Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 6 · Variant 2

1 On average, 1 in 50 000 people have a certain gene. Use a suitable approximating distribution to find the probability that more than 2 people in a random sample of 150 000 have the gene. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Poisson, any λ M1 Used 1 – e–3 23 1 3 2   + +     M1 Allow one end error = 0.577 (3sf) A1 SC Use of Binomial (or unsupported correct answer) scores B1 only 3

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2 A six-sided die has faces marked 1, 2, 3, 4, 5, 6. When the die is thrown 300 times it shows a six on 56 throws. (a) Calculate an approximate 96% confidence interval for the probability that the die shows a six on one throw. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Maroulla claims that the die is biased. Use your answer to part (a) to comment on this claim. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) 56 244 56 300 300 300 300 × ± × z z = 2.054 or 2.055 B1 0.14(0) to 0.233 (3sf) or 0.141 to 0.233 (3sf) A1 Must be an interval 3 2(b) 1 6 (= 0.167) This is within confidence interval, so no reason to believe die is biased. B1 FT Note if confidence interval set up with 1 56 , 6 300 it should be the value used here. FT their confidence interval. Not definite, e.g. not ‘Die not biased’. 1

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4 The areas, X cm2, of petals of a certain kind of flower have mean - cm2. In the past it has been found that - = 8.9. Following a change in the climate, a botanist claims that the mean is no longer 8.9. The areas of a random sample of 200 petals from this kind of flower are measured, and the results are summarized by Σx = 1850, Σx2 = 17 850. Test the botanist’s claim at the 2.5% significance level. 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Mark scheme: 4 est(μ) = 1850 200 or 9.25 B1 est(σ2) = 2 200 17850 1850 199 200 200     −           or 2 1 1850 199 200 17850         − M1 = 3.71 or 3.7060 or 1475 398 A1 H0: μ = 8.9 H1: μ ≠ 8.9 B1 Accept Population mean (not just mean) 1850 8.9 200 "3.706" 200 − M1 Use of biased variance (3.6875) still scores M1 = 2.57(3sf) (or using areas 0.00507 – 0.0051) A1 Accept 2.58 (3sf) or using areas 0.0049–0.005 where biased variance used. 2.24 < 2.57 or 0.00507 < 0.0125 M1 For valid comparison with 2.240 or 2.241 or valid comparison with 0.0125 Accept 2.24 < 2.58 or 0.00496 < 0.0125 where biased variance used (Reject H0) There is evidence that μ is not 8.9 A1 FT Not definite, e.g. NOT ‘μ ≠ 8.9’ Must be in context. No contradictions. (Accept cv method) (Note: Use of 1 tail test scores Max B1M1A1B0M1A1M1A0, max 6 out of 8) 8

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6 A biscuit manufacturer claims that, on average, 1 in 3 packets of biscuits contain a prize offer. Gerry suspects that the proportion of packets containing the prize offer is less than 1 in 3. In order to test the manufacturer’s claim, he buys 20 randomly selected packets. He finds that exactly 2 of these packets contain the prize offer. (a) Carry out the test at the 10% significance level. 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(b) Maria also suspects that the proportion of packets containing the prize offer is less than 1 in 3. She also carries out a significance test at the 10% level using 20 randomly selected packets. She will reject the manufacturer’s claim if she finds that there are 3 or fewer packets containing the prize offer. Find the probability of a Type II error in Maria’s test if the proportion of packets containing the prize offer is actually 1 in 7. 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(c) Explain what is meant by a Type II error in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) H0: P(contains offer) = 1 3 H1: P(contains offer) < 1 3 B1 Allow p for P(contains offer) but not just proportion P(0,1 or 2 offers in 20 | H0) = 20 2 3       + 20 19 2 3       1 3       + 20C2 18 2 3       2 1 3       M1 = 0.0176 (3sf) A1 ‘0.0176’ < 0.1 M1 For valid comparison. SC comparison of 0.982(4) > 0.9 scores M1 and recovers the previous M1 A1 (Reject H0 ) No evidence (at 10% level) to support manufacturers claim A1 FT In context. Not definite. No contradictions. (Note 2 tail test scores max B0M1A1M1A0, max 3 out of 5) Accept critical region method: M1 A1 for correctly finding critical region of < 4 ; 2 in critical region M1; A1 conclusion SC Use of Normal approximation 20 40 N , 3 9       scores B1 M1 A0 M1 A1 max; the first M1 for 20 2.5 3 40 9   −           requires use of correct continuity correction and the comparison 0.024 < 0.1 OE must be a valid comparison 5 Question Answer Marks Guidance 6(b) 1 – P(X ⩽ 3) M1 M1 for 1 – (one term omitted or extra or incorrect) or omit '1 ' − 20 19 18 2 17 3 20 20 2 3 6 6 1 6 1 6 1 1 20 C C 7 7 7 7 7 7 7              = − + + +                             A1 for all correct expression = 0.318 (3sf) A1 As final answer. 3 6(c) Concluding that prop is 1 in 3 when it is actually less(1 in 7) B1 OE, in context. 1

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7 Before a certain type of book is published it is checked for errors, which are then corrected. For costing purposes each error is classified as either minor or major. The numbers of minor and major errors in a book are modelled by the independent distributions N 380, 140 and N 210, 80 respectively. You should assume that no continuity corrections are needed when using these models. A book of this type is chosen at random. (a) Find the probability that the number of minor errors is at least 200 more than the number of major errors. 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The costs of correcting a minor error and a major error are 20 cents and 50 cents respectively. (b) Find the probability that the total cost of correcting the errors in the book is less than $190. 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Mark scheme: 7(a) P(S > L + 200) = P(S – L > 200) E(S – L) = 380–210 (=170) or E(S–L–200)= 380–210–200 (=–30) B1 These may be implied by next line Var(S – L) = 140+80 (=220) or Var(S–L–200)= 140+80 (=220) B1 200 "170" "220" − or 0 " 30" "220" −− (= 2.023) M1 Standardising with their values (must be from a combination attempt) Allow with attempted continuity correction. 1 – ɸ("2.023") M1 Area consistent with their values = 0.0216 (3sf) A1 (0.0234 with continuity correction) 5 Question Answer Marks Guidance 7(b) E(total cost) = 380 × 20 + 210 × 50 (= 18 100) B1 or $181 Var(total cost) = 140 × 202 + 80 × 502 (= 256 000) B1 or 25.6 (dollar2) These may be implied by next line 19000 "18100" "256000" − or 190 "181" "25.6" − (= 1.778) M1 Standardising with their values (must be from a combination attempt). No mixed methods. Allow with attempted continuity correction. ɸ("1.778") M1 Area consistent with their values = 0.962 or 0.963 (3 sf) A1 (0.953 or 0.954 with continuity correction) 5

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