Cambridge A Level Mathematics 9709 — 2019 May/June Paper 7 · Variant 2

2 The random variable X has mean 372 and standard deviation 54. (i) Describe fully the distribution of the mean of a random sample of 36 values of X. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The distribution in part (i) might be either exact or approximate. State a condition under which the distribution is exact. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Normal with mean 372 B1 sd = 54 36 M1 or variance = 2 54 36 M1 (= 9) A1 (= 81) A1 3 2(ii) Pop normal B1 Allow X is normal 1

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3 It is claimed that, on average, a particular train journey takes less than 1.9 hours. The times, t hours, taken for this journey on a random sample of 50 days were recorded. The results are summarised below. n = 50 Σt = 92.5 Σt2 = 175.25 (i) Calculate unbiased estimates of the population mean and variance. 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(ii) Test the claim at the 5% significance level. 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Mark scheme: 3(i) Est(µ) = 1.85 B1 Est(σ2) = 2 50 175.25 '1.85' 49 50   −     M1 Allow 2 50 175.25 '1.85' 49 150   −     or 0.0290 for M1 = 0.0842 (3 sf) or 33 392 A1 Cao If 50 49 omitted (giving var = 0.0825 or sd = 0.287) M0A0 3 3(ii) H0: Pop mean time = 1.9 (h) H1: Pop mean time < 1.9 (h) B1 Allow ‘µ’ but not just ‘mean’ 1.85 1.9 '0.0842' 50 − ± M1 ±1.85 1.9 '0.290' 50 − Accept totals method (92.5–95) / 4.21 = –1.22 A1 = –1.22 comp z = –1.645 M1 Or other valid comparison 0.888 or 0.889 < 0.95 OR 0.111 or 0.112>0.05 No evidence that mean time < 1.9 h A1 FT their z. Correct conclusion. No contradictions If 50 49 not used in (1): var = 0.8225, sd = 0.907, cr = 1.17 can score all marks in (ii) Note- 2 tail test can score B0 M1 A1 M1 (comparison with 1.96) A0 (no ft) max3/5 5

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4 The heights of a certain variety of plant are normally distributed with mean 110 cm and variance 1050 cm2. Two plants of this variety are chosen at random. Find the probability that the height of one of these plants is at least 1.5 times the height of the other. 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Mark scheme: 4 Use of 1.5X1 – X2 or similar B1 E(1.5X1–X2) =1.5(110) –110 (= 55) B1 or E(X1–1.5X2) =110 – 1.5(110) (= –55) Var(1.5X1–X2) =1.52×1050+1050 (or 3412.5) M1 Correct expression or result 0 55 '3412.5' − or 0 ( 55) '3412.5' −− (= ± 0.942) M1 Their ‘55’. Allow incorrect var (dep > 0 and ≠ 1050) 1 – Φ(‘0.942’) M1 Area consistent with their working = 0.173 A1 Ans 0.346 (3 sf) B1 FT double their prob (must be <1) 7

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5 The manufacturer of a certain type of biscuit claims that 10% of packets include a free offer printed on the packet. Jyothi suspects that the true proportion is less than 10%. He plans to test the claim by looking at 40 randomly selected packets and, if the number which include the offer is less than 2, he will reject the manufacturer’s claim. (i) State suitable hypotheses for the test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability of a Type I error. 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On another occasion Jyothi looks at 80 randomly selected packets and finds that exactly 6 include the free offer. (iii) Calculate an approximate 90% confidence interval for the proportion of packets that include the offer. 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(iv) Use your confidence interval to comment on the manufacturer’s claim. 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Mark scheme: 5(i) H0: p = 0.1 H1: p < 0.1 B1 1 5(ii) B(40, 0.1) stated or implied by use of B1 e.g. by 40Cx or 0.9p×0.1q (p + q = 40) 0.940 + 40×0.939×0.1 M1 Correct working (if seen). If working not seen, M1 may be implied by 0.0805 = 0.0805 A1 3 Question Answer Marks Guidance 5(iii) z = 1.645 B1 seen 6 80 ± z (80 6) 6 80 80 80 − × M1 Formula of correct form. Must be a ‘z’ = 0.0266 to 0.123 (3 sfs) A1 Allow 0.03 to 0.12 or better Must be an interval 3 5(iv) 10% (or manufacturer’s claim) is within CI Hence no reason to question claim B1 FT Allow ‘10% is within CI, accept claim’ oe Must include both parts. No contradictions. FT their CI Note if CI is centred on 0.1 allow ft 0.075 is within CI, accept claim 1

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