Cambridge A Level Mathematics 9709 — 2017 May/June Paper 6 · Variant 2

1 Rani and Diksha go shopping for clothes. (i) Rani buys 4 identical vests, 3 identical sweaters and 1 coat. Each vest costs $5.50 and the coat costs $90. The mean cost of Rani’s 8 items is $29. Find the cost of a sweater. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Diksha buys 1 hat and 4 identical shirts. The mean cost of Diksha’s 5 items is $26 and the standard deviation is $0. Explain how you can tell that Diksha spends $104 on shirts. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) M1 mean, x may be implied. 112 + 3x = 232 x = 40 A1 Correct complete unsimplified expression / calculation (Cost = $)40 A1 Units not required Total: 3 1(ii) sd = 0 so all cost the same M1 Must see comment interpreting sd = 0, OE shirts cost 4 × $26 = $104 AG A1 See 4 × $26, $130 – $26 OE. Must have a final value of $104 stated Total: 2 Accept 3.2 ± 0.05

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2 Anabel measured the lengths, in centimetres, of 200 caterpillars. Her results are illustrated in the cumulative frequency graph below. 200 160 120 frequency 80 Cumulative 40 0 0 1 2 3 4 5 Length in centimetres (i) Estimate the median and the interquartile range of the lengths. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Estimate how many caterpillars had a length of between 2 and 3.5 cm. [1] ........................................................................................................................................................ ........................................................................................................................................................ (iii) 6% of caterpillars were of length l centimetres or more. Estimate l. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) med = 3.2 B1 UQ = 3.65 ⩽ uq ⩽ 3.7 LQ = 2.55⩽ lq ⩽ 2.6 M1 UQ – LQ, UQ greater than their ‘median’, LQ less than their ‘median’ IQR = 1.05 ⩽ iqr ⩽ 1.15 A1 Correct answer from both LQ and UQ in given ranges Total: 3 2(ii) 134 – 24 = 110 B1 Accept 108 ⩽ n ⩽ 112, n an integer Total: 1 Question Answer Marks Guidance 2(iii) 200 – 12 = 188 less than length l M1 188 seen, can be implied by answer in range, mark on graph. l = 4.5 cm A1 Correct answer accept 4.4 ⩽ l ⩽ 4.5 Total: 2 k (–2)2 is the same as k (2)2 = 4k need to see –22 k, 22k and 4k, algebraically correct

