Cambridge A Level Mathematics 9709 — 2018 May/June Paper 6 · Variant 2

1 Each of a group of 10 boys estimates the length of a piece of string. The estimates, in centimetres, are as follows. 37 40 45 38 36 38 42 38 40 39 (i) Find the mode. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the median and the interquartile range. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) 38 B1 1 1(ii) Median = 38.5 B1 CAO IQR = 40 – 38 M1 39 < UQ < 45 – 36 < LQ ⩽ 38 = 2 A1 If M0 awarded SCB1 for both UQ = 40 or 40.5 and LQ = 38 or 37.75 seen 3

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2 In a group of students, 3 are male. The proportion of male students who like their curry hot is 3 and 4 5 the proportion of female students who like their curry hot is 45. One student is chosen at random. (i) Find the probability that the student chosen is either female, or likes their curry hot, or is both female and likes their curry hot. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Showing your working, determine whether the events ‘the student chosen is male’ and ‘the student chosen likes their curry hot’ are independent. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Method 1 P(M ∩ H) = 3 4 × 3 5 = 9 20 (0.45) B1 Seen, accept unsimplified P(F or M ∩ H) = 1 4 + 9 20 = 14 20 M1 Numerical attempt at P(F) + P(M ∩ H) A1 Correct unsimplified expression = 7 10 (0.7) OE A1 Correct final answer Method 2 P(M ∩ H′) = 3 4 × 2 5 = 6 20 (0.3) B1 Seen, accept unsimplified P(F or M ∩ H) = 1 ‒ P(M ∩ H′) M1 Numerical attempt at 1 ‒ P(M ∩ H′) = 1 ‒ 3 4 × 2 5 A1 Correct unsimplified expression = 7 10 (0.7) OE A1 Correct final answer Question Answer Marks Guidance 2(i) Method 3 P(F ∩ H′ or H) = 1 4 × 1 5 + 1 4 × 4 5 + 3 4 × 3 5 B1 3 4 × 3 5 ( 9 20 ) or 1 4 × 4 5 ( 4 20 ) or 3 4 × 3 5 + 1 4 × 4 5 ( 13 20 ) seen = 1 20 + 4 20 + 9 20 M1 Numerical attempt at P(F ∩ H′) + P(F ∩ H) + P(M ∩ H) A1 Correct unsimplified expression = 7 10 (0.7) oe A1 Correct final answer Method 4 – Venn diagram style approach P(F U H) = P(F) + P(H) – P(F ∩ H) B1 3 4 × 3 5 ( 9 20 ) or 1 4 × 4 5 ( 4 20 ) or 3 4 × 3 5 + 1 4 × 4 5 ( 13 20 ) seen = 1 4 + 1 4 × 4 5 + 3 4 × 3 5 – 1 4 × 4 5 M1 Numerical attempt at P(F) + P(H) – P(F ∩ H) = 1 4 + 4 20 + 9 20 – 4 20 A1 Correct unsimplified expression = 7 10 (0.7) oe A1 Correct final answer 4 Question Answer Marks Guidance 2(ii) Method 1 (P(M) × P(H) =) 3 4 × their 13 20 = 39 80 (P(M ∩ H) =) 3 4 × 3 5 = 0.45 M1 Unsimplified, or better, legitimate numerical attempt at P(M) × P(H) and P(M ∩ H) Descriptors P(M ∩ H) and P(M) × P(H) seen, correct numerical evaluation and comparison, conclusion stated 39 80 (0.4875) ≠ 0.45, not independent A1 Method 2 P( M H ) = 9 P( ) 20 = 13 P( ) their 20 ∩ M H H = 9 13 P(M) = 3 4 M1 Unsimplified, or better, numerical attempt at P(H) and P(M ∩ H), P(M) 9 13 ≠ 3 4 , not independent A1 Descriptors P(M ∩ H), P(H) and P(M) OR P(M|H) and P(M) seen, numerical evaluation and comparison, conclusion stated Any appropriate relationship can be used, the M is awarded for an unsimplified, or better, numerical attempt at the terms required, the A mark requires the correct descriptors, numerical evaluation and comparison and the conclusion 2

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3 (i) The volume of soup in Super Soup cartons has a normal distribution with mean - millilitres and standard deviation 9 millilitres. Tests have shown that 10% of cartons contain less than 440 millilitres of soup. Find the value of -. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) A food retailer orders 150 Super Soup cartons. Calculate the number of these cartons for which you would expect the volume of soup to be more than 1.8 standard deviations above the mean. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) z = –1.282 B1 –1.282 = 440 9 µ − M1 ±Standardisation equation with 440, 9 and µ, equated to a z-value, (not 1 – z-value or probability e.g. 0.1841, 0.5398, 0.6202, 0.8159) µ = 452 A1 Correct answer rounding to 452, not dependent on B1 3 3(ii) P(z > 1.8) = 1 – 0.9641 = 0.0359 B1 Number = 0.0359 × 150 = 5.385 M1 p × 150, 0 < p < 1 (Number of cartons = ) 5 A1FT Accept either 5 or 6, not indicated as an approximation, e.g. ~, about FT their p × 150, answer as an integer 3

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5 The lengths, t minutes, of 242 phone calls made by a family over a period of 1 week are summarised in the frequency table below. Length of phone 0 < t ≤1 1 < t ≤2 2 < t ≤5 5 < t ≤10 10 < t ≤30 call (t minutes) Frequency 14 46 102 a 40 (i) Find the value of a. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Calculate an estimate of the mean length of these phone calls. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) On the grid, draw a histogram to illustrate the data in the table. [4]

