Cambridge A Level Mathematics 9709 — 2021 May/June Paper 6 · Variant 1

1 Accidents at two factories occur randomly and independently. On average, the numbers of accidents per month are 3.1 at factory A and 1.7 at factory B. Find the probability that the total number of accidents in the two factories during a 2-month period is more than 3. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 λ = (3.1 + 1.7) × 2 M1 Attempt combined mean. Allow 3.1 + 1.7 for M1 = 9.6 A1 Correct mean 1 – e–9.6 (1 + 9.6 + 2 3 9.6 9.6 2 3! ) + M1 Allow incorrect mean. Allow one end error. = 0.986 (3 sf) A1 SC If 9.6 seen and unsupported 0.986 M1A1B1. SC Unsupported correct answer of 0.986 only if 9.6 also not seen scores B2 only. 4

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2 The time, in minutes, taken by students to complete a test has the distribution N 125, 36 . (a) Find the probability that the mean time taken to complete the test by a random sample of 40 students is less than 123 minutes. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Explain whether it was necessary to use the Central Limit theorem in the solution to part (a). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) ±123 125 6 40 − [= –2.108...] No standard deviation/variance mix. Ignore any continuity correction attempts for this mark. P(z < ‘–2.108’) = 1 – Φ(‘2.108’) M1 For correct probability area consistent with their working. = 0.0175 or 0.0176 (3 sf) A1 3 2(b) No, population is normal B1 Need both. 1

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4 100 randomly chosen adults each throw a ball once. The length, l metres, of each throw is recorded. The results are summarised below. n = 100 Σl = 3820 Σl2 = 182 200 Calculate a 94% confidence interval for the population mean length of throws by adults. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 3820 100 [= 38.2] B1 2 100 182200 '38.2' 99 100   −     or 2 1 3820 182200 99 100   −     M1 Use of biased (362.76) scores M0 = 12092 33 or 366.424 or 366 (3 sf) A1 Accept SD=19.1422 or 19.1(3sf) ‘38.2’ ± z× '366.424' 100 M1 Expression of the correct form must be a z-value. z = 1.881 or 1.882 B1 Seen. 34.6 to 41.8 (3 sf) A1 Allow use of biased giving (34.6,41.8) Must be an interval. 6

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5 On average, 1 in 75 000 adults has a certain genetic disorder. (a) Use a suitable approximating distribution to find the probability that, in a random sample of 10 000 people, at least 1 has the genetic disorder. 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(b) In a random sample of n people, where n is large, the probability that no-one has the genetic disorder is more than 0.9. Find the largest possible value of n. 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Mark scheme: 5(a) Po 2 15       P(X ⩾ 1) = 1 – 2 15 e − M1 Allow incorrect λ allow one end error = 0.125 (3 sf) A1 SC Partially unsupported final answer: Po 2 15       stated B1 then unsupported 0.125 B1 SC Use of Binomial (0.1248) B1 only Use of Normal scores M0 3 Question Answer Marks Guidance 5(b) λ = 75000 n B1 75000 e n − > 0.9 M1 Allow ‘=’ Allow incorrect λ – 75000 n > ln 0.9 [n < 7902.04] M1 Attempt ln both sides Largest value of n is 7902 A1 CWO. Must be an integer. Alternative method for Question 5(b) -e μ > 0.9 M1 Allow ‘=’ –μ > ln 0.9 [μ < 0.10536] M1 Attempt ln both sides n = μ × 75000 B1 Largest value of n is 7902 A1 CWO. Must be an integer. Alternative method for Question 5(b) 74999 75000 B1 74999 75000 n       > 0.9 M1 nln 74999 75000 > ln 0.9 M1 Attempt ln or log both sides Largest value of n is 7901 A1 CWO Must be an integer 4

