Cambridge A Level Mathematics 9709 — 2017 Feb/March Paper 6 · Variant 2

2 A bag contains 10 pink balloons, 9 yellow balloons, 12 green balloons and 9 white balloons. 7 balloons are selected at random without replacement. Find the probability that exactly 3 of them are green. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 12 28 3 4 40 7 × C C C M1 denom. M1 Correct numerator or denominator unsimplified = 0.242 A1 OR P(GGG) = 7 3 12 11 10 28 27 26 25 40 39 38 37 36 35 34 × × × × × × × C M1 Multiplying 3 green probs with 4 non-green probs, without replacement M1 Multiplying by 7C3 = 0.242 A1 Total: 3

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3 It is found that 10% of the population enjoy watching Historical Drama on television. Use an appropriate approximation to find the probability that, out of 160 people chosen randomly, more than 17 people enjoy watching Historical Drama on television. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 np = 160×0.1 (16) npq = 160×0.1×0.9 (14.4) B1 Correct unsimplified np and npq P(> 17) = P 17.5 16 14.4 −   >     z = P(z > 0.3953) M1 Standardising need √ M1 16.5 or 17.5 seen in standardised eqn for continuity correction = 1 – 0.6536 M1 Correct area from their mean (1 – Φ), final solution = 0.346 A1 Total: 5

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4 The weights in kilograms of packets of cereal were noted correct to 4 significant figures. The following stem-and-leaf diagram shows the data. 747 3 (1) 748 1 2 5 7 7 9 (6) 749 0 2 2 2 3 5 5 5 6 7 8 9 (12) 750 1 1 2 2 2 3 4 4 5 6 7 7 8 8 9 (15) 751 0 0 2 3 3 4 4 4 5 5 7 7 9 (13) 752 0 0 0 1 1 2 2 3 4 4 4 (11) 753 2 (1) Key: 748 5 represents 0.7485 kg. (i) On the grid, draw a box-and-whisker plot to represent the data. [5] (ii) Name a distribution that might be a suitable model for the weights of this type of cereal packet. Justify your answer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) LQ = 0.7495 Med = 0.7507 UQ = 0.7517 M1 Attempt to find all 3 quartiles can be implied, Condone LQ=0.7496, Med=0.7506, UQ=0.7515 B1 Correct median line in box using their scale A1 Correct quartiles in box B1 Correct end whiskers(not dots or boxes), lines not through box, B1 Correct uniform scale from at least 0.7473 to 0.7532, and label (wt) kg oe can be seen in title or scale Total: 5 0.747 0.748 0.749 0.750 0.751 0.752 0.753 Wt kg Question Answer Marks Guidance 4(ii) Normal B1 Symmetrical/peaks in middle or tails off quickly B1 Need symm + another reason Total: 2

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5 (i) A plate of cakes holds 12 different cakes. Find the number of ways these cakes can be shared between Alex and James if each receives an odd number of cakes. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Another plate holds 7 cup cakes, each with a different colour icing, and 4 brownies, each of a different size. Find the number of different ways these 11 cakes can be arranged in a row if no brownie is next to another brownie. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) A plate of biscuits holds 4 identical chocolate biscuits, 6 identical shortbread biscuits and 2 identical gingerbread biscuits. These biscuits are all placed in a row. Find how many different arrangements are possible if the chocolate biscuits are all kept together. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) M1 A1 Correct unsimplified answer (can be implied by final answer) = 2048 A1 Correct answer Total: 3 5(ii) 7! 8P4 B1 7! seen alone or multiplied only (cupcakes ordered) M1 multiplying by 8P4 o.e (placing brownies) = 8467200 A1 correct answer Total: 3 5(iii) 9! / (6! 2!) B1 9! oe seen alone or as numerator M1 dividing by at least one of 6!,2! (removing repeated shortbread or gingerbread biscuits) ignore 4! if present = 252 A1 correct answer Total: 3 × ×

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7 (a) The lengths, in centimetres, of middle fingers of women in Raneland have a normal distribution with mean - and standard deviation 3. It is found that 25% of these women have fingers longer than 8.8 cm and 17.5% have fingers shorter than 7.7 cm. (i) Find the values of - and 3. [5] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ The lengths, in centimetres, of middle fingers of women in Snoland have a normal distribution with mean 7.9 and standard deviation 0.44. A random sample of 5 women from Snoland is chosen. (ii) Find the probability that exactly 3 of these women have middle fingers shorter than 8.2 cm. [5] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) The random variable X has a normal distribution with mean equal to the standard deviation. Find the probability that a particular value of X is less than 1.5 times the mean. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a)(i) 8.8 0.674 σ = ⇒ 0.674σ = 8.8 – µ B1 ±0.674 seen 7.7 0.935 µ σ − − = ⇒-0.935σ = 7.7 – µ B1 ±0.935 seen (condone ±0.934) M1 An eqn with a z-value, µ and σ allow sq rt, sq cc M1 sensible attempt to eliminate µ or σ by substitution or subtraction σ = 0.684 µ = 8.34 A1 correct answers (from –0.935) Total: 5 7(a)(ii) P(< 8.2) = P 8.2 7.9 0.44 −   <     z M1 Standardising no cc no sq rt no sq M1 Correct area ie Φ, final solution = P(z < 0.6818) = 0.7524 A1 Correct prob rounding to 0.752 P(3) = 5C3 (0.7524)3(0.2476)2 M1 Binomial 5Cx powers summing to 5, any p, Σp = 1 = 0.261 A1 Total: 5 Question Answer Marks Guidance 7(b) P(< 1.5µ) = P 1.5µ µ µ   − <     z = P (z < 0.5) *M1 standardising with µ and σ (σ may be replaced by µ) DM1 just one variable = 0.692 A1 Total: 3

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