Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 6 · Variant 2

1 Each of a random sample of 80 adults gave an estimate, h metres, of the height of a particular building. The results were summarised as follows. n = 80 Σh = 2048 Σh2 = 52 760 (a) Calculate unbiased estimates of the population mean and variance. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Using this sample, the upper boundary of an !% confidence interval for the population mean is 26.0. Find the value of !. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 2048 128 B1 Est μ = 25.6 or or 80 5 80  52760  2048  2  1  20482  M1 Substitution into a correct formula. Est σ2 =  −    or  52760 −  Biased 4.14 scores M0.  79  80  80  79  80    1656 A1 = 4.19 (3 sf) or 395 3 1(b) '4.19' M1 Use of correct equation with their values. ‘25.6’ + z = 26.0 80 z = 1.748 or 1.747 A1 Accept 3sf. FT Biased z = 1.758. (ɸ(‘1.748’) = 0.960) ‘0.960’ – (1 – ‘0.960’) M1 Correct area using their values. α = 92.0 or 91.9 A1 Allow 92 . FT Biased 92.1. A final answer of 0.92 or 0.919 scores A0. 4

More questions on Sampling and estimation

2 In the past, the mean length of a particular variety of worm has been 10.3cm, with standard deviation 2.6cm. Following a change in the climate, it is thought that the mean length of this variety of worm has decreased. The lengths of a random sample of 100 worms of this variety are found and the mean of this sample is found to be 9.8cm. Assuming that the standard deviation remains at 2.6cm, carry out a test at the 2% significance level of whether the mean length has decreased. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Ho: Population mean length = 10.3 cm B1 or μ = 10.3 (not just ‘mean’). H1: Population mean length < 10.3 cm μ < 10.3 9.8 − 10.3 M1 If ± 1.923 (or 0.0272) seen allow M1 implied.  2.6 / 100 = –1.923 A1 Accept ± . Accept 3sf. –1.923 > –2.054 or –2.055 M1 OE For a valid comparison. Or compare 1 – ϕ(‘1.923’’) with 0.02 e.g. 0.0272 > 0.02 Use of CV 9.8 > 9.766 scores M1 A1 for 9.766 and M1 for comparison. [Not reject H0 ] No evidence that [mean] length has decreased A1 FT FT their z. No contradictions, not definite, in context. 5

More questions on Hypothesis tests

3 1.6% of adults in a certain town ride a bicycle. A random sample of 200 adults from this town is selected. (a) Use a suitable approximating distribution to find the probability that more than 3 of these adults ride a bicycle. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Justify your approximating distribution. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Use of Poisson. mean = 3.2 B1 B1 −3.2  3.2 2 3.23  M1 Allow any λ . 1 − e  1 + 3.2 + +  or 1 – e–3.2 (1 + 3.2 + 5.12 + 5.46133) Allow one end error.  2 3!  or 1 – (0.04076 + 0.1304 + 0.2087 + 0.2226) = 0.397 or 0.398 A1 SC Use of binomial: B1 for answer 0.398 (3 sf). 0.397 or 0.398 with no working scores SC B1. 4 3(b) [Binomial with] [n =] 200 > 50 B1 [np =][200 × 0.016 =] 3.2 < 5 or [p =]0.016 < 0.1 B1 If B0 B0 SC n large (or n > 50), and p small or p < 0.1 or np <5 : B1. 2

