Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 6 · Variant 2

1 (i) How many different arrangements are there of the 11 letters in the word MISSISSIPPI? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Two letters are chosen at random from the 11 letters in the word MISSISSIPPI. Find the probability that these two letters are the same. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) 11! 4!4!2! M1 11! 11! 4! 2! or k k × × , k a positive integer = 34650 A1 Correct final answer 2 1(ii) Method 1 P(SS) = 4 3 12 11 10 110 × = (= 0.10911) B1 One of P(SS), P(PP) or P(II) correct, allow unsimplified P(PP) = 2 1 2 11 10 110 × = (= 0.01818) P(II) = 4 3 12 11 10 110 × = (= 0.10911) 4 3 11 10 × M1 Sum of probabilities from 3 appropriate identifiable scenarios (either by labelling or of form 4 2 4 11 11 11 a c a b b b × + × + × where a = 4 or 3, b = 11 or 10, c = 2 or 1) Total = 26 13 110 55 = oe (0.236) A1 Correct final answer Method 2 Total number of selections = 11C2 = 55 Selections with 2 Ps = 1 B1 Seen as the denominator of fraction (no extra terms) allow unsimplified Selections with 2 Ss = 4C2 = 6 Selections with 2 Is = 4C2 = 6, M1 Sum of 3 appropriate identifiable scenarios (either by labelling or values, condone use of permutations. May be implied by 2,12,12) Total selections with 2 letters the same = 13 Probability of 2 letters the same = 13 55 oe (0.236) A1 Correct final answer, without use of permutations 3

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2 The following back-to-back stem-and-leaf diagram shows the reaction times in seconds in an experiment involving two groups of people, A and B. A B (4) 4 2 0 0 20 5 6 7 (3) (5) 9 8 5 0 0 21 1 2 2 3 7 7 (6) (8) 9 8 7 5 3 2 2 2 22 1 3 5 6 6 8 9 (7) (6) 8 7 6 5 2 1 23 4 5 7 8 8 9 9 9 (8) (3) 8 6 3 24 2 4 5 6 7 8 8 (7) (1) 0 25 0 2 7 8 (4) Key: 5 22 6 means a reaction time of 0.225 seconds for A and 0.226 seconds for B (i) Find the median and the interquartile range for group A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The median value for group B is 0.235 seconds, the lower quartile is 0.217 seconds and the upper quartile is 0.245 seconds. (ii) Draw box-and-whisker plots for groups A and B on the grid. [3]

Mark scheme: 2 2(i) median LQ = 0 IQR = 2(ii) A B n = 0.225; 0.215: UQ = 0.236 0.236 – 0.215 = 0.021 0.200 0 0.205 0 6 0.215 0.225 0.217 0.235 0.236 0 0.245 0 Time seconds B 0.250 0.258 M1 0.232 < UQ A1 www Omission o If M0 awar SCB1 for b 3 B1 Linear scale labelled (tim through box 1 ft Labelled co lines throug B1 Labelled co whiskers at SC If B0B0 SCB1 if bot Penalty MR align exactl 3 dian (Q2) Q (Q3) < 0.238 – 0 of all decimal poin ded both LQ = 0.215: U e between 0.20 to me and seconds), xes, whiskers not orrect graph for A gh boxes, whisker orrect graph for B, t corner of boxes 0 scored because g th ‘correct’ R-1 if graphs plott ly. 0.204 < LQ (Q1) < nts MR-1 UQ = 0.236 seen 0.26 (condone om at least one box p at corner of boxe , (ft their median/ rs at corner of box , condone lines th graphs not labelle ted on separate ax < 0.219 mission of 0.26) a plot attempted, no s /quartiles), condon xes hrough boxes, ed/labels reversed xes unless both sca axis lines ne ales

