Cambridge A Level Mathematics 9709 — 2023 Feb/March Paper 6 · Variant 2

1 Anita carried out a survey of 140 randomly selected students at her college. She found that 49 of these students watched a TV programme called Bunch. (a) Calculate an approximate 98% confidence interval for the proportion, p, of students at Anita’s college who watch Bunch. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Carlos says that the confidence interval found in (a) is not useful because it is too wide. (b) Without calculation, explain briefly how Carlos can use the results of Anita’s survey to find a narrower confidence interval for p. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a)  49 = 0.35  140  M1 Use of formula of correct form, ft their 14049 , any z (not a 0.35 ± z 0.35(1140− 0.35) probability). z = 2.326 B1 Accept 2.326 to 2.329 . Confidence interval = 0.256 to 0.444 (3 sf) A1 Must be an interval. 3 1(b) Find a smaller percentage confidence interval/ lower level of confidence B1 ISW if 2 reasons given. Just saying ‘use smaller z’ oe B0. Accept a correct example e.g. 90% (even if not qualified with statement). 1

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2 The number of orders arriving at a shop during an 8-hour working day is modelled by the random variable X with distribution Po 25.2 . (a) State two assumptions that are required for the Poisson model to be valid in this context. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) (i) Find the probability that the number of orders that arrive in a randomly chosen 3-hour period is between 3 and 5 inclusive. 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(ii) Find the probability that, in two randomly chosen 1-hour periods, exactly 1 order will arrive in one of the 1-hour periods, and at least 2 orders will arrive in the other 1-hour period. 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(c) The shop can only deal with a maximum of 120 orders during any 36-hour period. Use a suitable approximating distribution to find the probability that, in a randomly chosen 36-hour period, there will be too many orders for the shop to deal with. 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Mark scheme: 2(a) Orders arrive at constant mean rate (must say mean or rate) Must be in context (accept 25.2 as context). Orders arrive at random Orders arrive independently B1 Any one reason correctly stated. Orders arrive singly B1 A second reason correctly stated. SC B1: both correct, not in context. 2 2(b)(i) λ = 83 × 25.2 [= 9.45] B1 e–’9.45’( 9.45 3! 3 + 9.454! 4 + 9.455! 5 ) or e–‘9.45’ (140.65 + 332.29 + 628.03) or 0.01107 + M1 Allow any λ. Allow end errors. Expression must be seen. 0.02615 + 0.04942 = 0.0866 (3 sf) A1 If M0 allow SC B1 for 0.0866 no working seen. 3 2(b)(ii) e–3.15 ×3.15 or (1 – e–3.15(1 + 3.15)) or 0.135 or 0.822 (3 sf) B1 e–3.15 × 3.15 × (1 – e–3.15(1 + 3.15) ) M1 M1 for product of two Poisson probabilities P(1)  (1– P(0,1)) (no end errors accepted). Accept any λ. × 2 or 0.111 × 2 M1 M1 for their product of two Poisson probabilities (accept end errors) × 2. Accept any λ 0.222 (3 sf) A1 4 2(c) N(113.4, 113.4) B1 SOI 120.5−113.4 [= 0.667] M1 Standardise with their values. Allow wrong or no cc. 113.4 Must have √. 1 – ɸ(their ‘0.667’) M1 For probability area consistent with their values. = 0.252 (3 sf) A1 4

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4 The number of accidents per 3-month period on a certain road has the distribution Po , . In the past the value of , has been 5.7. Following some changes to the road, the council carries out a hypothesis test to determine whether the value of , has decreased. If there are fewer than 3 accidents in a randomly chosen 3-month period, the council will conclude that the value of , has decreased. (a) Find the probability of a Type I error. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability of a Type II error if the mean number of accidents per 3-month period is now actually 0.9. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) e-5.7(1 + 5.7 + 5.72 ) or e-5.7(1 + 5.7 + 16.245) or 0.003346 + 0.01907 + 0.05436 M1 Allow one end error. 2! Must see this expression. = 0.0768 (3 sf) A1 SC B1 for unsupported answer of 0.0768 . 2 4(b) e –0.9(1 + 0.9 + 0.92 ) M1 Attempted; allow one end error (must see expression). 2! = 1 – e –0.9(1 + 0.9 + 0.92 )= 1 – e –0.9(1 + 0.9 + 0.405) = 1 – (0.4066 + 3659 + A1 Correct expression P(X ⩾ 3) no end errors (must see 2! expression). 0.1647) = 0.0629 (3 sf) A1 SC B2 for unsupported answer of 0.0629 . 3

