Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 7 · Variant 2

2 The standard deviation of the volume of drink in cans of Koola is 4.8 centilitres. A random sample of 180 cans is taken and the mean volume of drink in these 180 cans is found to be 330.1 centilitres. (i) Calculate a 95% confidence interval for the mean volume of drink in all cans of Koola. Give the end-points of your interval correct to 1 decimal place. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Explain whether it was necessary to use the Central Limit theorem in your answer to part (i). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) z = 1.96 B1 seen 330.1 ± z × 4.8 180 M1 Must be of correct form. Any z = 329.4 to 330.8 (1 dp) A1 Must be to 1 dp. Must be an interval. 3 2(ii) Yes, because vol of all cans not stated to be normal B1 Or Yes, population not stated to be normal 1

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4 The time, X hours, taken by a large number of runners to complete a race is modelled by the probability density function given by t k 0 ≤x ≤a, f x = x + 1 2 0 otherwise, where k and a are constants. a + 1 (i) Show that k = . 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(ii) State what the constant a represents in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Three quarters of the runners take half an hour or less to complete the race. (iii) Find the value of a. 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Mark scheme: 4(i) 2 ( 1) 0 d + ∫ a k x x = 1 M1 – ( 1) 0 +     k x a = 1 –k( 1 1) 1 − + a = 1 M1 Attempt subst correct limits into correct integral k × 1 + a a = 1 and k = 1 + a a AG A1 No errors seen 3 Question Answer Marks Guidance 4(ii) Max time allowed by model (for runners to finish) B1 Allow: All runners finish in time a or less or Longest time (taken by any runner) oe 1 4(iii) 1 + a a 2 0.5 1 ( 1) 0 + ∫ x dx = 3 4 M1 Attempt integ f(x) and = 3 4 ; ignore limits oe. Condone missing / incorrect k – 1 + a a 1 ( 1) 0.5 0 +     x = 3 4 – 1 + a a ( 2 3 1 −) = 3 4 M1 Attempt subst correct limits into correct integral. Condone missing / incorrect k a = 0.8 oe A1 3

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5 The numbers of basketball courts in a random sample of 70 schools in South Mowland are summarised in the table. Number of basketball courts 0 1 2 3 4 >4 Number of schools 2 28 26 10 4 0 (i) Calculate unbiased estimates for the population mean and variance of the number of basketball courts per school in South Mowland. 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The mean number of basketball courts per school in North Mowland is 1.9. (ii) Test at the 5% significance level whether the mean number of basketball courts per school in South Mowland is less than the mean for North Mowland. 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(iii) State, with a reason, which of the errors, Type I or Type II, might have been made in the test in part (ii). 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Mark scheme: 5(i) ˆµ = 126 70 or 9 5 or 1.8 oe B1 Σx2f = 286 B1 Seen or implied 2 Est( ) σ = 2 2 70 69 70 ( '1.8' ) Σ − x f M1 oe attempted = 0.858 or 296 / 345 A1 Note: Final answer for var 0.846 (biased) and no working implies B1 for 286 4 Question Answer Marks Guidance 5(ii) H0: µ = 1.9 H1: µ < 1.9 B1 Or ‘pop mean’; not just ‘mean’ '0.858' 70 1.8 1.9 − M1 Standardise with their values from (i). Must have sqr 70. No SD / Var mix = –0.903 A1 Accept ± 0.903 < 1.645 M1 comp 1.645 allow comp 1.96 if H1: µ ≠ 1.9 or comp 1 – φ(‘0.903’)=0.182 or 0.183 with 0.05 (or 0.025 if H1: µ ≠ 1.9) No evidence that mean no courts in S is less than in N A1ft No contradictions. ft their 0.903, but not comp 1.96 i.e. no ft for a 2 tail test Accept cv method: cv = 1.718 M1A1 1.718 < 1.8 M1 conclusion A1 (cv centred on 1.8 gives 1.982 M1A1 and M1 for 1.982 > 1.9 A1 conclusion) 5 5(iii) Type II because H0 was not rejected B1ft ft their conclusion, i.e. if H0 rejected, ‘Type I because H0 rejected’ B1 Answer must be consistent with their conclusion. No conclusion in (ii) will score B0 1

More questions on Hypothesis tests

6 In the past, Angus found that his train was late on 15% of his daily journeys to work. Following a timetable change, Angus found that out of 60 randomly chosen days, his train was late on 6 days. (i) Test at the 10% significance level whether Angus’ train is late less often than it was before the timetable change. 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Angus used his random sample to find an !% confidence interval for the proportion of days on which his train is late. The upper limit of his interval was 0.150, correct to 3 significant figures. (ii) Calculate the value of ! correct to the nearest integer. 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Mark scheme: 6(i) Ho: p = 0.15 H1: p < 0.15 (N(60 × 0.15, 60 × 0.15 × 0.85) ) = N(9, 7.65) B1 H1: µ < 9 Use of Normal approximation: (N(0.15, 0.15 0.85 60 × )) = N(0.15, 0.002125) 6.5 '9' '7.65' − M1 For standardising (or 6 0.5 60 60 '0.15' '0.002125' + − = –0.904) Allow wrong or no cc = –0.904 A1 Accept ± ‘0.904’ < 1.282 M1 Valid comparison of z values or ɸ('–0.904')= 0.183 > 0.1 ft their 0.904 No evidence train late less often A1ft Use of Bin (60,0.15) to give Pr (< = 6) = 0.1848 M1A1 Valid comparison with 0.1 M1 Conclusion A1ft 5 6(ii) 0.1 + z × 0.1 0.9 60 × = 0.150 M1 For √ (0.1 × 0.9 / 60) seen M1 for 0.1 + z × ... = 0.150 or 2z… = 0.1 z = 1.291 A1 φ(‘1.291’) (= 0.90(16)) M1 for correct method to find α α = 80 A1ft ft their z. Must be a +ve non-zero integer < 100 5

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