Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 6 · Variant 1

1 9 people are to be divided into a group of 4, a group of 3 and a group of 2. In how many different ways can this be done? [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 B1 =126 × 10 × 1 B1 5 or 7C3 or 6 or 7C4 or 6 or 5C2 times an integer (2nd group) =1260 B1 Correct answer 3

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3 In an orchestra, there are 11 violinists, 5 cellists and 4 double bass players. A small group of 6 musicians is to be selected from these 20. (i) How many different selections of 6 musicians can be made if there must be at least 4 violinists, at least 1 cellist and no more than 1 double bass player? 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The small group that is selected contains 4 violinists, 1 cellist and 1 double bass player. They sit in a line to perform a concert. (ii) How many different arrangements are there of these 6 musicians if the violinists must sit together? 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Mark scheme: 3(i) Scenarios are: 4V + 1C + 1DB: 11C4 × 5C1 × 4C1 M1 4V + 2C: 11C4 × 5C2 5V + 1C: 11C5 × 5C1 B1 2 correct unsimplified options 6600 + 3300 + 2310 M1 Add 2 or 3 correct scenarios only = 12210 A1 Correct answer 4 3(ii) 4! × 3! M1 k multiplied by 3! or 4!, k an integer ⩾ 1 A1 Correct unsimplified expression = 144 A1 Correct answer 3

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5 At the Nonland Business College, all students sit an accountancy examination at the end of their first year of study. On average, 80% of the students pass this examination. (i) A random sample of 9 students who will take this examination is chosen. Find the probability that at most 6 of these students will pass the examination. 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(ii) A random sample of 200 students who will take this examination is chosen. Use a suitable approximate distribution to find the probability that more than 166 of them will pass the examination. 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(iii) Justify the use of your approximate distribution in part (ii). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 1 – (P(7) + P(8) + P(9)) = 1 – ( 9C7 7 2 0.8 0.2 × + 9C8 8 1 0.8 0.2 × + 9C9 9 0 0.8 0.2 ) × M1 Any binomial term of form 9Cxpx(1 – p)9 – x, x ≠ 0 M1 Correct unsimplified expression = 1 – (0.3019899 + 0.3019899 + 0.1342177) = 0.262 A1 Correct answer 3 Question Answer Marks Guidance 5(ii) Mean = 200 × 0.8 = 160: var = 200 × 0.8 × 0.2 = 32 B1 Both unsimplified P(X > 166) = P( 166.5 160 32 Z − > ) M1 Standardise, 1 60 32 x their z their − = ± with square root M1 166.5 or 165.5 seen in attempted standardisation expression = P(Z > 1.149) = 1 – 0.8747 M1 1 – a Φ -value, correct area expression, linked to final answer = 0.125 A1 Correct final answer 5 5(iii) np = 160, nq = 40: both > 5 (so normal approx. holds) B1 Both parts required 1

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6 The daily rainfall, x mm, in a certain village is recorded on 250 consecutive days. The results are summarised in the following cumulative frequency table. Rainfall, x mm x ≤20 x ≤30 x ≤40 x ≤50 x ≤70 x ≤100 Cumulative frequency 52 94 142 172 222 250 (i) On the grid, draw a cumulative frequency graph to illustrate the data. [2] (ii) On 100 of the days, the rainfall was k mm or more. Use your graph to estimate the value of k. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Calculate estimates of the mean and standard deviation of the daily rainfall in this village. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6 6(i) 6(ii) 42 1 1 2 2 3 CUMULATIVE FREQUENCY 0 50 100 150 200 250 300 0 20 4 40 60 80 RAINFALL (MM) 100 120 Appropriate axes la B1 Correct gra 2 M1 Read off fro A1 Correct ans 2 e linear scales star abelled cf and Rai aph, points plotted om increasing gra swer (41 ⩽ r ⩽ 43 rting at (0,0), infall, mm d at ucb, allow stra aph at cf = 150 3) aight lines or curv ve Question Answer Marks Guidance 6(iii) Frequencies 52, 42, 48, 30, 50, 28 B1 Correct frequencies Mean age = (10 52 25 42 35 48 45 30 60 50 85 28) / 250 × + × + × + × + × + × B1 Correct midpoints (allow one error) =9980/250 M1 Using Σfx/250 with mid-points attempt, not cf, cw, lb, ub = 39.9(2) oe A1 Correct answer Variance = 2 2 2 2 2 2 10 52 25 42 35 48 45 30 60 50 85 28) / 250 × + × + × + × + × + × − mean 2 = 539.59 M1 Attempt at variance using their midpoints and their mean σ = 23.2 A1 Correct answer for sd 6

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7 In a group of students, the numbers of boys and girls studying Art, Music and Drama are given in the following table. Each of these 160 students is studying exactly one of these subjects. Art Music Drama Boys 24 40 32 Girls 15 12 37 (i) Find the probability that a randomly chosen student is studying Music. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Determine whether the events ‘a randomly chosen student is a boy’ and ‘a randomly chosen student is studying Music’ are independent, justifying your answer. 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(iii) Find the probability that a randomly chosen student is not studying Drama, given that the student is a girl. 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(iv) Three students are chosen at random. Find the probability that exactly 1 is studying Music and exactly 2 are boys. 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Mark scheme: 7(i) 52/160 = 13/40, 0.325 B1 oe 1 7(ii) P(boy) = 96/160: P(Music) = 52/160 P(boy and Music) = 40/160 M1 Use of P(B) × P(M) = P(B∩M), appropriate probabilities used 96/160 × 52/160 ≠ 40/160: Not independent A1 Numerical comparison and conclusion stated 2 Question Answer Marks Guidance 7(iii) Method 1 P(not Music/girl) = P(not Music and girl)/P(girl) (27/160) / (64/160) M1 Appropriate probabilities in a fraction = 27 64 A1 Correct answer www implies method Method 2 Direct from table M1 27/a or b/64, a ≠ 160 27 64 A1 Correct answer www implies method 2 7(iv) P(B M) × P(B NM) × P(G NM) or P(G M) × P(B NM) ×P(B NM) M1 One scenario identified with 3 probs multiplied 40/160 × 56/159 × 52/158 or 12/160 × 56/159 × 55/158 A1 One scenario correct (ignore multiplying factor) × 3! × 3!/2! B1 Both multiplying factors correct 0.17387 0.02759 P = 0.17387 + 0.02759 M1 Both cases attempted and added (multiplying factor not required), accept unsimplified = 0.201 Note: If score in this part is 0, award SCB1 for 1 1 1 160 159 158 k × × × , for positive integer k, seen A1 Correct answer, oe Question Answer Marks Guidance 7(iv) Method 2 40 56 52 12 56 1 1 1 1 2 160 3        × × + ×                     M1 One scenario identified with 2 or 3 combination multiplied A1 One scenario correct B1 Denominator correct 116480 18480 669920 + M1 Both scenarios attempted, and added, seen as a numerator of a fraction 1687 8374 A1 Correct answer, oe 5

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