Cambridge A Level Mathematics 9709 — 2021 May/June Paper 6 · Variant 3

1 The number of goals scored by a team in a match is independent of other matches, and is denoted by the random variable X, which has a Poisson distribution with mean 1.36. A supporter offers to make a donation of $5 to the team for each goal that they score in the next 10 matches. Find the expectation and standard deviation of the amount that the supporter will pay. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 λ = 10 × 1.36 [= 13.6] M1 E(amount) = 5 × 13.6 = [$]68 A1 Var(amount) = 52 × 13.6 [= 340] M1 52 × … M1 … × their λ Standard Deviation = [$]18.4(4) (3 s.f.) A1 CAO condone 2 85 5

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2 In the past, the time, in hours, for a particular train journey has had mean 1.40 and standard deviation 0.12. Following the introduction of some new signals, it is required to test whether the mean journey time has decreased. (a) State what is meant by a Type II error in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) The mean time for a random sample of 50 journeys is found to be 1.36 hours. Assuming that the standard deviation of journey times is still 0.12 hours, test at the 2.5% significance level whether the population mean journey time has decreased. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) State, with a reason, which of the errors, Type I or Type II, might have been made in the test in part (b). [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Conclude (mean) (journey) time has not decreased when in fact it has. B1 OE in context 1 2(b) H0: Pop mean (or μ) = 1.4 H1: Pop mean (or μ) < 1.4 B1 May be seen in (a) 1.36 1.4 0.12 50 − M1 Accept totals method 68 70 50 0.12 − × No mixed methods or no standard deviation/variance mixes –2.357 or – 2.36 A1 Correct z or correct area if used –2.357 < –1.96 or 0.0092 < 0.025 or 0.9908 > 0.975 Or CV method 1.36 < 1.367 M1 valid comparison There is evidence that (mean) (journey) times have decreased A1 FT in context not definite no contradictions NB use of two tail test scores max B0M1A1M1A0 no ft for two tail test 5 2(c) H0 was rejected OE *B1 FT FT H0 was accepted OE Type I DB1 FT FT Type II 2

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3 The local council claims that the average number of accidents per year on a particular road is 0.8. Jane claims that the true average is greater than 0.8. She looks at the records for a random sample of 3 recent years and finds that the total number of accidents during those 3 years was 5. (a) Assume that the number of accidents per year follows a Poisson distribution. (i) State null and alternative hypotheses for a test of Jane’s claim. [1] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Test at the 5% significance level whether Jane’s claim is justified. [4] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Jane finds that the number of accidents per year has been gradually increasing over recent years. State how this might affect the validity of the test carried out in part (a)(ii). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a)(i) H0: λ = 2.4 H1: λ > 2.4 B1 Accept λ or µ Accept 2.4 or 0.8 (per year) 1 3(a)(ii) 1 – e–2.4(1 + 2.4 + 2 3 4 2.4 2.4 2.4 2 3! 4! + + ) M1 Any λ; allow one end error 0.0959 (3 sf) A1 SC unsupported answer 0.0959 scores B1 only not M1A1 0.0959 > 0.05 M1 Valid comparison Use of 0.9041 < 0.95 can recover either M1A1 or B1 There is evidence that Jane’s claim not justified or There is insufficient evidence to support Jane’s claim A1 FT OE. In context, not definite, e.g. not ‘Jane is wrong’, no contradictions. Condone omission of Jane. 4 3(b) Mean not constant so Poisson model not valid B1 1

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4 The masses, m kilograms, of flour in a random sample of 90 sacks of flour are summarised as follows. n = 90 Σm = 4509 Σm2 = 225 950 (a) Find unbiased estimates of the population mean and variance. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Calculate a 98% confidence interval for the population mean. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Explain why it was necessary to use the Central Limit theorem in answering part (b). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Find the probability that the confidence interval found in part (b) is wholly above the true value of the population mean. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) 4509 90 [= 50.1] B1 2 225950 90 ( '50.1' ) 89 90 − or 2 1 4509 225950 89 90   −     M1 Attempted. Use of biased = 0.5455 scores M0A0 491 890 or 0.552 (3 sf) A1 3 4(b) ‘50.1’ ± z 491 ' ' 890 90 M1 Expression of the correct form, allow any z-value but must be a z-value z = 2.326 B1 Accept 2.326 to 2.329 49.9 to 50.3 (3 sf) A1 FT from biased variance. Must be an interval. 3 4(c) Population of masses is unknown B1 Accept population of masses is not normal 1 4(d) 1 – 0.98 M1 0.02 seen 0.02 ÷ 2 = 0.01 A1 As final answer 2

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