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5 The masses, in grams, of large andsmall packets of Maxwheat cereal have the independent distributions N 410.0, 3.62 and N 206.0, 3.72 respectively. (a) Find the probability that a randomly chosen large packet has a mass that is more than double the mass of a randomly chosen small packet. 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The packets are placed in boxes. The boxes are identical in appearance. 60% of the boxes contain exactly 10 randomly chosen large packets. 40% of the boxes contain exactly 20 randomly chosen small packets. (b) Find the probability that a randomly chosen box contains packets with a total mass of more than 4080 grams. 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Mark scheme: 5(a) D = L – 2S E(D) = 410 – 2(206) = –2 B1 SOI. OE using 2S–L. Var(D) = 3.62 + 4 × 3.72 [= 67.72] B1 SOI 0 −−( 2) [= 0.243] M1 For standardising using their values. '67.72' 1 – ɸ(their ‘0.243’) M1 For probability area consistent with their values. = 0.404 (3 sf) A1 As final answer. 5 5(b) B1 One of N(4100, 129.6) or N(4120, 273.8) USED TL ~ N(4100, 10×3.62) TS ~ N(4120, 20×3.72) (unchanged) in a standardising equation. M1 Standardising with either their N(4100, 129.6) or 4080 − 4100 (= –1.757) 4080 − 4120 (= –2.417) '129.6' '273.8' N(4120, 273.8) or their N(…,…) (could be from a combination). M1 One area consistent with their working (could be from a 1 – ɸ(‘–1.757’) = ɸ(1.757) 1 – ɸ(‘–2.417’) = ɸ(2.417) combination). Do not ISW. A1 Both of these correct. Do not ISW. = 0.9605 or 0.961 = 0.9921 or 0.9922 or 0.992 0.6 × ‘their 0.9605’ + 0.4 × ‘their 0.9921’ M1 Must be using probabilities. = 0.973 (3 sf) A1 6

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6 Last year, the mean time taken by students at a school to complete a certain test was 25 minutes. Akash believes that the mean time taken by this year’s students was less than 25 minutes. In order to test this belief, he takes a large random sample of this year’s students and he notes the time taken by each student. He carries out a test, at the 2.5% significance level, for the population mean time, - minutes. Akash uses the null hypothesis H0: - = 25. (a) Give a reason why Akash should use a one-tailed test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Akash finds that the value of the test statistic is z = −2.02. (b) Explain what conclusion he should draw. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ In a different one-tailed hypothesis test the z-value was found to be 2.14. (c) Given that this value would lead to a rejection of the null hypothesis at the !% significance level, find the set of possible values of !. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The population mean time taken by students at another school to complete a test last year was m minutes. Sorin carries out a one-tailed test to determine whether the population mean this year is less than m, using a random sample of 100 students. He assumes that the population standard deviation of the times is 3.9 minutes. The sample mean is 24.8 minutes, and this result just leads to the rejection of the null hypothesis at the 5% significance level. (d) Find the value of m. 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Mark scheme: 6(a) He is expecting a decrease (in μ) B1 OE 1 6(b) −2.02 < −1.96 M1 For valid comparison. Allow 2.02 > 1.96 or 0.0217 < 0.025 or 0.9783 > 0.975 (Reject H0 ) A1 OE (such as evidence to support Akash’s belief), in There is evidence to suggest that this year’s (mean) time is less than 25 context, not definite. No contradictions. 2 6(c) 1 – ɸ(2.14) [= 0.0162] M1 1.62 A1 Allow 1.62% or 1.6 or 1.6%. α ⩾ 1.62 (3 sf) A1ft FT their 1.62 . Allow α ⩾ 1.62% or 1.6 or 1.6%. Condone >. 3 6(d) 24.8 − m M1 For standardising. 3.9 10 24.8 − m = −1.645 M1 Equate their standardised value to −1.645 (signs must be 3.9  10 consistent). m = 25.4 (3 sf) A1 3

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