Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 7 · Variant 2

2 The number of words in History essays by students at a certain college has mean - and standard deviation 1420. (i) The mean number of words in a random sample of 125 History essays was found to be 4820. Calculate a 98% confidence interval for -. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Another random sample of n History essays was taken. Using this sample, a 95% confidence interval for - was found to be 4700 to 4980, both correct to the nearest integer. Find the value of n. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 1420 4820 125 z ± × z = 2.326 B1 Accept 2.326 - 2.329 4524/4525 to 5115/5116 or 4520 to 5120 (3 sf) A1 Must be an interval 3 Question Answer Marks Guidance 2(ii) 4840 x = B1 or width = 280 or half width = 140 4840 + 1.96 × n 1420 = 4980 OE M1 or 140 = 1.96 × n 1420 OE n = 395 A1 CAO must be an integer 3

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3 The masses, m kg, of packets of flour are normally distributed. The mean mass is supposed to be 1.01 kg. A quality control officer measures the masses of a random sample of 100 packets. The results are summarised below. n = 100 Σm = 98.2 Σ m2 = 104.52 (i) Test at the 5% significance level whether the population mean mass is less than 1.01 kg. 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(ii) Explain whether it was necessary to use the Central Limit theorem in your answer to part (i). 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Mark scheme: 3(i) m = 98.2 100 = 0.982 s = 2 982 .0 100 52 . 104 99 100 − × (= 0.28582) or var = 0.08169 M1 H0: Pop mean mass = 1.01 H1: Pop mean mass < 1.01 B1 not just ‘mean’, but allow just ‘µ’ 0.28582 100 0.982 1.01 − ± M1 0.284387 100 0.982 1.01 − ± M1 = −0.980 (3 sf) accept ± A1 = –0.985 (3 sfs) accept ± A1 Comp with z = − 1.645 (or areas 0.1635 > 0.05) M1 Valid comparison of z’s or area’s No evidence that (mean) mass is less than 1.01 A1 FT Correct conclusion FT their z 7 Question Answer Marks Guidance 3(ii) Distr of X normal (so distr of X normal) Must state or imply No B1 X/parent population 1

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4 The random variable X has probability density function given by t k 0 < x ≤a, f x = x 0 otherwise, where k and a are constants. It is given that E X = 3. (i) Find the value of a and show that k = 16. 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(ii) Find the median of X. 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Mark scheme: 4(i) k x a xd 0 1 ∫ = 1 (2k[x0.5] 0 a = 1) 2ka0.5 = 1 or a = 2 4 1 k A1 OE; a correct eqn in k & a after sub limits k x a x x d 0 ∫ = 3 M1 Attempt int xf(x) and = 3 e.g. 3 2 ka1.5 = 3 or a3 = 2 4 81 k A1 OE; a correct eqn in k and a after sub limits e.g. a2 = 81 or e.g. k2 = 3 9 4 81 × M1 Attempt eliminate one letter a = 9 A1 Convincingly obtained e.g. k = 54 9 k = 6 1 AG A1 7 Question Answer Marks Guidance 4(ii) 6 1 x m xd 0 1 ∫ = 0.5 OE M1 Attempt int f(x), unknown limit and = 0.5 3 1 m0.5 = 0.5 A1 a correct equn in m after sub limits m = 2.25 A1 3

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5 The marks in paper 1 and paper 2 of an examination are denoted by X and Y respectively, where X and Y have the independent continuous distributions N 56, 62 and N 43, 52 respectively. (i) Find the probability that a randomly chosen paper 1 mark is more than a randomly chosen paper 2 mark. 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(ii) Each candidate’s overall mark is M where M = X + 1.5Y. The minimum overall mark for grade A is 135. Find the proportion of students who gain a grade A. 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Mark scheme: 5(i) E(X − Y) = 56-43 (= 13) B1 Var(X− Y) = 62 + 52 (= 61) M1 ' 61 ' 13 0− (= −1.664) M1 Ignore any attempted cc/no SD/var mixes. var must be attempt at a combination 1 – ɸ('−1.664') = ɸ('1.664') M1 For area consistent with their working = 0.952 (3 sf) A1 Similar scheme for use of Y – X 5 Question Answer Marks Guidance 5(ii) E(M) = 56 +1.5(43) (= 120.5) B1 Var(M) = 62 + 1.52×52 (= 92.25) M1 25 . 92 ' 5. 120 135− (= 1.510) M1 Ignore any attempted cc/no SD/var mixes. var must be attempt at a combination 1 − ɸ('1.510') M1 For area consistent with their working = 0.0655 or 0.0656 or 6.55% or 6.56% (3 sf) As final answer A1 Allow 6.6% or 6.5% or 7% if correct working seen 5

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6 In a certain factory the number of items per day found to be defective has had the distribution Po 1.03 . After the introduction of new quality controls, the management wished to test at the 10% significance level whether the mean number of defective items had decreased. They noted the total number of defective items produced in 5 randomly chosen days. It is assumed that defective items occur randomly and that a Poisson model is still appropriate. (i) Given that the total number of defective items produced during the 5 days was 2, carry out the test. 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(ii) Using another random sample of 5 days the same test is carried out again, with the same significance level. Find the probability of a Type I error. 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(iii) Explain what is meant by a Type I error in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) H0: Pop mean no. defectives = 5.15 H1: Pop mean no. defectives < 5.15 B1 or ‘= 1.03 (per day)’ not just ‘mean’, but allow just ‘λ’ or ‘µ’ P(X ⩽ 2) M1 Attempted. Any one term error/end error/incorrect λ/expression 1–… = e−5.15 (1 + 5.15 + 2 15 .5 2 ) M1 Correct expression attempted = 0.113 A1 Comp with 0.1 M1 Valid comparison No evidence to believe mean no. of defectives has decreased A1 FT Correct conclusion (FT their value) No contradictions 6 Question Answer Marks Guidance 6(ii) BOTH P(X ⩽ 1) = e−5.15 (1 + 5.15) (= 0.0357) AND P(X ⩽ 2) = = e−5.15 (1 + 5.15 + 2 15 .5 2 )= (0.113) B1* (Could be seen in (i)) Comp either with 0.1 DB1 One comparison with 0.01 (could be seen in (i)) P(Type I error) = 0.0357 (3 sf) B1 3 6(iii) Actually mean = 1.03 but conclude that mean < 1.03 B1 Mean no. of defectives not reduced, but conclude that it is reduced. 1

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