Cambridge A Level Mathematics 9709 — 2020 Feb/March Paper 6 · Variant 2

1 The booklets produced by a certain publisher contain, on average, 1 incorrect letter per 30 000 letters, and these errors occur randomly. A randomly chosen booklet from this publisher contains 12 500 letters. Use a suitable approximating distribution to find the probability that this booklet contains at least 2 errors. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 (λ =) 5 12 = 0.417 or better B1 1 – 5 12 e −(1 + 5 12 ) M1 1 – P(X = 0 or 1), by Poisson, using any λ, allow 1 – P(X = 0 or 1 or 2) for M1 = 0.0661 or 0.0662 (3 sf) A1 Final answer SC use of Binomial (from 0.06607…) B1 only 3

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2 Lengths of a certain species of lizard are known to be normally distributed with standard deviation 3.2 cm. A naturalist measures the lengths of a random sample of 100 lizards of this species and obtains an !% confidence interval for the population mean. He finds that the total width of this interval is 1.25 cm. Find !. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 × z × 3.2 10 = 1.25 z = 1.953 A1 SOI ɸ(‘their 1.953’) (= 0.9746) M1 = 1 – 2(1 – ‘0.9746’) = 0.9492 M1 OE α = 94.9 or 95 A1 CWO 5

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3 In the past, the mean time taken by Freda for a particular daily journey was 39.2 minutes. Following the introduction of a one-way system, Freda wishes to test whether the mean time for the journey has decreased. She notes the times, t minutes, for 40 randomly chosen journeys and summarises the results as follows. n = 40 Σt = 1504 Σt2 = 57 760 (a) Calculate unbiased estimates of the population mean and variance of the new journey time. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Test, at the 5% significance level, whether the population mean time has decreased. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) est (μ) = 37.6 or 1504 40 or 188 5 B1 est (σ2) = 2 40 57760 39 40 37.6   −  = 31.0154 = 2016 65 M1 Correct substitution in any correct formula 2 1504 1 39 40 57760   −   = 31.(0) (3 sf) A1 Accept 2016 65 or 1 65 31 3 3(b) H0: Pop mean (or μ) = 39.2 H1: Pop mean (or μ) < 39.2 B1 Both. Not just ‘mean’ 40 ' 0154 . 31 ' 2. 39 '6. 37 ' − M1 Allow use of biased variance (30.2), must have √40 = –1.817 A1 SC FT use of biased = –1.840 for A1 ‘1.817’ > 1.645 OE M1 Valid comparison ‘their 1.817’ with 1.645 or valid area comparison 0.0346 < 0.05 OE There is evidence that mean time has decreased A1FT FT their 1.817; in context, not definite, no contradictions SC For 2 tail test: H1: μ ≠ 39.2 and comp 1.96, max B0M1A1M1A0 (no FT for final mark) 5

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4 The number of accidents on a certain road has a Poisson distribution with mean 0.4 per 50-day period. (a) Find the probability that there will be fewer than 3 accidents during a year (365 days). [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) The probability that there will be no accidents during a period of n days is greater than 0.95. Find the largest possible value of n. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) λ (= 0.4 × 365 ÷ 50) = 2.92 B1 e–2.92(1 + 2.92 + 2 2.92 2 ) M1 Any λ. Allow one end error = 0.441 (3 sf) A1 3 4(b) e–λ > 0.95 M1 Allow ‘=’ throughout –λ > ln 0.95 or λ < 0.051293 OE M1 Attempt ln both sides ‘0.051293’ × 50 ÷ 0.4 (= 6.411) M1 Largest n is 6 (3 sf) Allow n = 6 or n ⩽ 6 (NOT n < 6 or n ⩾ 6 as final answer) A1 SC Trial and Improvement M1 for e–λ > 0.95 SOI; M1 for 0.4 50 n λ = × ; M1 for use of both n = 6 giving 0.9531 and n = 7 giving 0.9455; A1 n = 6 4

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6 The volumes, in millilitres, of large and small cups of tea are modelled by the distributions N 200, 30 and N 110, 20 respectively. (a) Find the probability that the total volume of a randomly chosen large cup of tea and a randomly chosen small cup of tea is less than 300 ml. 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(b) Find the probability that the volume of a randomly chosen large cup of tea is more than twice the volume of a randomly chosen small cup of tea. 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Mark scheme: 6(a) N(310, 50) B1 SOI ' 50 ' ' 310 ' 300− (= –1.414) M1 Standardise using their values Φ(‘–1.414’) = 1 – ɸ(‘1.414’) M1 Area consistent with their values = 0.0786 or 0.0787 (3 sf) A1 As final answer 4 Question Answer Marks Guidance 6(b) P(L – 2S > 0 ) M1 OE SOI E(X) = 200-2x110 or = – 20 B1 OE seen Var = 30 + 22 × 20 or = 110 B1 Seen N(–20, 110) ' 110 ' )' 20 (' 0 − − (= 1.907) M1 Standardising with their values. Mean and variance must come from a combination attempt. 1 – Φ(‘1.907’) M1 Correct area consistent with their working = 0.0283 (3 sf) A1 Final answer 6

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7 A national survey shows that 95% of year 12 students use social media. Arvin suspects that the percentage of year 12 students at his college who use social media is less than the national percentage. He chooses a random sample of 20 students at his college and notes the number who use social media. He then carries out a test at the 2% significance level. (a) Find the rejection region for the test. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability of a Type I error. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Jimmy believes that the true percentage at Arvin’s college is 70%. Assuming that Jimmy is correct, find the probability of a Type II error. 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Mark scheme: 7(a) M1 OE P(X ⩽ 17) or P(X ⩽ 16) attempted, using B(20, 0.95) M1 OE (P(X ⩽ 17)) = 0.0755 and (P(X ⩽ 16)) = 0.0159 A1 OE (0.925 and 0.984) both correct Rej region is X ⩽ 16 or X < 17 A1 Dependent on M1M1 and previous answers correct to at least 0.075/0.076 and 0.016 or 0.92/0.93 and 0.98 Correct unsupported answers of 0.0755 and 0.0159 OE scores M1 M1 A0 4 Question Answer Marks Guidance 7(b) 0.0159 B1 FT their rejection region, from Binomial in a, if P(X in rejection region) < 0.025 1 7(c) Use of B(20, 0.7) M1 P(X > 16 | p = 0.7) M1 Correct method using B(20, 0.7) = 0.107 A1 3

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