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3 Jake attempts the crossword puzzle in his daily newspaper every day. The probability that he will complete the puzzle on any given day is 0.75, independently of all other days. (i) Find the probability that he will complete the puzzle at least three times over a period of five days. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Kenny also attempts the puzzle every day. The probability that he will complete the puzzle on a Monday is 0.8. The probability that he will complete it on a Tuesday is 0.9 if he completed it on the previous day and 0.6 if he did not complete it on the previous day. (ii) Find the probability that Kenny will complete the puzzle on at least one of the two days Monday and Tuesday in a randomly chosen week. 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Mark scheme: 3(i) Method 1 P(3) + P(4) + P(5) = 5C3 3 2 0.75 0.25 × + M1 One binomial term 5Cxpx(1 – p)5-x , x ≠ 0 or 5, any p 5C4 4 1 0.75 0.25 × + 5C5 5 0 0.75 0.25 × M1 Correct unsimplified expression = 0.26367 + 0.39551 + 0.23730 = 0.896 (459/512) A1 Correct final answer, allow 0.8965 (isw) but not 0.897 alone Method 2 1 – P(0) − P(1) – P(2) = 1−5C0 0 5 0.75 0.25 × M1 One binomial term 5Cxpx(1 – p)5-x , x ≠ 0 or 5, any p − 5C1 1 4 0.75 0.25 × − 5C2 2 3 0.75 0.25 × M1 Correct simplified expression = 1 – 0.00097656 – 0.014648 – 0.087891 = 0.896 (459/512) A1 Correct final answer, allow 0.8965 (isw) but not 0.897 alone 3 Question Answer Marks Guidance 3(ii) Method 1 P(C,C) + P(C,C′) + P(C′,C) 0.8 × 0.9 B1 Unsimplified prob completed on both days 0.8 × 0.1 + 0.2 × 0.6 M1 Unsimplified prob 0.8 × a + 0.2 × b, a = 0.1or 0.4, b = 0.6 or 0.9 = 0.92 oe A1 Correct final answer Method 2 1 – P(C′,C′) = 1 – 0.2 × 0.4 B1 Unsimplified prob completed on no days M1 1 – 0.2 × a, a=0.1or 0.4 allow unsimplified = 0.92 A1 Correct final answer 3

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4 (i) Find the number of different ways that 5 boys and 6 girls can stand in a row if all the boys stand together and all the girls stand together. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the number of different ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) 5! × 6! ×2 B1 n × 5! × 6! (n integer, n ⩾ 1), no inappropriate addition = 172800 B1 Correct final answer, isw rounding (www scores B3) All marks based on their final answer 3 Question Answer Marks Guidance 4(ii) … G … G … G … G … G … G … No. ways girls placed × No. ways boys placed in gaps = M1 k × 6! or k × 7P5 (k is an integer, k ⩾ 1) no inappropriate add. (7P5 ≡ 7 × 6 × 5 × 4 × 3 or 7C5 × 5!) 6! × 7P5 M1 Correct unsimplified expression = 1814400 A1 Correct exact final answer (ignore subsequent rounding) 3

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5 The Quivers Archery club has 12 Junior members and 20 Senior members. For the Junior members, the mean age is 15.5 years and the standard deviation of the ages is 1.2 years. The ages of the Senior members are summarised by Σy = 910 and Σy2 = 42 850, where y is the age of a Senior member in years. (i) Find the mean age of all 32 members of the club. 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(ii) Find the standard deviation of the ages of all 32 members of the club. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 15.5 12 910 12 20 + or 2) =34.25 or 34¼ (years) A1 Correct exact answer (isw rounding), oe (34 years 3 months) 2 5(ii) Considering Juniors: variance = 2 2 15.5 12 x ∑ − = 1.22 M1 2 2 15.5 x k ∑ − = 1.22 , k = 12 or 20 2 2900.28 x ∑ = A1 Answer wrt 2900 Considering whole group: 2 2 2 2900.28 42850 45750 z x y ∑ = ∑ + ∑ = + = Variance = 2 2 32 z µ ∑ − = ( ) 2 45750 34.25 12 20 their their − + (= 256.63) M1 Their 45750 > 42850 (not 85700 or rounding to 1.8 × 109) in correct variance or std deviation formula (Σx2 and addition may not be seen) s d = 16.0(2) A1 Correct final answer, condone 16.03 4

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