Mark scheme: 5(i) a = 40 B1 1 5(ii) Mean = 0.5 14 1.5 46 3.5 102 7.5 40 20 40 242 × + × + × + × + × their = 1533 242 M1 Numerator: 5 products with at least 3 acceptable mid-points × appropriate frequency FT (i). Denominator: 242 CAO 1533 242 implies M1, but if FT an unsimplified expression required = 81 6 242 or 6.33 A1 CAO (6.3347… rounded to 3 or more SF) 2 5(iii) fd = 14, 46, 34, ( ( ) 5 their i =) 8, 2 M1 Attempt at fd [f/(attempt at cw)] or scaled freq fd 50 40 30 20 10 5 10 15 20 25 30 Length phone call /mins A1FT Correct heights seen on diagram with linear vertical scale from (x, 0) FT their 5 a only B1 Correct bar widths (1:1:3:5:20) at axis, visually no gaps, with linear horizontal scale from (0, y), first bar starting at (0,0) B1 Labels (time, mins, and fd(OE) seen, some may be as a title) and a linear scale with at least 3 values marked on each axis. (Interval notation not acceptable) 4

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6 (a) Find the number of ways in which all 9 letters of the word AUSTRALIA can be arranged in each of the following cases. (i) All the vowels (A, I, U are vowels) are together. 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(ii) The letter T is in the central position and each end position is occupied by one of the other consonants (R, S, L). 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(b) Donna has 2 necklaces, 8 rings and 4 bracelets, all different. She chooses 4 pieces of jewellery. How many possible selections can she make if she chooses at least 1 necklace and at least 1 bracelet? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a)(i) (AAAIU) * * * * Arrangements of vowels/repeats × arrangements of (consonants & vowel group) = M1 5!× 5! 3! M1 ! 3 m (m is an integer, m ⩾ 1) Both Ms can only be awarded if expression is fully correct = 2400 A1 Correct answer 3 6(a)(ii) E.g. R * * * T * * * L . Arrangements of consonants RL, RS, SL = 3P2 = 6 Arrangements of remaining letters = 6! 3! = 120 M1 k × 6! 3! or k × 3P2 or k × 3C2 or k × 3! or k × 3 × 2 (k is an integer, k ⩾ 1), no irrelevant addition Total 120 × 6 M1 Correct unsimplified expression or 6! 3! × 3C2 = 720 ways A1 Correct answer 3 Question Answer Marks Guidance 6(b) Method 1 N(2) R(8) Br(4) 1 2 1 = 2×8C2×4 = 224 M1 Multiply 3 combinations, 2Cx×8Cy×4Cz. Accept 2C1 = 2 etc. 2 1 1 = 1×8C1×4 = 32 1 1 2 = 2×8×4C2 = 96 A1 3 or more options correct unsimplified 2 0 2 = 1×1×4C2 = 6 1 0 3 = 2×1×4 = 8 M1 Summing their values of 4 or 5 legitimate scenarios (no extra scenarios) Total = 366 ways A1 Correct answer Method 2 14C4 – (2N2R or 1N3R or 4R or 3R1B or 2R2B or 1R3B or 4B) M1 ‘14C4 – k’ seen, k an integer from an expression containing 8Cx 1001 – (1×8C2 + 2×8C3 + 8C4 + 8C3×4 + 8C2×4C2 + 8×4 +1) A1 4 or more ‘subtraction’ options correct unsimplified, may be in a list 1001 – (28 + 112 + 70 + 224 + 168 + 32 + 1) M1 Their 14C4 – [their values of 6 or more legitimate scenarios] (no extra scenarios, condone omission of final bracket) = 366 A1 Correct answer 4

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7 In a certain country, 60% of mobile phones sold are made by Company A, 35% are made by Company B and 5% are made by other companies. (i) Find the probability that, out of a random sample of 13 people who buy a mobile phone, fewer than 11 choose a mobile phone made by Company A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use a suitable approximation to find the probability that, out of a random sample of 130 people who buy a mobile phone, at least 50 choose a mobile phone made by Company B. 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(iii) A random sample of n mobile phones sold is chosen. The probability that at least one of these phones is made by Company B is more than 0.98. Find the least possible value of n. 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Mark scheme: 7(i) Method 1 P(< 11) = 1 – P(11, 12, 13) M1 = 1 – 13C11(0.6)11(0.4)2 – 13C12(0.6)12(0.4) – (0.6)13 M1 Correct unsimplified answer = 0.942 A1 CAO Method 2 P(< 11) = P(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10) M1 Binomial expression of form 13Cx (p)x(1–p)13–x 0 < x < 13, 0 < p < 1 = (0.4)13 + 13C1(0.4)12(0.6) + … + 13C10(0.4)3(0.6)10 M1 Correct unsimplified answer = 0.942 A1 CAO 3 7(ii) µ = 130 × 0.35 = 45.5 var = 130 × 0.35 × 0.65 = 29.575 B1 Correct unsimplified mean and var (condone 2 σ = 29.6, σ = 5.438) P( ⩾ 50) = P 49.5 45.5 29.575 −   >     z = P (z > 0.7355) M1 Standardising, using mean σ −   ±    x their their , x = value to standardise 49.5 or 50.5 seen in ± standardisation equation = 1 – Φ(0.7355) M1 Correct final area = 1 – 0.7691 M1 = 0.231 A1 Correct final answer 5 Question Answer Marks Guidance 7(iii) 1 – (0.65)n > 0.98 or 0.02 > (0.65)n M1 Eqn or inequality involving, 0.65n and 0.02 or 0.35n and 0.98 n > 9.08 M1 Attempt to solve their eqn or inequality by logs or trial and error n = 10 A1 CAO 3

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