Cambridge A Level Mathematics 9709 — 2023 May/June Paper 6 · Variant 2

1 In a survey of 200 randomly chosen students from a certain college, 23% of the students said that they owned a car. Calculate an approximate 93% confidence interval for the proportion of students from the college who own a car. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 0.23 ± z × 0.23 (1 0.23) 200  M1 Expression of correct form. Any z, but z = 0.8328 scores B0M0. z = 1.811 or 1.812 B1 0.176 to 0.284 (3 sf) A1 Must be an interval. 3

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3 The masses, in kilograms, of newborn babies in country A are represented by the random variable X, with mean - and variance 32. The masses of a random sample of 500 newborn babies in this country were found and the results are summarised below. n = 500 Σx = 1625 Σx2 = 5663.5 (a) Calculate unbiased estimates of - and 32. 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A researcher wishes to test whether the mean mass of newborn babies in a neighbouring country, B, is different from that in country A. He chooses a random sample of 60 newborn babies in country B and finds that their sample mean mass is 2.95kg. Assume that your unbiased estimates in part (a) are the correct values for - and 32. Assume also that the variance of the masses of newborn babies in country B is the same as in country A. (b) Carry out the test at the 1% significance level. 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Mark scheme: 3(a) Est (μ) = 3.25 = 13/4 or 1625/500 B1 Est(σ2) = 2 500 5663.5 ( "3.25" ) 499 500  or 2 1 1625 5663.5 499 500        M1 Expression of correct form. = 0.766 (3 sf) or 1529/1996 A1 Biased variance of 0.7645 scores M0A0. 3 Question Answer Marks Guidance 3(b) H0: Pop mean (or μ) = ‘3.25’ H1: Pop mean (or μ) ≠ ‘3.25’ B1FT Not just ‘mean’. FT their 3.25 . 2.95 "3.25" "0.766" 60   M1 Standardising with their values. Must have √60. = –2.655 A1 Or P(𝑋ത < 2.95) = 0.0039 or 0.00396 or 0.00397 . SC FT their biased est(σ2), i.e. 0.7645 to give z = 2.658 A1. ‘2.655’ > 2.576 or ‘–2.655’ < –2.576 M1 For valid comparison, e.g. 0.0039 or 0.00396 or 0.00397 < 0.005, or 0.0078 < 0.01, or 0.00792 < 0.01 . [Reject H0] There is evidence that (mean) mass in (country B) is different (from country A). A1FT OE. Must be in context and not definite, e.g., not ‘Mean mass is not different’, No contradictions. Context needs either ‘mass’ or ‘countries’ OE. SC, Use of one-tail test. ‘2.655’ > 2.326 or 0.0039 < 0.01 M1A0 (Max B0M1A1M1A0 3/5). Accept critical value method. Either: Xcrit=2.959 M1A1 2.95<2.959 M1A1FT with correct conclusion, or Xcrit=3.241 M1A1 3.25>3,241 M1A1FT with correct conclusion. 5

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4 The number, X, of books received at a charity shop has a constant mean of 5.1 per day. (a) State, in context, one condition for X to be modelled by a Poisson distribution. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Assume now that X can be modelled by a Poisson distribution. (b) Find the probability that exactly 10 books are received in a 3-day period. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use a suitable approximating distribution to find the probability that more than 180 books are received in a 30-day period. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The number of DVDs received at the same shop is modelled by an independent Poisson distribution with mean 2.5 per day. (d) Find the probability that the total number of books and DVDs that are received at the shop in 1 day is more than 3. 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Mark scheme: 4(a) Books received independently or singly or randomly. B1 OE. Must be in context. If more than one condition given, ignore extras. 1 Question Answer Marks Guidance 4(b) 10 15.3 15.3 10! e  M1 Allow incorrect λ. = 0.0439 (3sf) A1 SC No working shown but correct answer seen scores B1. 2 4(c) N(153, 153) B1 Seen or implied. 180.5 153 153  [= 2.223] M1 For standardising with their values (can be implied). Allow with wrong or missing continuity correction. 1 ̶ ɸ(‘2.223’) M1 For correct probability area consistent with their values. = 0.0131 (3sf) A1 4 4(d) (λ =) 5.1 + 2.5 [= 7.6] B1 Give at early stage (seen or implied). 1 – 2 3 7.6 7.6 7.6 2 3! e (1 7.6 )     = 1– e-7.6(1+7.6+28.88+73.16) = 1 – (0.0005005+0.003803+0.01445+0.03661) M1 Allow incorrect λ. Allow one end error. Must see an expression (accept correct sigma notation). = 0.945 (3sf) A1 SC No working, 0.945 B1(could be implied) SC B1. 3

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6 When a child completes an online exercise called a Mathlit, they might be awarded a medal. The publishers claim that the probability that a randomly chosen child who completes a Mathlit will be awarded a medal is 13. Asha wishes to test this claim. She decides that if she is awarded no medals while completing 10 Mathlits, she will conclude that the true probability is less than 13. (a) Use a binomial distribution to find the probability of a Type I error. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The true probability of being awarded a medal is denoted by p. (b) Given that the probability of a Type II error is 0.8926, find the value of p. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) (1 – 1 3 )10 M1 = 0.0173 (3 sf) A1 No working scores SC B1. 2 6(b) 1 – (1 – p)10 = 0.8926 M1 Accept 1 – q10 = 0.8926 . Equation must be in p or in q but not both. 1 – p = 0.10740.1 [= 0.800] M1 For valid attempt to solve their (binomial) equation in p10 or q10. p = 0.200 (3 sf) or 0.2 A1 3

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