Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 6 · Variant 1

1 Ashok has 3 green pens and 7 red pens. His friend Rod takes 3 of these pens at random, without replacement. Draw up a probability distribution table for the number of green pens Rod takes. [4]

Mark scheme: 1 P(0) = 7/10× 6/9× 5/8 = 210/720 B1 Finding P(0, 1, 2, 3) P(1) = 3/10× 7/9× 6/8 × 3C1 = 378/720 B1 1 or 2 correct P(2) = 3/10× 2/9× 7/8 × 3C2 = 126/720 B1 3 correct P(3) = 3/10× 2/9× 1/8 = 6/720 (1/120) B1 [4] All correct

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2 The amounts of money, x dollars, that 24 people had in their pockets are summarised by Σ(x −36) = −60 and Σ(x −36)2 = 227.76. Find Σ x and Σx2. [5]

Mark scheme: 2 Σx –Σ36 = – 60 M1 Expanding brackets ie mult by 24 and subt 60 Σx = 24× 36 – 60 = 804 A1 [2] Correct answer OR x = 36 − 60 / 24 = 335. M1 Dividing by 24 and subt from 36 Σx = 335. × 24 = 804 A1 Correct answer M1 Expanding brackets with 36Σx and Σ362 Σx2 – 2.36Σx + Σ362 = 227.6 M1 min Σx2 – 2×36Σx + Σ362 = 227.6 seen Σx2 = 27011.76 (27000) A1 [3] Correct answer M1 227.76/24 – (their coded mean)2 seenOR 227.76/24 – (–2.5)2 = sd2 = 3.24 M1 Σx2/24 – ( x )2 = their var if +ve seen o.e. Σx2/24 – (33.5)2 = 3.24 A1 Σx2 = 27011.76 (27000) Correct answer 73 75 B1 ± correct z value accept ± 1 037

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3 Lengths of rolls of parcel tape have a normal distribution with mean 75 m, and 15% of the rolls have lengths less than 73 m. (i) Find the standard deviation of the lengths. [3] Alison buys 8 rolls of parcel tape. (ii) Find the probability that fewer than 3 of these rolls have lengths more than 77 m. [3]

Mark scheme: 73 − 75 B1 ± correct z value accept ± .10373 (i) z = –1.036 = σ M1 Equation with 73, 75, σ and a z value σ = 1.93 A1 [3] Rounding to correct answer (ii) P(> 77) = 0.15 M1 Prob rounding to 0.15 and 0.85 P(< 3) = P(0, 1, 2) M1 8Cxpx(1–p)8–x seen any p, 0<p<1 = (0.85)8 + 8C1(0.15)(0.85)7 + 8C2(0.15)2(0.85)6 = 0.895 A1 [3] Correct answer

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4 Prices in dollars of 11 caravans in a showroom are as follows. 16 800 18 500 17 700 14 300 15 500 15 300 16 100 16 800 17 300 15 400 16 400 (i) Represent these prices by a stem-and-leaf diagram. [3] (ii) Write down the lower quartile of the prices of the caravans in the showroom. [1] (iii) 3 different caravans in the showroom are chosen at random and their prices are noted. Find the probability that 2 of these prices are more than the median and 1 is less than the lower quartile. [3]

Mark scheme: 4 (i) 14 3 B1 Correct stem 15 3 4 5 16 1 4 8 8 17 3 7 B1 Correct leaves 18 5 Key: 143 represents 14300 dollars B1 [3] Key need dollars (ii) LQ = 15400 B1 [1] Correct answer (iii) 5/11× 4/10× 2/9 × 3C2 = 4/33 (0.121) B1 Mult 3 diff fractions or (5C2 or 2C1) 5C 2 × 2 C1 B1 seen in num OR 11C3 B1 [3] Mult by 3C2 o.e. or correct denom Correct answer 20 19 1 20

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5 A company set up a display consisting of 20 fireworks. For each firework, the probability that it fails to work is 0.05, independently of other fireworks. (i) Find the probability that more than 1 firework fails to work. [3] The 20 fireworks cost the company $24 each. 450 people pay the company $10 each to watch the display. If more than 1 firework fails to work they get their money back. (ii) Calculate the expected profit for the company. [4]

Mark scheme: 5 (i) P(> 1) = 1 – (0.95)20 – (0.95)19(0.05)120C1 M1 Binomial term 20Cx(0.05)x(0.95)20–x M1 Correct unsimplified expression = 0.264 A1 [3] Correct answer GCE AS/A LEVEL – October/November 2012 9709 61 (ii) Profit 19 or 20 work = 450×10 – 480 B1 4020 seen = 4020 Profit < 19 work = – 480 M1 Multiplying 4020 by their (i) or their 4020 × 1( − .0264) − M1 (1 – (i)) Expected profit = 480 × .0264 Multiplying 480 by [1 – their (i)] and subtracting A1 [4] Rounding to correct answer = $2830 ($2832) Or –480 + 4500 (1 – 0.264) = 2830 6 (i) p = 0.2 µ= 96× 0.2 = 19.2 σ2 = 96× 0.2× 0.8 =15.36 B1 96× 0.2 and 96× 0.2× 0.8 seen M1 d di i h

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