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7 The masses, in kilograms, of large and small sacks of flour have the distributions N 55, 32 and N 27, 2.52 respectively. (a) Some sacks are loaded onto a boat. The maximum load of flour that the boat can carry safely is 340 kg. Find the probability that the boat can carry safely 3 randomly chosen large sacks of flour and 6 randomly chosen small sacks of flour. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that the mass of a randomly chosen large sack of flour is greater than the total mass of two randomly chosen small sacks of flour. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) E(T)= [ ] 3 55 6 27 327 × + × = B1 OE. Accept unsimplified. Var(T) = 3×32 + 6×2.52 [= 64.5] B1 Accept unsimplified. 340 '327' '64.5' − [= 1.619] M1 Must have √ P(z < '1.619') = Φ('1.619') M1 Correct probability area consistent with their working. 0.947 (3 sf) A1 5 7(b) E(L–S1–S2) = 55 – 2 × 27 [=1] B1 OE e.g. E(S1+S2 – L)= –1. Accept unsimplified. Var(L– S1–S2) = 32 + 2 × 2.52 [= 21.5] B1 Accept unsimplified. 0 '1' '21.5' − [= –0.216] M1 Standardising with their values. Must come from a combination attempt. P(L–S1–S2 > 0) = Φ(‘0.216’) M1 Correct probability area consistent with their working. 0.586 or 0.585 (3 sf) A1 5

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8 At a certain large school it was found that the proportion of students not wearing correct uniform was 0.15. The school sent a letter to parents asking them to ensure that their children wear the correct uniform. The school now wishes to test whether the proportion not wearing correct uniform has been reduced. (a) It is suggested that a random sample of the students in Grade 12 should be used for the test. Give a reason why this would not be an appropriate sample. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ A suitable sample of 50 students is selected and the number not wearing correct uniform is noted. This figure is used to carry out a test at the 5% significance level. (b) State suitable null and alternative hypotheses. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use a binomial distribution to find the probability of a Type I error. You must justify your answer fully. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) In fact 4 students out of the 50 are not wearing correct uniform. State the conclusion of the test, explaining your answer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (e) State, with a reason, which of the errors, Type I or Type II, may have been made. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Not representative (of all students in the school) B1 OE idea of ‘not being representative’ e.g. different grades in the school have different characteristics/proportions.... Don’t accept ‘not random’ or ‘biased’ without further explanation. 1 8(b) H0: P(not correct uniform) = 0.15 H1: P(not correct uniform) < 0.15 B1 Allow "p" 1 8(c) Any two probs attempted using B(50,0.15) M1 P(X ⩽ 3) = 0.8550 + 50 × 0.8549 × 0.15 + 50C2 × 0.8548 × 0.152 + 50C3 × 0.8547 × 0.153 M1 Attempt the tail probability P(0,1,2,3) with B(50,0.15) must be added. P(X ⩽ 4) = 0.04605 + 50C4×0.8546×0.154 M1 OE. Their P(X ⩽ 3) + P(X = 4) or P(0,1,2,3,4) with B(50,0.15) must be added. P(X ⩽ 3) = 0.0460 or 0.0461 [<0.05] P(X ⩽ 4) = 0.112 or [>0.05] A1 Both correct. OR if P(X ⩽ 4) not seen; P(4)=0.06606 and 0.06606>0.05 and P(X ⩽ 3)=0.0460 scores M1 A1 P(Type I) = 0.0460 or 0.0461 (3 sf) A1 Dependent on second M1. SC If M1M1M1A0 scored allow A1FT for incorrect P(X ⩽ 3) as long as <0.05 5 Question Answer Marks Guidance 8(d) 4 is outside critical region (⩽3) OE or P(X ⩽ 4) = 0.112 which is > 0.05 M1 FT working from (c). No evidence that proportion not wearing the correct uniform has decreased (Accept Ho) A1 In context not definite, e.g. not ‘Proportion has not decreased’. No contradiction. 2 8(e) Not rejected H0 *B1 FT FT If Reject H0 in (d) Type II DB1 FT FT Type I 2

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