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5 The lengths of videos of a certain popular song have a normal distribution with mean 3.9 minutes. 18% of these videos last for longer than 4.2 minutes. (i) Find the standard deviation of the lengths of these videos. 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(ii) Find the probability that the length of a randomly chosen video differs from the mean by less than half a minute. 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The lengths of videos of another popular song have a normal distribution with the same mean of 3.9 minutes but the standard deviation is twice the standard deviation in part (i). The probability that the length of a randomly chosen video of this song differs from the mean by less than half a minute is denoted by p. (iii) Without any further calculation, determine whether p is more than, equal to, or less than your answer to part (ii). You must explain your reasoning. 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Mark scheme: 5(i) 4.2 3.9 ( ) z σ = M1 z = 0.916 or 0.915 B1 Accept 0.915 ⩽ ± z ⩽0.916 seen σ = 0.328 A1 Correct final answer (allow 20/61 or 75/229) Total: 3 Question Answer Marks Guidance 5(ii) z = 4.4 – 3.9/their 0.328 or z = 3.4 – 3.9/their 0.328 = 1.5267 = –1.5267 M1 Standardising attempt with 3.4 or 4.4 only, allow square root of σ, or σ2 Φ = 0.9364 A1 0.936 ⩽ Φ ⩽ 0.937 or 0.063 ⩽ Φ ⩽ 0.064 seen Prob = 2Φ – 1 = 2(0.9364) – 1 M1 Correct area 2Φ – 1OE i.e. Φ = – (1 – Φ), linked to final solution = 0.873 A1 Correct final answer from 0.9363 ⩽ Φ ⩽0.9365 Total: 4 5(iii) dividing (0.5) by a larger number gives a smaller z-value or more spread out as sd larger or use of diagrams *B1 No calculations or calculated values present e.g. (σ = )0.656 seen Reference to spread or z value required Prob is less than that in (ii) DB1 Dependent upon first B1 Total: 2 6(i) EITHER: Route 1 A*********A in 9! / 2!2!5! = 756 ways (*M1 Considering AA and BB options with values B*********B in 9! / 4!5! = 126 ways A1 Any one option correct 756 + 126 DM1 Summing their AA and BB outcomes only Total = 882 ways A1) Question Answer Marks Guidance OR1: Route 2 A*********A in 9C5 × 4C2 = 756 ways (M1 Considering AA and BB options with values B*********B in 9C4 × 5C5= 126 ways A1 Any one option correct 756 + 126 DM1 Summing their AA and BB outcomes only Total = 882 A1) Total: 4 Question Answer Marks Guidance 6(ii) EITHER: (The subtraction method) As together, no restrictions 8! / 2!5! = 168 (*M1 Considering all As together – 8! seen alone or as numerator – condone × 4! for thinking A’s not identical As together and Bs together 7! / 5! = 42 M1 Considering all As together and all Bs together – 7! seen alone or numerator M1 Removing repeated Bs or Cs – Dividing by 5! either expression or 2! 1st expression only – OE Total 168 – 42 DM1 Subt their 42 from their 168 (dependent upon first M being awarded) = 126 A1) OR1: As together, no restrictions 8C5 x 3C1 = 168 (*M1 8C5 seen alone or multiplied M1 7C5 seen alone or multiplied As together and Bs together 7C5 x 2C1 = 42 M1 First expression x 3C1 or second expression x 2C1 Total 168 – 42 DM1 Subt their 42 from their 168 (dependent upon first M being awarded) = 126 A1) OR2: (The intersperse method ) (M1 Considering all “As together” with Cs – Mult by 6! (AAAA)CCCCC then intersperse B and another B M1 Removing repeated Cs – Dividing by 5!– [Mult by 6 implies M2] *M1 Considering positions for Bs – Mult by 7P2 oe –

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7 During the school holidays, each day Khalid either rides on his bicycle with probability 0.6, or on his skateboard with probability 0.4. Khalid does not ride on both on the same day. If he rides on his bicycle then the probability that he hurts himself is 0.05. If he rides on his skateboard the probability that he hurts himself is 0.75. (i) Find the probability that Khalid hurts himself on any particular day. 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(ii) Given that Khalid hurts himself on a particular day, find the probability that he is riding on his skateboard. 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(iii) There are 45 days of school holidays. Show that the variance of the number of days Khalid rides on his skateboard is the same as the variance of the number of days that Khalid rides on his bicycle. 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(iv) Find the probability that Khalid rides on his skateboard on at least 2 of 10 randomly chosen days in the school holidays. 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Mark scheme: 7(i) M1 Summing two 2-factor probs using 0.6 with 0.05 or 0.95, and 0.4 with 0.75 or 0.25 = 0.330 or 33 100 A1 Correct final answer accept 0.33 Total: 2 7(ii) P( ) S H = ( ) ( ) P S H P H ∩ = 0.4 0.75 0.33 × = 0.3 0.33 M1 FT Their ( ) ( ) P S H P H ∩ unsimplified, FT from (i) = 10 11 or 0.909 A1 Total: 2 7(iii) Var (B) = 45×0.6×0.4 Var (S)= 45×0.4×0.6 B1 One variance stated unsimplified Variances same B1 Second variance stated unsimplified and at least one variance clearly identified, and both evaluated or showing equal or conclusion made SR B1 – Standard Deviation calculated Fulfil all the criteria for the variance method but calculated to Standard Deviation Total: 2 Question Answer Marks Guidance 7(iv) 1 – P(0, 1) = 1 – [(0.6)10 + 10C1(0.4)(0.6)9] = 1 – 0.0464 OR P(2,3,4,5,6,7,8,9,10) = 10C2(0.4)2(0.6)8 + … + 10C9(0.4)9(0.6) + (0.4)10 M1 M1 Bin term 10Cx px(1 – p)10 – x 0 < p < 1 Correct unsimplified answer = 0.954 A1 Total: 3

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