Cambridge A Level Mathematics 9709 — 2017 May/June Paper 7 · Variant 3

1 A residents’ association has 654 members, numbered from 1 to 654. The secretary wishes to send a questionnaire to a random sample of members. In order to choose the members for the sample she uses a table of random numbers. The first line in the table is as follows. 1096 4357 3765 0431 0928 9264 The numbers of the first two members in the sample are 109 and 643. Find the numbers of the next three members in the sample. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 573, 43 (or 043), 289 B1B1B1 Ignore incorrect numbers. But allow other correct use of table (i.e. 573, 650, 431) Total: 3

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2 In a random sample of 200 shareholders of a company, 103 said that they wanted a change in the management. (i) Find an approximate 92% confidence interval for the proportion, p, of all shareholders who want a change in the management. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State the probability that a 92% confidence interval does not contain p. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) z = 1.751 B1 103 200 ± z 103 103 200 200 (1 ) 200 × − oe M1 all correct except for recognisable value of z, allow for one side only = 0.453 to 0.577 (3 sf) as final answer A1 must be an interval Total: 3 2(ii) 0.08 oe 8%, 8/100 B1 SOI

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3 The mass, in tonnes, of iron ore produced per day at a mine is normally distributed with mean 7.0 and standard deviation 0.46. Find the probability that the total amount of iron ore produced in 10 randomly chosen days is more than 71 tonnes. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 10 × 0.462 (= 2.116) or 0.46 10 B1 Total mass of ore ~ N(70, 2.116) or ~N 2 0.46 7, 10               B1 71 "70" "2.116" − ± or 7.1 "7.0" 0.46 / 10 − ± (= 0.687) M1 correct, using their sd or √(their var) e.g. allow 71 "70" 4.6 − for M1 1 – ɸ("0.687") M1 for correct area consistent with their working = 0.246 (3 sf) A1 Total: 5

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4 Last year the mean level of a certain pollutant in a river was found to be 0.034 grams per millilitre. This year the levels of pollutant, X grams per millilitre, were measured at a random sample of 200 locations in the river. The results are summarised below. n = 200 Σx = 6.7 Σx2 = 0.2312 (i) Calculate unbiased estimates of the population mean and variance. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Test, at the 10% significance level, whether the mean level of pollutant has changed. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) x = 6.7/200 (= 67/2000 = 0.0335) B1 s2 = 2 200 0.2312 "0.0335" 199 200   × −     M1 s2 = 2 0.2312 0.0335 200 − M0 = 0.0000339(2) = 27/796000 A1 = 0.00003375 A0 Total: 3 4(ii) H0: Pop mean level = 0.034 H1: Pop mean level ≠ 0.034 B1 not just "mean", but allow just “µ” "030335" 0.034 "0.00003392" 200 − M1 must have 200 "0.00003375" 200 0.0335 0.034 − M1 = –1.21(4) (3 sfs) (–1.22 ↔–1.21) A1 = –1.217 (3 sfs) A1 Comp with z = −1.645 (or 0.1124>0.05) M1 0.112 > 0.05 valid comparison z or areas No evidence that (mean) pollutant level has changed, accept H0 (if correctly defined) A1FT correct conclusion no contradictions SR: One tail test: B0, M1A1 as normal, M1 (comparison with 1.282 consistent signs) A0 Total: 5

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7 In the past the number of accidents per month on a certain road was modelled by a random variable with distribution Po 0.47 . After the introduction of speed restrictions, the government wished to test, at the 5% significance level, whether the mean number of accidents had decreased. They noted the number of accidents during the next 12 months. It is assumed that accidents occur randomly and that a Poisson model is still appropriate. (i) Given that the total number of accidents during the 12 months was 2, carry out the test. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Explain what is meant by a Type II error in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given that the mean number of accidents per month is now in fact 0.05. (iii) Using another random sample of 12 months the same test is carried out again, with the same significance level. Find the probability of a Type II error. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) H0: Pop mean no. accidents = 5.64 H1: Pop mean no. accidents < 5.64 B1 not just "mean", but allow just "λ" or “µ” Use of λ = 5.64 B1 used in a Poisson calculation = e−5.64 (1 + 5.64 + 2 5.64 2 ) M1 Allow incorrect λ in otherwise correct = 0.08(0) A1 Comp with 0.05 M1 Valid comparison (Poisson only), no contradictions. No evidence to believe mean no. of accidents has decreased; accept H0 (if correctly defined) A1FT Normal distribution: M0M0 Total: 6 7(ii) Mean < 0.47 but conclude that this is not so B1 (Mean) no. of accidents reduced, but conclude not reduced. Must be in context. Total: 1 7(iii) (Need greatest x such that P(X ⩽ x) < 0.05 ) P(X ⩽ 1) = e−5.64 (1 + 5.64) = 0.024 P(X ⩽ 2) = 0.08 B1 Both, could be seen in (i) Hence rejection region is X ⩽ 1 B1 Can be implied With λ = 12× 0.05 = 0.6, 1 ̶ P(X ⩽ 1) = 1 ̶ e−0.6(1+ 0.6) M1 λ=0.6 and 1 ̶ P(X ⩽ 1) = 0.122 (3 sf) A1 Normal scores 0 Total: 4

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