Cambridge A Level Mathematics 9709 — 2018 May/June Paper 6 · Variant 1

2 In a survey 55 students were asked to record, to the nearest kilometre, the total number of kilometres they travelled to school in a particular week. The results are shown below. 5 5 9 10 13 13 13 15 15 15 15 16 18 18 18 19 19 20 20 20 20 21 21 21 21 23 25 25 27 27 29 30 33 35 38 39 40 42 45 48 50 50 51 51 52 55 57 57 60 61 64 65 66 69 70 (i) On the grid, draw a box-and-whisker plot to illustrate the data. [5] An ‘outlier’ is defined as any data value which is more than 1.5 times the interquartile range above the upper quartile, or more than 1.5 times the interquartile range below the lower quartile. (ii) Show that there are no outliers. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 0 20 40 60 80 Distance km B1 LQ and UQ correct B1 Quartiles and median plotted as box graph with linear scale min 3 values B1ft Whiskers drawn to correct end points with linear scale, not thr’ box, not joining at top or bottom of box. Ft their UQ and LQ. Whiskers must be with ruler If scale non-linear or non-existent SCB1if all 5 data values (quartiles and end points) have values shown and all are correct numerically and fulfil the ‘box’ and ‘whiskers ruled line’ requirements B1 Label to include ‘distance or travelled’ and ‘km,’ allow ‘total km’, linear scale, numbered at least 5 – 70. 5 Question Answer Marks Guidance 2(ii) 1.5 × IQR = 48 Method 1 LQ – 48 = –ve, (i.e. < 0) UQ + 48 = 98 (i.e. > 70) M1 Attempt to find 1.5 × their IQR and add to UQ or subt from LQ hence no outliers A1 Correct conclusion from correct working, need both ends. No need to state comparisons. Method 2 LQ – 5 = 13 (< 48) 70 – UQ = 20 (< 48) M1 Compare their 1.5 × IQR (= 48) > gap (20) between UQ and max 70 or LQ and min 5 Hence no outliers A1 Correct conclusion from correct working, need both ends. No need to state comparisons 2

More questions on Representation of data

5 In Pelmerdon 22% of families own a dishwasher. (i) Find the probability that, of 15 families chosen at random from Pelmerdon, between 4 and 6 inclusive own a dishwasher. 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(ii) A random sample of 145 families from Pelmerdon is chosen. Use a suitable approximation to find the probability that more than 26 families own a dishwasher. 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Mark scheme: 5(i) M1 15C6(0.22)6(0.78)9 A1 Correct unsimplified expression = 0.398 A1 Correct answer 3 5(ii) µ = 145 × 0.22 = 31.9 σ2 = 145 × 0.22 × 0.78 = 24.882 B1 Correct unsimplified mean and variance P(x > 26) = P 26.5 31.9 24.882 z −   >     = P(z > –1.08255) M1 Standardising must have sq rt M1 25.5 or 26.5 seen as a cc = Φ(1.08255) M1 Correct area Φ, must agree with their µ = 0.861 A1 Correct final answer accept 0.861, or 0.860 from 0.8604 not from 0.8599 5

More questions on The normal distribution

6 Vehicles approaching a certain road junction from town A can either turn left, turn right or go straight on. Over time it has been noted that of the vehicles approaching this particular junction from town A, 55% turn left, 15% turn right and 30% go straight on. The direction a vehicle takes at the junction is independent of the direction any other vehicle takes at the junction. (i) Find the probability that, of the next three vehicles approaching the junction from town A, one goes straight on and the other two either both turn left or both turn right. 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(ii) Three vehicles approach the junction from town A. Given that all three drivers choose the same direction at the junction, find the probability that they all go straight on. 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Mark scheme: 6(i) P(SLL) = (0.3)(0.55)(0.55) = 0.09075 ( 363 4000 ) M1 P(SRR) = (0.3)(0.15)(0.15) = 0.00675 ( 27 4000 ) A1 Two correct options 0.09075 or 0.00675 can be unsimplified Total = 3C1 × P(SLL) + 3C1 × P(SRR) = 0.27225 + 0.02025 M1 Summing 6 prob options not all identical Prob = 0.293 accept 0.2925 ( 117 400 ) A1 Correct answer 4 6(ii) n P(SSS all samedir ) = ( ) ( ) n P SSS and same dir P same direction B1 (0.3)3 oe seen on its own as num or denom of a fraction M1 Attempt at P(SSS+LLL+RRR) seen anywhere = 3 3 3 0.3 0.3 0.3 (0.15) (0.55) (0.3) × × + + A1 (0.15)3 + (0.55)3 + (0.3)3 oe seen as denom of a fraction = 0.137 ( 108 787 ) A1 Correct answer 4

More questions on Probability

7 Find the number of different ways in which all 9 letters of the word MINCEMEAT can be arranged in each of the following cases. (i) There are no restrictions. 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(ii) No vowel (A, E, I are vowels) is next to another vowel. 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(iii) Find the number of possible selections which contain exactly 1 M and exactly 1 E. 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(iv) Find the number of possible selections which contain at least 1 M and at least 1 E. 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Mark scheme: 7(i) 9! 2!2! = 90720 B1 1 7(ii) Method 1 * * * * * A B1 5! seen multiplied (arrangement of consonants allowing repeats) No. arrangements of consonants × ways of inserting vowels = B1 6P4 oe (i.e. 6 × 5 × 4 × 3, 6C4 × 4!) seen mult (allowing repeats) no extra terms 5 2 ! ! × 6 4P 2! B1 Dividing by at least one 2! (removing at least one set of repeats) Answer 6 4P 5 2! 2 × = 10 800 B1 Correct final answer 4 7(iii) 5C3 = 10 M1 5Cx or 5Px seen alone, x = 2 or 3 A1 Correct final answer not from 5C2 2 Question Answer Marks Guidance 7(iv) Method 1 Considering separate groups M1 Considering two scenarios of MME or EEM or MMEE with attempt, may be probs or perms MME** = 5C2 = 10 MEE** = 5C2 = 10 MMEE* = 5C1 = 5 M1 Summing three appropriate scenarios from the four need 5Cx seen in all of them ME*** = 5C3 = 10 see (iii) Total = 35 A1 Correct final answer Method 2 Considering criteria are met if ME are chosen M1 7Cx only seen, no other terms M1 xC3 only seen, no other terms ME *** = 7C3 = 35 A1 Correct final answer 3

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