Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 7 · Variant 3

1 The random variable X has mean 2.4 and variance 3.1. (i) The random variable Y is the sum of four independent values of X. Find the mean and variance of Y. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The random variable Z is defined by Z = 4X −3. Find the mean and variance of Z. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) 9.6, 12.4 B1 B1 2 1(ii) 6.6, 49.6 B1 B1 2

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2 Cars arrive at a filling station randomly and at a constant average rate of 2.4 cars per minute. (i) Calculate the probability that fewer than 4 cars arrive in a 2-minute period. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use a suitable approximating distribution to calculate the probability that at least 140 cars arrive in a 1-hour period. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 2 2.4 4.8 λ = × = 2 3 4.8 4.8 4.8 e 1 4 2 3! −   + + +     M1 Any λ 0.294 (3 sf) A1 2 2(ii) ( ) ( ) 60 2.4 144 λ = × = N(‘144’, ‘144’) M1 N and σ2=µ SOI ( ) 139.5 '144' 0.375 '144' − = − M1 Allow with no continuity correction φ(‘0.375’) M1 Correct area consistent with their working 0.646 (3 sf) A1 4

More questions on The Poisson distribution

3 The times, in minutes, taken by competitors to complete a puzzle have mean - and standard deviation 3. The times taken by a random sample of 10 competitors are noted and the results are given below. 25.2 26.8 18.5 25.5 30.1 28.9 27.0 26.1 26.0 24.9 (i) Stating a necessary assumption, calculate a 97% confidence interval for -. 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(ii) Two more random samples, each of 10 competitors, are taken. Their times are used to calculate two more 97% confidence intervals for -. Find the probability that neither of these intervals contains the true value of -. 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Mark scheme: 3(i) Assume population is normally distributed B1 25.9 = x B1 Allow 259 10 z =2.17 B1 3 '25.9' 10 ± × z M1 Must have correct form and z. 23.8 to 28.0 (3 sf) A1 CWO 5 3(ii) 0.032 (=0.0009) B1 1

More questions on Sampling and estimation

4 A train company claims that 92% of trains on a particular line arrive on time. Sanjeep suspects that the true percentage is less than 92%. He chooses a random sample of 20 trains on this line and finds that exactly 16 of them arrive on time. Making an assumption that should be stated, test at the 5% significance level whether Sanjeep’s suspicion is justified. 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Mark scheme: 4 Assume trains are independent OR probability of being on time is constant H0: P(on time)=0.92 H1: P(on time)<0.92 B1 Both. Allow ‘p’ or π ( ) 20 17 3 20 18 2 19 20 17 18 1 C 0.92 0.08 C 0.92 0.08 20 0.92 0.08 0.92 − × × + × × + × × + M1 Allow one end error Must have 1 – … =0.0706 (3 sf) A1 Compare with 0.05 M1 Valid comparison needed No evidence that percentage less than 92% A1FT OE No contradictions. Method using normal approximation: If the first B1B1 is earned then: 16.5 20 0.92 CV 1.566 from ,withcontinuity correction 20 0.92 0.08 − ×   −   × ×   or CV=1.978 (without continuity correction) comp z=1.645 No evidence that % decreased (1.566) or evidence that % decreased (1.978) is awarded SC2 after B marks 6

More questions on Hypothesis tests

7 Bob is a self-employed builder. In the past his weekly income had mean $546 and standard deviation $120. Following a change in Bob’s working pattern, his mean weekly income for 40 randomly chosen weeks was $581. You should assume that the standard deviation remains unchanged at $120. (i) Test at the 2.5% significance level whether Bob’s mean weekly income has increased. 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Bob finds his mean weekly income for another random sample of 40 weeks and carries out a similar test at the 2.5% significance level. (ii) Given that Bob’s mean weekly income is now in fact $595, find the probability of a Type II error. 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................

Mark scheme: 7(i) H0: Pop mean=546 H1: Pop mean>546 581 546 120 40 − M1 Standardising. Need 120 40 =1.845 allow 1.844 A1 Allow 1.84 or 1.85 AWRT 1.845<1.96 M1 OE. Or area comparison 0.0325>0.025 or large probabilities No evidence that mean weekly income has increased A1FT No contradictions. If H1: ≠, and 2.241 used, max B0M1A1M1A0 5 7(ii) 546 1.96 120 40 − = a M1 Standardise to find a. Need 120 40 and 546 and a value of z a = 583.19 A1 Allow 583 to 3sf ( ) '583.19' 595 0.622 120 40 − = − M1 Standardise. Need 120 40 and 595 φ(‘–0.622’)=1 – φ(‘0.622’) M1 Consistent area 0.267 A1 5

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