Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 6 · Variant 2

1 The following are the times, in minutes, taken by 11 runners to complete a 10 km run. 48.3 55.2 59.9 67.7 60.5 75.6 62.5 57.4 53.4 49.2 64.1 Find the mean and standard deviation of these times. [3]

Mark scheme: 1 x¯ = 59.4 B1 M1 Correct method (can be implied by correct answer) σ = 7.68 A1 [3] Correct answer 12 12

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2 Twelve coins are tossed and placed in a line. Each coin can show either a head or a tail. (i) Find the number of different arrangements of heads and tails which can be obtained. [2] (ii) Find the number of different arrangements which contain 7 heads and 5 tails. [1]

Mark scheme: 12 seen2 (i) each in 2 ways = 212 M1 2 = 4096 A1 [2] Correct answer 12! (ii) B1 [1] 7!5! = 792

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3 (a) Geoff wishes to plant 25 flowers in a flower-bed. He can choose from 15 different geraniums, 10 different roses and 8 different lilies. He wants to have at least 11 geraniums and also to have the same number of roses and lilies. Find the number of different selections of flowers he can make. [4] (b) Find the number of different ways in which the 9 letters of the word GREENGAGE can be arranged if exactly two of the Gs are next to each other. [3]

Mark scheme: 3 (a) G R L 11 7 7 = 15C11 × 10C7 × 8C7 = 1310400 M1 Multiplying 3 combinations 13 6 6 = 15C13 × 10C6 × 8C6 = 617400 A1 One of 1310400, 617400, 14112 seen 15 5 5 = 15C15 × 10C5 × 8C5 = 14112 M1 Adding 3 options Total = 1941912 (1940000) A1 [4] Correct answer (b) e.g. * E * R * E (GG) N * A * E * gives 6 ways for G B1 7! / 3! Or 7!/3!3! seen oe 7! 6 or 8!/3! – 2 × 7!/3! B1 Multiplying by 6 (gaps) oe 3!× = 5040 ways. B1 [3] Correct final answer

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4 The weights of 220 sausages are summarised in the following table. Weight (grams) <20 <30 <40 <45 <50 <60 <70 Cumulative frequency 0 20 50 100 160 210 220 (i) State which interval the median weight lies in. [1] (ii) Find the smallest possible value and the largest possible value for the interquartile range. [2] (iii) State how many sausages weighed between 50 g and 60 g. [1] (iv) On graph paper, draw a histogram to represent the weights of the sausages. [4]

Mark scheme: 4 (i) 45 – 50 g B1 [1] (ii) LQ in 40 – 45 UQ in 50 – 60 M1 Considering groups containing LQ and Smallest IQ range could be 5 UQ (can be implied) Largest IQ range could be 20 A1 [2] Correct answer (iii) 50 B1 [1] (iv) freqs 0, 20, 30, 50, 60, 50, 10 fd 0, 2, 3, 10, 12, 5, 1 M1 Attempt at frequencies and fd B1 Correct labels and scales with a histogram-type shape B1 Correct bar widths starting at 20 A1 [4] Correct heights of bars

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5 A triangular spinner has one red side, one blue side and one green side. The red side is weighted so that the spinner is four times more likely to land on the red side than on the blue side. The green side is weighted so that the spinner is three times more likely to land on the green side than on the blue side. (i) Show that the probability that the spinner lands on the blue side is 8.1 [1] (ii) The spinner is spun 3 times. Find the probability that it lands on a different coloured side each time. [3] (iii) The spinner is spun 136 times. Use a suitable approximation to find the probability that it lands on the blue side fewer than 20 times. [5]

Mark scheme: 5 (i) 4p + p + 3p = 1 so P(blue) = 1/8 AG B1 [1] Must show something (ii) P(R) = ½, P(B) = 1/8, P(G) = 3/8 M1 [3] Multiplying P (R, B, G) together P(all different) = ½ × 1/8 × 3/8 × 3! M1 Mult by 3! =9/64 (0.141) A1 Correct answer GCE AS/A LEVEL – October/November 2011 9709 62 (iii) mean = 136 × 1/8 = 17, var = 14.875 B1 Unsimplified mean and variance  19.5 − 17  correct P(<20) = P  z <  M1 Standardising, need sq rt  14.875  M1 Cont correction 19.5 or 20.5 = Ф(0.648) M1 Correct area, > 0.5 legit = 0.742 A1 [5] Correct answer

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6 There are a large number of students in Luttley College. 60% of the students are boys. Students can choose exactly one of Games, Drama or Music on Friday afternoons. It is found that 75% of the boys choose Games, 10% of the boys choose Drama and the remainder of the boys choose Music. Of the girls, 30% choose Games, 55% choose Drama and the remainder choose Music. (i) 6 boys are chosen at random. Find the probability that fewer than 3 of them choose Music. [3] (ii) 5 Drama students are chosen at random. Find the probability that at least 1 of them is a boy. [6]

Mark scheme: 6 (i) P(0, 1, 2) B1 0.15 and 0.85 seen =(0.85)6 + (0.15)(0.85)56C1 + M1 Any binomial expression Σpowers = 6, (0.15)2(0.85)46C2 Σ p = 1 = 0.953 A1 [3] Correct answer (ii) P(D) = 0.6 × 0.1 + 0.4 × 0.55 = 0.28 M1 Attempt to find P(D) P ( B ∩ D ) A1 0.28 seen P(B|D) = M1 Using cond prob formula to find P ( D ) P(B|D) 0.06/0.28 = 0.2143 √A1 Correct unsimplified answer P(> 1) = 1 – P(0) M1 Binomial expression 1 –P(0) or 1 –P(0, = 1 – (0.7857)5 1) Σ p = 1 = 1 – 0.7078 A1 [6] Correct answer accept 0.700 = 0.701 12 − 8 M1 Standardising any one sq rt no cc

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7 (i) In a certain country, the daily minimum temperature, in ◦C, in winter has the distribution N(8, 24). Find the probability that a randomly chosen winter day in this country has a minimum temperature between 7 ◦C and 12 ◦C. [3] The daily minimum temperature, in ◦C, in another country in winter has a normal distribution with mean µ and standard deviation 2µ. (ii) Find the proportion of winter days on which the minimum temperature is below zero. [2] (iii) 70 winter days are chosen at random. Find how many of these would be expected to have a minimum temperature which is more than three times the mean. [3] (iv) The probability of the minimum temperature being above 6 ◦C on any winter day is 0.0735. Find the value of µ. [3]

Mark scheme: 12 8 M1 Standardising any one, sq rt, no cc 7 (i) z1 = = 0.816 Ф1(0.816) = 0.7926 24 7 − 8 z2 = = –0.204 Ф2(–0.204) = 1 – 0.5808 M1 Correct area Ф1 + Ф2 – 1 24 Prob = 0.7926 – (1 – 0.5808) = 0.373 A1 [3] Correct answer 0 −µ (ii) z = = –0.5 M1 Standardising, no cc no sq rt, one 2µ variable P(z < –0.5) = 1 – 0.6915 = 0.309 or 30.9% A1 [2] Correct answer oe 3µ −µ M1 Standardising and eliminating µ (iii) z = = 1 2µ M1 Subt from 1 and multiplying by 70 P(z > 1) = 1 – 0.8413 = 0.1587 A1 [3] Correct answer accept 11 or 12 70 × 0.1587 = 11.1 (iv) z = 1.45 B1 ± 1.45 seen 6 −µ 1.45 = M1 Solving forµ with 6, 2µ , µ and their z 2µ µ = 1.54 A1 [3] Correct answer

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