Cambridge A Level Mathematics 9709 — 2022 May/June Paper 6 · Variant 2

1 (a) A javelin thrower noted the lengths of a random sample of 50 of her throws. The sample mean was 72.3m and an unbiased estimate of the population variance was 64.3m2. Find a 92% confidence interval for the population mean length of throws by this athlete. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) A discus thrower wishes to calculate a 92% confidence interval for the population mean length of his throws. He bases his calculation on his first 50 throws in a week. Comment on this method. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 72.3 ± z 64.3 50 M1 Expression of correct form (allow only one side for M1). Must be a z value z = 1.751 B1 Accept 1.75 if nothing better seen CI is 70.3 to 74.3 metres (3 s.f.) A1 Allow without units Must be an interval 3 1(b) Not random sample B1 Need ‘random’ or ‘not representative/biased because…’ OE 1

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2 In the past, the mean height of plants of a particular species has been 2.3m. A random sample of 60 plants of this species was treated with fertiliser and the mean height of these 60 plants was found to be 2.4m. Assume that the standard deviation of the heights of plants treated with fertiliser is 0.4m. Carry out a test at the 2.5% significance level of whether the mean height of plants treated with fertiliser is greater than 2.3m. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 H0: Pop mean height = 2.3 H1: Pop mean height > 2.3 B1 Not just ‘mean’ Allow μ 2.4 2.3 0.4 60  M1 For standardising, must have 60 1.936 or 1.937 or 1.94 A1 ‘1.936’ < 1.96 M1 Valid comparison with 1.96 Or 2.64% > 2.5% OE Accept 1.936 < 2.24 or 2.64% > 1.25% OE if H1µ ≠ 2.3 [Do not reject H0] No evidence that (mean) height (with fertiliser) is more than without A1 FT FT their z In context, not definite. E.g. not ‘Mean height is not greater’ with no contradictions No FT for 2 tail test (max B0 M1 A1 M1 A0 3/5) Accept critical values method 2.401 (M1 A1) 2.4 < 2.401 (M1) Condone 2.299 (M1 A1) < 2.3 (M1) A1 conclusion 5

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3 It is known that 1.8% of children in a certain country have not been vaccinated against measles. A random sample of 200 children in this country is chosen. (a) Use a suitable approximating distribution to find the probability that there are fewer than 3 children in the sample who have not been vaccinated against measles. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Justify your approximating distribution. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Poisson B1 SOI Mean = 3.6 B1 Can be awarded for N(3.6, …) e-3.6(1 + 3.6 + 2 3.6 2 ) M1 Allow any λ Allow one end error Expression must be seen 0.303 (3 s.f.) A1 If M0 awarded allow SC B1 for 0.303 SC Use of binomial: B1 for answer 0.300 (3 sf) 4 3(b) [Binomial with] 200 > 50 B1 [200  0.018 =] 3.6 < 5 or [p =] 0.018 < 0.1 B1 If B0 B0 then SC n large, p small: B1 or n large np < 5: B1 or n > 50 and either np < 5 or p < 0.1: B1 2

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4 The number of cars arriving at a certain road junction on a weekday morning has a Poisson distribution with mean 4.6 per minute. Traffic lights are installed at the junction and a council officer wishes to test at the 2% significance level whether there are now fewer cars arriving. He notes the number of cars arriving during a randomly chosen 2-minute period. (a) State suitable null and alternative hypotheses for the test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the critical region for the test. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) H0: Pop mean = 4.6 [or 9.2] H1: Pop mean < 4.6 [or 9.2] B1 or λ = 4.6 or μ (Not just ‘mean’) or λ < 4.6 1 Question Answer Marks Guidance 4(b) Use of Poisson with λ = 9.2 B1 SOI P(X ⩽ 3) = e–9.2(1 + 9.2 + 2 9.2 2 + 3 9.2 3! ) = 0.0184 or 0.018 [< 0.02] P(X ⩽ 4) = 0.0184 + e–9.2  4 9.2 4! = 0.0486 or 0.049 [> 0.02] M1 At least one of these attempted correct λ (with Poisson expression seen not implied) *A1 Both correct SC Use of λ = 4.6 scores B1 for P(X = 0) = 0.01[0][1] and P(X ⩽ 1) = 0.056[3]only CR is X ⩽ 3 DA1 From CWO and at least one comparison seen SC If M0 awarded allow *B1 for both 0.018 and 0.049 or better and DB1 for correct critical region from CWO and at least one comparison seen. 4 4(c) 5 is not in critical region OR P(X ⩽ 5) =0.104 > 0.02 so [not reject H0] no evidence that number of cars arriving is now fewer M1 A1 FT For a comparison (i.e. 5 > 3) OE In context, not definite No contradictions e.g. not ‘No. of cars arriving is not fewer’ ft their critical region if used (but must be from Poisson and integers) 2 4(d) No, because H0 was not rejected B1 FT OE, FT their (c) 1 Question Answer Marks Guidance 4(e) N(276, 276) B1 SOI 300.5 276 276  [= 1.475] M1 Standardising with their values Allow with wrong or no continuity correction 1 – ɸ(‘1.475’) = 0.0701 (3 s.f.) A1 SC Use of Poisson: B1 for answer 0.0727 (3 sf) 3

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6 The masses, in kilograms, of large and small sacks of grain have the distributions N 53, 11 and N 14, 3 respectively. (a) Find the probability that the mass of a randomly chosen large sack is greater than four times the mass of a randomly chosen small sack. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) A lift can safely carry a maximum mass of 1000kg. Find the probability that the lift can safely carry 12 randomly chosen large sacks and 25 randomly chosen small sacks. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) B1 OE Give at early stage Var(D) = 11 + 42  3 [= 59] B1 or ( ) 2 11 4 3 + ´ (= 7.68 (3 s.f.)) Give at early stage 0 ( 3) 59  [= 0.391] M1 For standardising with their values (var must be from a combination attempt) Ignore continuity correction attempts 1 – Φ(‘0.391’) M1 For area consistent with their values 0.348 (3 s.f.) A1 As final answer 5 6(b) E(T) = 12  53 + 25  14 [= 986] B1 Give at early stage (N.B. accept E(T – 1000)= –14) Var(T) = 12  11 + 25  3 [= 207] B1 Or ( ) 12 11 25 3 ´ + ´ (= 14.4 (3sf)) Give at early stage 1000 986 207  [= 0.973] M1 For standardising with their values (var must be from a combination attempt) Ignore continuity correction attempts Φ(‘0.973’) M1 For area consistent with their values 0.835 (3 sf) A1 As final answer 5

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