Cambridge A Level Mathematics 9709 — 2018 May/June Paper 7 · Variant 3

2 Amy has to choose a random sample from the 265 students in her year at college. She numbers the students from 1 to 265 and then uses random numbers generated by her calculator. The first two random numbers produced by her calculator are 0.213 165 448 and 0.073 165 196. (i) Use these figures to find the numbers of the first four students in her sample. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ There were 25 students in Amy’s sample. She asked each of them how much money, $x, they earned in a week, on average. Her results are summarised below. n = 25 Σ x = 510 Σ x2 = 13 225 (ii) Find unbiased estimates of the population mean and variance. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Explain briefly what is meant by ‘population’ in this question. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 213, 165, 73, 196 Allow 073 B1 For 3-digit no, < 265, consisting of three consecutive integers from given digits, backwards or forward. (73 or 073 counts as a 3-digit no.) B1 For another three such. Other answers may be valid. If other method used, method must be clear 2 Question Answer Marks Guidance 2(ii) 510 25 = 102 5 or 20.4 B1 2 25 13225 102 24 25 5       −        M1 2 1 510 13225 24 25     −     118 (3 sf) or 2821 24 A1 3 2(iii) (Average) weekly earnings of all students in Amy’s year B1 Not ‘All students in Amy’s year’ 1

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3 A researcher wishes to estimate the proportion, p, of houses in London Road that have only one occupant. He takes a random sample of 64 houses in London Road and finds that 8 houses in the sample have only one occupant. Using this sample, he calculates that an approximate !% confidence interval for p has width 0.130. Find ! correct to the nearest integer. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 8 8 (1 ) 64 64 64 × − (= 7 4096 or 0.00171) M1 OE, e.g. 7 1 8 8 64 × 2 × z 7 " " 4096 = 0.130 M1 Correct equation using their variance z = 1.572 A1 ɸ("1.572") (= 0.942) (0.942 – (1 – 0.942) = 0.884) M1 2ɸ(their z) -1 α = 88 A1 CAO 5

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4 The numbers, M and F, of male and female students who leave a particular school each year to study engineering have means 3.1 and 0.8 respectively. (i) State, in context, one condition required for M to have a Poisson distribution. [1] ........................................................................................................................................................ ........................................................................................................................................................ Assume that M and F can be modelled by independent Poisson distributions. (ii) Find the probability that the total number of students who leave to study engineering in a particular year is more than 3. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Given that the total number of students who leave to study engineering in a particular year is more than 3, find the probability that no female students leave to study engineering in that year. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) No of males leaving (to do eng) each yr has const mean or Males leave (to do eng) indep of other males leaving (to do eng) or Males leave (to do eng) at random B1 1 4(ii) λ = 3.9 B1 1 – e–3.9(1 + 3.9 + 2 3 3.9 3.9 2! 3! + ) M1 Any λ. Allow one end error or extra term. 0.546753 or 0.547 (3 sf) A1 3 4(iii) P(F = 0 and M > 3) = 2 3 3.1 3.1 0.8 3.1 e 1 1 3.1 2! 3! e     − −     × − + + +         (= 0.16857) M1 Attempt P(F = 0) × P(M > 3) allow one end error for P(M > 3) provided λ = 3.1 P(F=0 and M>3) P(M+F>3) "0.16857" "0.54675" M1 Attempted, allow any probability/their (ii) provided the answer is <1 = 0.308 (3 sf) A1 3

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5 The time taken for a particular train journey is normally distributed. In the past, the time had mean 2.4 hours and standard deviation 0.3 hours. A new timetable is introduced and on 30 randomly chosen occasions the time for this journey is measured. The mean time for these 30 occasions is found to be 2.3 hours. (i) Stating any assumption(s), test, at the 5% significance level, whether the mean time for this journey has changed. 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(ii) A similar test at the 5% significance level was carried out using the times from another randomly chosen 30 occasions. (a) State the probability of a Type I error. [1] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) State what is meant by a Type II error in this context. 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Mark scheme: 5(i) Assume (pop) sd same (0.3) H0: Pop mean = 2.4 B1 H1: Pop mean ≠ 2.4 B1 Allow ‘µ’ but not just ‘mean’ ± 2.3 2.4 0.3 30 − M1 Must have 30 , Critical region approach (2.293, 2.507) or (2.193, 2.407) = ±1.826 A1 comp z = ±1.96 M1 Valid comparison (e.g. compare 0.034 with 0.025) No evidence that mean time changed A1f In context, allow accept H0 if correctly defined, no contradictions. One-tail test can score B1, B0, M1, A1, M1, A0 Max 4/6 6 5(ii)(a) 0.05 B1 1 5(ii)(b) Concluding mean time has not changed when it has. B1 OE, must have e.g. conclude/accept SR Allow mean has decreased if a one tailed test in Part (i) 1

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6 The times, in minutes, taken to complete the two parts of a task are normally distributed with means 4.5 and 2.3 respectively and standard deviations 1.1 and 0.7 respectively. (i) Find the probability that the total time taken for the task is less than 8.5 minutes. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that the time taken for the first part of the task is more than twice the time taken for the second part. 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Mark scheme: 6(i) E(T) = 4.5 + 2.3 (= 6.8) Var(T) = 1.12 + 0.72 (= 1.7) M1 8.5 "6.8" "1.7" − (= 1.304) M1 Correct stand'n using their µ and σ 2 must be a combination of the two variables ɸ("1.304") M1 Area consistent with their working = 0.904 (3 sf) A1 4 6(ii) E(D) = 4.5 – 2 × 2.3 or –0.1 M1 Var(D) = 1.12 + 22×0.72 or 3.17 M1 Both can seen or implied 0 (' 0.1') '3.17' −− (= 0.056) M1 Correct stand'n using their µ and σ 2 must be a Combination of the two variables 1 – ɸ("0.056") M1 Area consistent with their working = 0.478 (3 sf) A1 5

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