Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 6 · Variant 2

2 The circumferences, c cm, of some trees in a wood were measured. The results are summarised in the table. Circumference (c cm) 40 < c ≤50 50 < c ≤80 80 < c ≤100 100 < c ≤120 Frequency 14 48 70 8 (i) On the grid, draw a cumulative frequency graph to represent the information. [3] (ii) Estimate the percentage of trees which have a circumference larger than 75 cm. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) points (50, 14), (80, 62), (100, 132), (120, 140) B1 Correct cfs values seen listed, in or by table or on graph, 0 not required cf 200. 100 0 20 40 60 80 100 120 Circumference cm B1 Axes labelled ‘cumulative frequency’ (or cf) and ‘circumference [or cir or c etc.] (in) cm’. Linear scales – c.f. 0–140 circumference 40–120 (ignore <40 on circ.) At least 3 values stated on each axis, but (0,0) can be implied without stating. B1 All points plotted accurately 3 2(ii) 140 – 54 = 86 M1 Finding correct value from graph (checked ±1 mm) or linear interpolation. Subtraction from 140 can be implied Percentage = 61.4% A1 60.5% ⩽ Ans ⩽ 64.5% 2

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4 A fair tetrahedral die has faces numbered 1, 2, 3, 4. A coin is biased so that the probability of showing a head when thrown is 1 The die is thrown once and the number n that it lands on is noted. The 3. biased coin is then thrown n times. So, for example, if the die lands on 3, the coin is thrown 3 times. (i) Find the probability that the die lands on 4 and the number of times the coin shows heads is 2. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that the die lands on 3 and the number of times the coin shows heads is 3. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the probability that the number the die lands on is the same as the number of times the coin shows heads. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) P(4, 2H) = 1 4 ×4C2×( 1 3 )2( 2 3 )2 M1 M1 Remaining factor is ( 1 3 )2( 2 3 )2 [or 4 81 ] multiplied by integer value k ⩾ 1 OE = 2 27 (0.0741) A1 3 4(ii) P(3, 3H) = 1 4 × ( 1 3 )3 = 1 108 (0.00926) B1 1 4(iii) P(1, 1H) = 1 4 × 1 3 = 1 12 (0.08333) P(2, 2H) = 1 4 × ( 1 3 )2 = 1 36 (0.02778) P(3, 3H) = 1 4 × ( 1 3 )3 = 1 108 (0.009259) P(4, 4H) = 1 4 × ( 1 3 )4 = 1 324 (0.003086) M1 Correct expression for 1 of P(1, 1H), P(2, 2H), P(4, 4H) Unsimplified (or better) M1 Summing their values for 3 or 4 appropriate outcomes for the ‘game’ with no additional outcomes. Prob = 10 81 (0.123) A1 3

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5 Blank CDs are packed in boxes of 30. The probability that a blank CD is faulty is 0.04. A box is rejected if more than 2 of the blank CDs are faulty. (i) Find the probability that a box is rejected. 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(ii) 280 boxes are chosen randomly. Use an approximation to find the probability that at least 30 of these boxes are rejected. 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Mark scheme: 5(i) EITHER: P(> 2) = 1 – P(0, 1, 2) (M1 = 1 – (0.96)30 – 30C1(0.04)(0.96)29 – 30C2(0.04)2(0.96)28 ( = 1 – 0.2938… – 0.3673… – 0.2219… ) A1 Correct unsimplified answer = 1-0.883103 = 0.117 (0.116896) A1) OR: P(> 2) = P(3,4,5,6,….30) (M1 Binomial term of form 30Cxpx(1 – p)30 – x , 0 < p < 1 any p = 30C3(0.04)3(0.96)27+ 30C4(0.04)4(0.96)26 + … +(0.04)30 A1 Correct unsimplified answer = 0.117 A1) 3 Question Answer Marks Guidance 5(ii) np = 280 × 0.1169 = 32.73, npq = 280 × 0.1169 × 0.8831 = 28.9 M1 FT Correct unsimplified np and npq, FT their p from (i), P(⩾ 30) = P 29.5 32.73 28.9 −   >     z = P(z > – 0.6008) M1 Substituting their µ and σ (√npq only) into the Standardisation Formula M1 Using continuity correction of 29.5 or 30.5 M1 Appropriate area Φ from standardisation formula P(z >….) in final solution = 0.726 A1 5 Question Answer Marks Guidance 6(a)(i) EITHER: 3**, 4**, 6**, 8** (M1 5P2 or 5C2 × 2! or 5 × 4 OE (considering final 2 digits) options 4 × 5 × 4 = 80 M1 Mult by 4 or summing 4 options (considering first digit) A1) Correct final answer OR: Total number of values: 6 × 5 × 4 = 120 (M1 Calculating total number of values (with subtraction seen) Number of values less than 300: 2 × 5 × 4 = 40 M1 Calculating number of unwanted values Number of evens = 120 – 40 = 80 A1) Correct final answer 3 Question Answer Marks Guidance 6(a)(ii) 3**, 4**, 6**, 8** EITHER: options 4 × 6 × 4 (last) (M1 6 linked to considering middle digit e.g. multiplied or in list M1 Multiply an integer by 4 × 4 (condone × 16) (No additional figures present for both M’s to be awarded) = 96 A1) OR: Total number of values 4 × 6 × 6 = 144 (M1 Calculating total number of values (with subtraction seen) Number of odd values 4 × 6 × 2 = 48 M1 Calculating number of unwanted values Number of evens = 144 – 48 = 96 A1) 3 6(b)(i) 252 B1 1

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7 In Jimpuri the weights, in kilograms, of boys aged 16 years have a normal distribution with mean 61.4 and standard deviation 12.3. (i) Find the probability that a randomly chosen boy aged 16 years in Jimpuri weighs more than 65 kilograms. 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(ii) For boys aged 16 years in Jimpuri, 25% have a weight between 65 kilograms and k kilograms, where k is greater than 65. Find k. 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In Brigville the weights, in kilograms, of boys aged 16 years have a normal distribution. 99% of the boys weigh less than 97.2 kilograms and 33% of the boys weigh less than 55.2 kilograms. (iii) Find the mean and standard deviation of the weights of boys aged 16 years in Brigville. 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Mark scheme: 7(i) P(> 65) = P 65 61.4 12.3 −   >     z = P (z > 0.2927) M1 condone ± standardisation formula M1 Correct area (< 0.5) = 1 – 0.6153 = 0.385 A1 3 Question Answer Marks Guidance 7(ii) P (< 65) = 0.6153 so P(< k) = 0.25 + 0.6153 = 0.8653 B1 z = 1.105 B1 z = ± 1.105 seen or rounding to 1.1 1.105 = 61.4 12.3 − k M1 standardising allow ±, cc, sq rt, sq. Need to see use of tables backwards so must be a z-value, not 1 – z value. k = 75.0 A1 Answers which round to 75.0. Condone 75 if supported. 4 7(iii) 2.326 = B1 ± 2.326 seen (Use of critical value) –0.44 = B1 ± 0.44 seen M1 An equation with a z-value, µ, σ and 97.2 or 55.2, allow √σ or σ2 M1 Algebraic elimination µ or σ from their two simultaneous equations µ = 61.9 σ = 15.2 A1 both correct answers 5 σ µ − 2. 97 σ µ − 2. 55

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