Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 7 · Variant 2

1 On average, 1 in 150 components made by a certain machine are faulty. The random variable X denotes the number of faulty components in a random sample of 500 components. (i) Describe fully the distribution of X. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State a suitable approximating distribution for X, giving a justification for your choice. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Use your approximating distribution to find the probability that the sample will include at least 3 faulty components. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) Binomial B1 n = 500 and 1 150 p = or 0.00667 B1 Or 1 B 500,150       for B1B1 2 1(ii) Poisson B1 n large and mean = 10 3 or 3.3 or better, which is < 5 B1 Accept n > 50 2 1(iii) 2 10 3 10 10 3 1 1 3 2 e −             − × + +         M1 1-P(X = 0, 1, 2) = 1 – 0.353 A1 Correct expression with λ=3.3 or better = 0.647 (3 sf) A1 SC Use of Binomial scores B1 for 0.648. Use of Normal scores B1 for 0.67(0) to 0.677 3

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2 The heights of a certain species of animal have been found to have mean 65.2 cm and standard deviation 7.1 cm. A researcher suspects that animals of this species in a certain region are shorter on average than elsewhere. She takes a large random sample of n animals of this species from this region and finds that their mean height is 63.2 cm. She then carries out an appropriate hypothesis test. (i) She finds that the value of the test statistic z is −2.182, correct to 3 decimal places. (a) Stating a necessary assumption, calculate the value of n. 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(b) Carry out the hypothesis test at the 4% significance level. 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(ii) Explain why it was necessary to use the Central Limit theorem in carrying out the test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i)(a) Assume standard deviation for the region is 7.1 B1 Or standard deviation is same as for whole population OE 7.1 63.2 65.2 2.182 n − = − M1 Attempt to find correct equation (accept +2.182) n = {–2.182 × 7.1 ÷ (–2)}2 A1 Any correct expression for n or n . SOI n = 60 A1 CWO. Must be an integer 4 2(i)(b) H0: population mean (or µ) = 65.2 H1: population mean (or µ) < 65.2 B1 Not just ‘mean’ 2.182 > 1.751 M1 Or valid area comparison. There is evidence that animals are shorter in this region A1 CWO. No contradictions 3 2(ii) Population unknown or population not given as normal B1 Allow population not normal. Accept distribution of X unknown. 1

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3 The masses, in grams, of bags of flour are normally distributed with mean -. The masses, m grams, of a random sample of 50 bags are summarised by Σm = 25 110 and Σm2 = 12 610 300. (i) Calculate a 96% confidence interval for -, giving the end-points correct to 1 decimal place. 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Another random sample of 50 bags of flour is taken and a 99% confidence interval for - is calculated. (ii) Without calculation, state whether this confidence interval will be wider or narrower than the confidence interval found in part (i). Give a reason for your answer. 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Mark scheme: 3(i) est(µ) = 25110 50 (= 502.2) ( ) 2 2 50 12610300 25110 50 58 est 1.1836 49 50 50 49 50 σ    = − = × =       M1 OE 1.18 (3 sf) or 58 49 A1 Accept SD = 1.0879 z = 2.054 or 2.055 B1 '1.1836' 502.2 50 z ± × M1 Must be of correct form. 501.9 to 502.5 (1dp) A1 CWO. Must be in interval. SC accept use of biased variance (1.16) for M1 A1 6 3(ii) More confident or z would be greater, Hence wider. B1 OE Reason needed 1

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5 The masses, in grams, of large boxes of chocolates and small boxes of chocolates have the distributions N 325, 6.1 and N 167, 5.6 respectively. (i) Find the probability that the total mass of 10 randomly chosen large boxes of chocolates is less than 3240 g. 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(ii) Find the probability that the mass of a randomly chosen large box of chocolates is more than twice the mass of a randomly chosen small box of chocolates. 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Mark scheme: 5(i) B1 Or mean = 325 var. 6.1 10 = ( ) 3240 3250 1.280 61 − = − M1 Standardise with their values (no mixed methods) φ('–1.280') = 1 – φ('1.280) M1 Area consistent with their figures 0.100 A1 Allow 0.1 4 5(ii) E(D) = 325 – 2 × 167 = –9 B1 Accept ±9 Var(D) = 6.1 + 22 × 5.6 (= 28.5) B1 ( )( ) 0 9 1.686 28.5 −− = M1 Standardising with their values. Must have a combination attempt on denominator and 1 −φ('1.686') M1 Area consistent with their figures 0.0459 A1 5

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6 The number of accidents per month, X, at a factory has a Poisson distribution. In the past the mean has been 1.1 accidents per month. Some new machinery is introduced and the management wish to test whether the mean has increased. They note the number of accidents in a randomly chosen month and carry out a hypothesis test at the 1% significance level. (i) Show that the critical region for the test is X ≥5. Given that the number of accidents is 6, carry out the test. 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Later they carry out a similar test, also at the 1% significance level. (ii) Explain the meaning of a Type I error in this context and state the probability of a Type I error. 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(iii) Given that the mean is now 7.0, find the probability of a Type II error. 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................

Mark scheme: 6(i) H0: Pop mean (or λ or µ) is 1.1 H1: Pop mean (or λ or µ) is more than 1.1 P(X ⩾ 4) = 1 −e–1.1 2 3 1.1 1.1 1 1.1 2 3!   + + +     M1 Correct expression for either P(X ⩾ 4) or P(X ⩾ 5) 0.0257 A1 Correct value of either P(X ⩾ 4) or P(X ⩾ 5) ( ) 4 1.1 1.1 P 0.0257 e 0.00544 4! 5 X − = − × = . B1 B1 for the other value (Note use of P(X < 4) = 0.9743 and P(X < 5) = 0.99456 can score only if comparison with 0.99 seen) 0.00544 < 0.01 < 0.0257 M1 OE stated (valid comparison) There is evidence mean has increased B1 SC P(X ⩾ 6) = 0.000968 M1A1 Conclusion B1 6 6(ii) Concluding mean has increased when it has not B1 In context ‘0.00544’ B1FT FT their P(X ⩾ 5), dep < 0.01 2 6(iii) 2 3 4 7.0 7 7 7 e 1 7 2 3! 4! −   + + + +     M1 Correct expression for P(X ⩽ 4 | λ = 7.0) 0.173 (3 sf) A1 2

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