Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 6 · Variant 3

1 There are 300 students at a music college. All students play exactly one of the guitar, the piano or the flute. The numbers of male and female students that play each of the instruments are given in the following table. Guitar Piano Flute Female students 62 35 43 Male students 78 40 42 (i) Find the probability that a randomly chosen student at the college is a male who does not play the piano. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Determine whether the events ‘a randomly chosen student is male’ and ‘a randomly chosen student does not play the piano’ are independent, justifying your answer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) 120 300 = 0.4 B1 1 1(ii) P(male) × P(not piano) = 160 225 300 300 × 8 3 2 15 4 5 × =       M1 P(M) × P(P ') seen Can be unsimplified but the events must be named in a product As P(male ∩ not piano) also = 120 300 2 5 = The events are Independent A1 Numerical comparison and correct conclusion Alternative method for question 1(ii) P(male ∩ not piano) = 120 300 ; P(not piano) = 225 300 M1 P(M|P ') or P(P '|M) unsimplified seen with their probs with correctly named events P(M | not piano) = 120 300 225 300 = 120 225 = 8 15 = P(male) or P(not piano | M) = 120 300 160 300 =120 160 = 3 4 = P(not piano) Therefore the events are Independent A1 Numerical comparison with P(M) or P(P ') and correct conclusion 2

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2 (i) How many different arrangements are there of the 9 letters in the word CORRIDORS? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) How many different arrangements are there of the 9 letters in the word CORRIDORS in which the first letter is D and the last letter is R or O? [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 9! 2!3! = 30240 B1 9! Divided by at least one of 2! or 3! B1 Exact value 2 2(ii) D _ _ _ _ _ _ _ R: 7! 2!2!= 1260 D _ _ _ _ _ _ _ O: 7! 3! = 840 B1 7! Seen alone or as numerator in a term, can be multiplied not + or – B1 One term correct, unsimplified Total = 2100 B1 Final answer 3

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3 A sports team of 7 people is to be chosen from 6 attackers, 5 defenders and 4 midfielders. The team must include at least 3 attackers, at least 2 defenders and at least 1 midfielder. (i) In how many different ways can the team of 7 people be chosen? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The team of 7 that is chosen travels to a match in two cars. A group of 4 travel in one car and a group of 3 travel in the other car. (ii) In how many different ways can the team of 7 be divided into a group of 4 and a group of 3? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 4A 2D 1M : 6C4 × 5C2 × 4C1 (= 600) 3A 3D 1M : 6C3 × 5C3 × 4C1 (= 800) M1 A1 2 correct products, allow unsimplified M1 Summing their totals for 3 correct scenarios only Total = 2600 A1 Correct answer SC1 6C3 × 5C2 × 4C1 × 9C1 = 7200 4 Question Answer Marks Guidance 3(ii) 7C4 × 1 B1 7C3 or 7C4 seen anywhere 35 B1 2

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4 The heights of students at the Mainland college are normally distributed with mean 148 cm and standard deviation 8 cm. (i) The probability that a Mainland student chosen at random has a height less than h cm is 0.67. Find the value of h. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 120 Mainland students are chosen at random. (ii) Find the number of these students that would be expected to have a height within half a standard deviation of the mean. 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Mark scheme: 4(i) P(h < 148) = 0.67 B1 z = ±0.44 seen 148 8 − h = 0.44 M1 z-value = ± ( 148) 8 − h 151.52 ≈ 152 A1 CAO 3 4(ii) P(144 < X < 152) = P 144 148 152 148 8 8 − −   < <     Z M1 Using ± standardisation formula for either 144 or 152, µ = 148, σ = 8 and no continuity correction, allow σ2 or √σ = P 1 1 2 2   − < <     Z = 0.6915 – (1 – 0.6915) = 2 × 0.6915 – 1 M1 Correct final area legitimately obtained from phi(their z2) – phi(their z1) = 0.383 A1 Final probability answer 0.383 × 120 = 45.96 Accept 45 or 46 only B1FT Their prob (to 3 or 4 sf) × 120, rounded to a whole number or truncated 4

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5 Last Saturday, 200 drivers entering a car park were asked the time, in minutes, that it had taken them to travel from home to the car park. The results are summarised in the following cumulative frequency table. Time (t minutes) t ≤10 t ≤20 t ≤30 t ≤50 t ≤70 t ≤90 Cumulative frequency 16 50 106 146 176 200 (i) On the grid, draw a cumulative frequency graph to illustrate the data. [2] (ii) Use your graph to estimate the median of the data. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) For 80 of the drivers, the time taken was at least T minutes. Use your graph to estimate the value of T. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) Calculate an estimate of the mean time taken by all 200 drivers to travel to the car park. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) Correct labels and scales B1 Axes labelled ‘cumulative frequency’ (or cf) and ‘time (or t) [in] min(utes)’, linear scales from 0 to 90 and 0 to 200 with at least 3 values marked on each axis. 7 correctly plotted points above upper boundaries joined in a curve or line segments B1 (0, 0); (10, 16); (20, 50); (30, 106); (50, 146); (70,176); (90,200) 2 5(ii) 29 B1 28 ⩽ median ⩽ 30 1 5(iii) 120 seen M1 For seeing 120 in a calculation or marked on the graph 37 A1FT 36 ⩽ Ans ⩽ 39 or FT from their graph SC1 unsupported answer in range 2 5(iv) Frequencies 16 34 56 40 30 24 B1 Seen. Allow unsimplified Est. Mean = 5 16 15 34 25 56 40 40 60 30 80 24 200 × + × + × + × + × + × M1 At least 4 correct midpoints (5, 15, 25, 40, 60, 80) used in a calculation 7310 200 M1 Summing products of their 6 mid-points (not lower or upper bound or class width) × their frequencies / 200 (or their ∑f), unsimplified 36.55 A1 Accept 36.6 4

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7 A competition is taking place between two choirs, the Notes and the Classics. There is a large audience for the competition. ³ 30% of the audience are Notes supporters. ³ 45% of the audience are Classics supporters. ³ The rest of the audience are not supporters of either of these choirs. ³ No one in the audience supports both of these choirs. (i) A random sample of 6 people is chosen from the audience. (a) Find the probability that no more than 2 of the 6 people are Notes supporters. [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Find the probability that none of the 6 people support either of these choirs. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) A random sample of 240 people is chosen from the audience. Use a suitable approximation to find the probability that fewer than 50 do not support either of the choirs. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i)(a) M1 any p, x ≠ 6,0 0.1176 ... + 0.3025 ... + 0.3241 ... A1 Correct unsimplified answer 0.744 A1 Correct final answer 3 Question Answer Marks Guidance 7(i)(b) P(support neither choir) = 1 – (0.3 + 0.45) = 0.25 M1 0.25n seen alone, 1 < n ⩽ 6 P(6 support neither choir) = 0.256 = 0.000244 or 1 4096 A1 Correct final answer 2 7(ii) Mean = 240 × 0.25 = 60 Variance = 240 × 0.25 × 0.75 = 45 B1FT Correct unsimplified 240p and 240pq where p =their P(support neither choir) or 0.25 P(X < 50) = P 49.5 60 45 −   <     Z = P(Z < –1.565) M1 Substituting their µ and σ (condone σ2) into the ±Standardisation Formula with a numerical value for ‘49.5’. M1 Using continuity correction 49.5 or 50.5 within a standardisation expression 1 – 0.9412 M1 Appropriate area Φ from standardisation formula P(z< …) in final solution, (< 0.5 if z is –ve, > 0.5 if z is +ve) 0.0588 A1 Correct final answer 5

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