 77917.5 mean 224.57 100 − = A1 Standard deviation = 15.0 (FT their variance) A1 FT 4

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