Cambridge A Level Mathematics 9709 — 2019 May/June Paper 7 · Variant 1

1 At an internet caf´e, the charge for using a computer is 5 cents per minute. The number of minutes for which people use a computer has mean 23 and standard deviation 8. (i) Find, in cents, the mean and standard deviation of the amount people pay when using a computer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Each day, 15 people use computers independently. Find, in cents, the mean and standard deviation of the total amount paid by 15 people. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) Mean = 115 B1 SD = 40 B1 2 1(ii) Mean = 15 × ‘115’ = 1725 B1ft 15 × ‘40’2 (= 24000) M1 or SD = √15 × ‘40’. ft their (i) SD = √24000 SD = 155 (cents) (3 sf) A1 Accept √24000 SC: Allow correct answers in dollars 3

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2 The time, in minutes, that John takes to travel to work has a normal distribution. Last year the mean and standard deviation were 26.5 and 4.8 respectively. This year John uses a different route and he finds that the mean time for his first 150 journeys is 27.5 minutes. (i) Stating a necessary assumption, test at the 1% significance level whether the mean time for his journey to work has increased. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State, with a reason, whether it was necessary to use the Central Limit theorem in your answer to part (i). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Assume sd still 4.8 or is unchanged B1 or Assume the 150 times can be treated as a random sample / are independent H0: Pop mean = 26.5 H1: Pop mean > 26.5 B1 Allow ‘µ ’ but not just ‘mean’ 4.8 150 27.5 26.5 − M1 Standardise, with √ Accept CV method = 2.552 A1 Comp with z-value ‘2.552’ > 2.326 M1 or comp 1 – Φ(‘2.552’) with 0.01 1 – 0.9946 = 0.0054 < 0.01 There is evidence time has increased A1ft oe No contradictions (2 tail test scores max. B1 B0 M1 A1 M1 (for comparison with 2.576) A0 no ft) 6 Question Answer Marks Guidance 2(ii) No because pop is normal so distr of X is normal B1 Condone just ‘No because pop is normal’ 1

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3 Sumitra has a six-sided die. She suspects that it is biased so that it shows a six less often than it would if it were fair. She decides to test the die by throwing it 30 times and noting the number of throws on which it shows a six. (i) It shows a six on exactly 2 throws. Use a binomial distribution to carry out the test at the 5% significance level. 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(ii) Later, Sumitra repeats the test at the 5% significance level by throwing the die 30 times again. Find the probability of a Type I error in this second test. 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Mark scheme: 3(i) H0: P(6) = 1 6 H1: P(6) < 1 6 B1 ( 5 6 )30 + 30( 1 6 ) × ( 5 6 )29 + 30C2( 1 6 )2 × ( 5 6 )28 M1 Allow one term incorrect, omitted or extra = 0.103 A1 ‘0.103’ > 0.05 M1 No evidence (at 5% level) that die biased A1ft oe No contradictions 5 3(ii) ( 5 6 )30 + 30( 1 6 ) × ( 5 6 )29 M1 P(Type I) = 0.0295 A1 2

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6 Ramesh plans to carry out a survey in order to find out what adults in his town think about local sports facilities. He chooses a random sample from the adult members of a tennis club and gives each of them a questionnaire. (i) Give a reason why this will not result in Ramesh having a random sample of adults who live in the town. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Describe briefly a valid method that Ramesh could use to choose a random sample of adults in the town. 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Ramesh now uses a valid method to choose a random sample of 350 adults from the town. He finds that 47 adults think that the local sports facilities are good. (iii) Calculate an approximate 90% confidence interval for the proportion of all adults in the town who think that the local sports facilities are good. 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(iv) Ramesh calculates a confidence interval whose width is 1.25 times the width of this 90% confidence interval. Ramesh’s new interval is an x% confidence interval. Find the value of x. 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Mark scheme: 6(i) Biased towards people who like tennis Excludes people who don't like tennis B1 1 6(ii) Obtain a list of all people in the town B1 Use random numbers B1 or, e.g. pick numbers from a hat or other sensible 2 6(iii) Var(p) = 47 47 350 350 (1 ) 350 − (= 0.000332152) M1 z = 1.645 B1 47 47 350 350 (1 ) 47 350 350 z − ± M1 Must be a z value 0.104 to 0.164 (3 sf) A1 Must be an interval 4 6(iv) 1.25 × 1.645 (= 2.056) M1 or 1.25 × their width ÷ 2 ÷ their 47 47 350 350 (1 ) 350 − (Complete method) Φ(‘2.056’) (= 0.980) M1 Attempt Φ(their z) x = 96 (2 sf) A1 Allow 0.96 (2 sf) CWO 3

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