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4 The number of faults in cloth made on a certain machine has a Poisson distribution with mean 2.4 per 10m2. An adjustment is made to the machine. It is required to test at the 5% significance level whether the mean number of faults has decreased. A randomly selected 30m2 of cloth is checked and the number of faults is found. (a) State suitable null and alternative hypotheses for the test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability of a Type I error. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Exactly 3 faults are found in the randomly selected 30m2 of cloth. (c) Carry out the test at the 5% significance level. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Later a similar test was carried out at the 5% significance level, using another randomly selected 30m2 of cloth. (d) Given that the number of faults actually has a Poisson distribution with mean 0.5 per 10m2, find the probability of a Type II error. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) H0: Population mean = 7.2 or 2.4 B1 or λ or μ = 7.2 or 2.4 (Not just ‘mean’). H1: Population mean < 7.2 or 2.4 or λ or μ < 7.2 or 2.4 1 4(b) λ = 7.2 B1 SOI −7.2  7.2 2  M1 Both expressions needed, allow any λ  e  1 + 7.2 +  or e–7.2(1 + 7.2 + 25.92) If λ ≠ 7.2 allow P(X ⩽ n) for 2 consecutive values of n  P ( X 2 ) =  2  with P(X ⩽ n) < 0.05 and P(X ⩽ n + 1) > 0.05. or 0.0007465 + 0.0053754 + 0.01935 [= 0.0255] −7.2 7.23  '0.0255'+ e  or ‘0.0255’ +e–7.2 (62.21)  P ( X 3 ) = 3! or ‘0.0255’ +0.04644 [= 0.0719] P(Type I) = 0.02547 or 0.0255 (3 sf) B1 3 4(c) 3 > 2 or P(X ⩽ 3) > 0.05 or ‘0.0719’ > 0.05 M1 For a valid comparison or 3 outside critical region. FT their CR in (b). [Not reject H0] A1 FT No contradictions. In context, not definite. No evidence that [mean] number of faults has decreased 2 4(d) 1 – e–1.5(1 + 1.5 + 1.52 / 2) or 1 – e–1.5(1 + 1.5 + 1.125) M1 Must see expression. FT their CR in (b). or 1 – (0.2231 + 0.3347 + 0.2510) = 0.191 (3 sf) A1 2

More questions on Hypothesis tests

5 X is a random variable with distribution B 10, 0.2 . A random sample of 160 values of X is taken. (a) Find the approximate distribution of the sample mean, including the values of the parameters. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the probability that the sample mean is less than 1.8. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) For X, μ = 2 σ2 = 1.6 Mean = 2 B1 1.6 1 B1 Accept Var = 0.12 (accept sd=0.1 if clearly identified). Variance = or or 0.01 160 100 Normal B1 3 5(b) 1.8 − 3201 − '2' 1.8 − '2' M1 Allow with wrong continuity correction.  or  [= –2.03 or –2] M1 can be implied by correct final answer or for '0.01' '0.01' –2.03 / –2.0 or 0.9788 / 0.9772 seen. or ± (287.5 – ’320’) / '256' or ± (288 – ‘320’) / '256' [= –2.03 or –2] ɸ(‘–2.03’) = 1 – ɸ(‘2.03’) M1 Correct area consistent with their values. M1 can be implied by correct final answer. = 0.0212 or 0.0228 (3 sf) A1 3

More questions on Sampling and estimation

6 The masses, in grams, of small and large bags of flour have the distributions N 510, 100 and N 1015, 324 respectively. Andr´e selects 4 small bags of flour and 2 large bags of flour at random. (a) Find the probability that the total mass of these 6 bags of flour is less than 4130g. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that the total mass of the 4 small bags is more than the total mass of the 2 large bags. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) E(T) = 4  510 + 2  1015 [= 4070] B1 Var(T) = 4 × 100 + 2 × 324 [= 1048] B1 or ( 4  100 + 2  324 ) = 32.4 ( 3sf ) . 4130 − 4070 M1 Standardising with their values. [= 1.853] Variance must be from a combination attempt. '1048' M1 can be implied by correct final answer. Φ(‘1.853’) M1 For area consistent with their values. M1 can be implied by correct final answer. = 0.968 (3 sf) A1 As final answer. 5 6(b) E(D) = 4  510 – 2  1015 [= 10] B1 Var(D) = 4  100 + 2  324 [= 1048] B1 Or ( 4  100 + 2  324 ) = 32.4 ( 3sf ) . 0 − '10' M1 Standardising with their values. [= –0.309] Variance must be from a combination attempt. '1048' M1 can be implied by correct final answer. 1 – Φ(‘–0.309’) = Φ(‘0.309’) M1 For area consistent with their values. M1 can be implied by correct final answer. = 0.621 A1 As final answer. 5

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