Cambridge A Level Mathematics 9709 — 2018 May/June Paper 7 · Variant 2

3 The management of a factory wished to find a range within which the time taken to complete a particular task generally lies. It is given that the times, in minutes, have a normal distribution with mean - and standard deviation 6.5. A random sample of 15 employees was chosen and the mean time taken by these employees was found to be 52 minutes. (i) Calculate a 95% confidence interval for -. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Later another 95% confidence interval for - was found, based on a random sample of 30 employees. (ii) State, with a reason, whether the width of this confidence interval was less than, equal to or greater than the width of the previous interval. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 52 ± z × 6.5 15 M1 z = 1.96 B1 Seen or used 48.7 to 55.3 (3 sf) A1 Must be an interval 3 Question Answer Marks Guidance 3(ii) Narrower because more information or because σ n smaller B1 oe Accept ‘sample size is larger’ ‘more employees’ ‘width inversely proportional to sq root of n’ ‘if n increases width decreases’ ‘95% CI is 49.7 to 54.3’ or similar. No contradictions 1

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4 The mean mass of packets of sugar is supposed to be 505 g. A random sample of 10 packets filled by a certain machine was taken and the masses, in grams, were found to be as follows. 500 499 496 495 498 490 492 501 494 494 (i) Find unbiased estimates of the population mean and variance. 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The mean mass of packets produced by this machine was found to be less than 505 g, so the machine was adjusted. Following the adjustment, the masses of a random sample of 150 packets from the machine were measured and the total mass was found to be 75 660 g. (ii) Given that the population standard deviation is 3.6 g, test at the 2% significance level whether the machine is still producing packets with mean mass less than 505 g. 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(iii) Explain why the use of the normal distribution is justified in carrying out the test in part (ii). 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Mark scheme: 4(i) Est(µ) = 495.9 B1 Accept 496 Est(σ2) = 2 10 2459283 9 10 ( "495.9" ) − M1 Attempt Σx2 and subst in correct formula (1/9(“2459283” – “4959”2/10)). May be implied by correct answer = 12.8 (3 sf) or 383/30 A1 (Note: Biased var “11.49” scores M0 A0) 3 4(ii) H0: µ = 505 H1: µ < 505 75660 505 150 3.6 150 − ÷ B1 Allow ‘Pop mean’ but not just ‘mean’ = –2.04 M1 Correct stand'n; must have √150. No sd/var mixes. Condone sample SD (3.58/3.39) Accept standardisation of totals ((75660-75750)/44.091) Accept CV method A1 Accept +2.04 (Note: if valid area comparison done 0.0207/0.0206 or 0.979 needed for A1) comp z = –2.054 M1 Valid comparison of z’s or area (0.0207/6>0.02; 0.979(3)<0.98) No evidence (at 2%) that machine pkts mean mass < 505 A1ft oe No contradictions. SC Two tail test can score B0 M1 A1 M1 for comparison with 2.326 A0 (max 3/5) 5 Question Answer Marks Guidance 4(iii) Large sample, so sample mean approx normally distr'd B1 Allow just ‘Sample is large’ or ‘n is large’ n>30 1

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6 Accidents on a particular road occur at a constant average rate of 1 every 4.8 weeks. (i) State, in context, one condition for the number of accidents in a given period to be modelled by a Poisson distribution. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Assume now that a Poisson distribution is a suitable model. (ii) Find the probability that exactly 4 accidents will occur during a randomly chosen 12-week period. 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(iii) Find the probability that more than 3 accidents will occur during a randomly chosen 10-week period. 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(iv) Use a suitable approximating distribution to find the probability that fewer than 30 accidents will occur during a randomly chosen 2-year period (1042 weeks). 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Mark scheme: 6(i) Accidents occur independently or randomly B1 In context. Allow ‘singly’. 1 6(ii) e–2.5 × 4 2.5 4! M1 Poisson P(4), allow any λ = 0.134 (3 sfs) A1 2 Question Answer Marks Guidance 6(iii) λ = 25 12 or 2.08(333) B1 1 – 25 12 e − (1 + 25 12 + 2 25 12 2! + 3 25 12 3! ) M1 1 – Poisson P(0, 1, 2, 3), allow any λ allow one end error = 0.158 (3 sfs) A1 As final answer 3 6(iv) N(1825 84 ,1825 84 ) or N(21.7(26), 21.7(26)) B1 Stated or implied 1825 29.5 84 1825 84 − M1 Allow with wrong or no cc with their mean/sd Φ(“1.668”) M1 Correct area consistent with their working = 0.952 ( 3 sfs) A1 4

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7 A ten-sided spinner has edges numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Sanjeev claims that the spinner is biased so that it lands on the 10 more often than it would if it were unbiased. In an experiment, the spinner landed on the 10 in 3 out of 9 spins. (i) Test at the 1% significance level whether Sanjeev’s claim is justified. 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(ii) Explain why a Type I error cannot have been made. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ In fact the spinner is biased so that the probability that it will land on the 10 on any spin is 0.5. (iii) Another test at the 1% significance level, also based on 9 spins, is carried out. Calculate the probability of a Type II error. 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Mark scheme: 7(i) H0: P(10) = 0.1 H1: P(10) > 0.1 B1 B(9,0.1) P(X ⩾ 3) = 1 – (0.99 + 9×0.98 × 0.1 + 9C2 × 0.97 × 0.12) M1 Allow one extra term in bracket = 0.05297... or 0.053(0) A1 comp 0.01 M1 Valid comparison. (comparison with 0.99 can recover previous M1 A1 for 0.9470) No evidence (at 1% level) to reject H0 Claim not justified A1ft No contradictions 5 7(ii) H0 not rejected oe B1 1 7(iii) P(X ⩾ 4) = "0.05297" – 9C3×0.96×0.13 M1 or 1–(0.99 + 9 × 0.98 × 0.1 + 9C2 × 0.97 × 0.12 + 9C3 × 0.96 × 0.13) = 0.00833 A1 Note: 0.05297 and 0.00833 both needed in (i) or (iii) to justify CV Hence crit value is 4 B1 Allow without working. Or in (i) May be implied by attempt at P(X < 4) below B(9,0.5) P(X < 4) M1 stated or implied = 0.59 + 9 × 0.58 × 0.5 + 9C2 × 0.57 × 0.52 + 9C3×0.56×0.53 M1 Attempt P(X < 4) with p = 0.5 P(Type II) = 0.254 (3 sf) A1 6

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