Cambridge A Level Mathematics 9709 — 2019 Feb/March Paper 7 · Variant 2

1 The masses of a certain variety of plums are known to have standard deviation 13.2 g. A random sample of 200 of these plums is taken and the mean mass of the plums in the sample is found to be 62.3 g. (i) Calculate a 98% confidence interval for the population mean mass. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State with a reason whether it was necessary to use the Central Limit theorem in the calculation in part (i). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) z = 2.326 B1 62.3 ± z 13.2 200 M1 Any z. Expression of correct form. Must be a ‘z’ 60.1 to 64.5 (3 sfs) A1 Must be an interval 3 1(ii) Yes, because pop not (given to be) normal, or pop distribution unknown B1 No contradictions 1

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3 At factory A the mean number of accidents per year is 32. At factory B the records of numbers of accidents before 2018 have been lost, but the number of accidents during 2018 was 21. It is known that the number of accidents per year can be well modelled by a Poisson distribution. Use an approximating distribution to test at the 2% significance level whether the mean number of accidents at factory B is less than at factory A. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 H0: λ = 32 H1: λ < 32 B1 Accept ‘population mean’ (µ) X ~ N(32, 32) B1 seen or implied 21.5 32 32 − M1 Standardise with their values. Allow with no or wrong cc = –1.856 cv of z = –2.054 (or –2.055 or –2.053) A1 ‘1.856’ < 2.054 M1 Valid comparison or comp ɸ(“1.856”) with 0.98 i.e. 0.9682 < 0.98 oe No evidence that fewer accidents at B than at A A1f No contradictions Note Use of CV method x = 20.38 M1 A1 comparison 21.5 > 20.38 M1 conc A1 6

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5 The number of eagles seen per hour in a certain location has the distribution Po 1.8 . The number of vultures seen per hour in the same location has the independent distribution Po 2.6 . (i) Find the probability that, in a randomly chosen hour, at least 2 eagles are seen. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that, in a randomly chosen half-hour period, the total number of eagles and vultures seen is less than 5. 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Alex wants to be at least 99% certain of seeing at least 1 eagle. (iii) Find the minimum time for which she should watch for eagles. 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Mark scheme: 5(i) M1 Accept any λ. Accept 1–P(0,1,2) = 0.537 (3 sf) A1 2 5(ii) λ = 2.2 B1 e–2.2(1 + 2.2 + 2 3 4 2.2 2.2 2.2 2! 3! 4! + + ) M1 Attempt expr'n for P(X ⩽ 4),allow one end error, allow any λ = 0.928 (3 sf) or 0.927 A1 3 Question Answer Marks Guidance 5(iii) 1 – e–1.8t ⩾ 0.99 or 1 – e–λ ⩾ 0.99 M1 Condone = signs/incorrect inequality signs e–1.8t ⩽ 0.01 or e–λ ⩽ 0.01 –1.8t ⩽ ln0.01 M1 Valid attempt take logs (must have single term on each side) t ⩾ 2.56 She must watch for at least 2.56 (hours) A1 or 2 hours, 34 mins or better. No errors seen 3

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6 The time taken by volunteers to complete a certain task is normally distributed. In the past the time, in minutes, has had mean 91.4 and standard deviation 6.4. A new, similar task is introduced and the times, t minutes, taken by a random sample of 6 volunteers to complete the new task are summarised by Σt = 568.5. Andrea plans to carry out a test, at the 5% significance level, of whether the mean time for the new task is different from the mean time for the old task. (i) Give a reason why Andrea should use a two-tail test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State the probability that a Type I error is made, and explain the meaning of a Type I error in this context. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ You may assume that the times taken for the new task are normally distributed. (iii) Stating another necessary assumption, carry out the test. 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Mark scheme: 6(i) Test is for “difference” oe B1 Test is not for ‘increase’ or ‘decrease’ oe No contradictions 1 6(ii) 0.05 B1 Conclude mean time is different when it is not B1 oe, in context 2 Question Answer Marks Guidance 6(iii) Assume σ = 6.4 B1 H0: pop mean = 91.4 H1: pop mean ≠ 91.4 B1 Allow µ, but not ‘mean’ x = 568.5 6 (= 94.75) B1 6.4 6 '94.75' 91.4 − M1 Must have √6 = 1.282 cv of z = 1.96 A1 ‘1.282’ < 1.96 M1 Valid comparison or comp ɸ(“1.282”) with 0.975 0.9(001) < 0.975 or 0.0999 (or 0.1) > 0.025 consistent use of one tail test can score M1 for comparison with 1.645oe but not A1ft oe. No contradictions. ft their z. No evidence mean time different A1 ft CV method x = 96.52 M1 A1 94.75 < 96.52 M1 Conc A1 7

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