Cambridge A Level Mathematics 9709 — 2022 May/June Paper 6 · Variant 1

1 The diameters, x millimetres, of a random sample of 200 discs made by a certain machine were recorded. The results are summarised below. n = 200 Σx = 2520 Σx2 = 31852 (a) Calculate a 95% confidence interval for the population mean diameter. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Jean chose 40 random samples and used each sample to calculate a 95% confidence interval for the population mean diameter. How many of these 40 confidence intervals would be expected to include the true value of the population mean diameter? [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) Est(μ) = 2520 200 [= 12.6] B1 OE Est(σ2) =   2 200 31582 ‘12.6’ 199 200  or 2 2520 1 31852 199 200        M1 Allow M1 if 200 199 omitted = 0.5025 or 0.503 or 100 199 A1 CWO or σ = 0.7088 or 0.709 z = 1.96 B1 ‘12.6’ ± z  ‘0.5025’ 200  M1 For expression of correct form Any z but must be z CI = 12.5 to 12.7 (3 sf) A1 CWO Must be an interval Note: Use of biased can score maximum B1 M1 A0 B1 M1 A0 6 1(b) 0.95  40 [= 38] B1 Give at early stage 1

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2 Arvind uses an ordinary fair 6-sided die to play a game. He believes he has a system to predict the score when the die is thrown. Before each throw of the die, he writes down what he thinks the score will be. He claims that he can write the correct score more often than he would if he were just guessing. His friend Laxmi tests his claim by asking him to write down the score before each of 15 throws of the die. Arvind writes the correct score on exactly 5 out of 15 throws. Test Arvind’s claim at the 10% significance level. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 H0: P(correct) = 1 6 H1: P(correct) > 1 6 B1 Allow p = 1 6 Allow p > 1 6 1 – (15C4  ( 5 6 )11  ( 1 6 )4 + 15C3  ( 5 6 )12  ( 1 6 )3 + 15C2  ( 5 6 )13  ( 1 6 )2 + 15  ( 5 6 )14  1 6 + ( 5 6 )15) M1 Expression must be seen Allow one end error 0.0898 or 0.0897 (3 sf) A1 SC if M0 scored allow SCB1 for 0.0898 or 0.0897 0.0898 < 0.1 M1 Valid comparison For valid comparison with 0.9 (0.9102 > 0.9 seen the previous M1and A1 can be recovered [Reject H0] There is evidence (at the 10% level) that Arvind can predict scores FTA1 Not definite, e.g. not ‘He can predict’ or ‘Claim true’ In context and no contradictions 5

More questions on Probability

5 Cars arrive at a fuel station at random and at a constant average rate of 13.5 per hour. (a) Find the probability that more than 4 cars arrive during a 20-minute period. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use an approximating distribution to find the probability that the number of cars that arrive during a 12-hour period is between 150 and 160 inclusive. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Independently of cars, trucks arrive at the fuel station at random and at a constant average rate of 3.6 per 15-minute period. (c) Find the probability that the total number of cars and trucks arriving at the fuel station during a 10-minute period is more than 3 and less than 7. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) λ = 4.5 B1 1 – e–4.5 (1 + 4.5 + 2 4.5 2! + 3 4.5 3! + 4 4.5 4! ) M1 Allow one end error Allow any λ. Poisson expressions must be seen = 0.468 (3 sf) A1 If M0 awarded allow SC B1 for 0.468 3 5(b) λ = 162 (X ~ Po(162) X ~ N(162, 162)) B1 149.5 ‘162’ ‘162’  and 160.5 ‘162’ ‘162’  (= –0.982 and –0.118) M1 One of these; allow with incorrect or no continuity correction Φ(‘0.982’) – ɸ(‘0.118’) oe M1 Area consistent with their values (both standardisations must be seen) = 0.290 (3 sf) A1 Allow 0.29 4 Question Answer Marks Guidance 5(c) λ = 13.5 2 3.6 6 3   OE or 4.65 M1 Attempt to find λ e–4.65( 4 5 6 4.65 4.65 4.65 4! 5! 6!   ) M1 Allow any λ Allow one end error Poisson terms not be seen 0.494 (3 sf) A1 If M0 allow SC B1 for 0.494 3

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7 In the past, the mean time for Jenny’s morning run was 28.2 minutes. She does some extra training and she wishes to test whether her mean time has been reduced. After the training Jenny takes a random sample of 40 morning runs. She decides that if the sample mean run time is less than 27 minutes she will conclude that the training has been effective. You may assume that, after the training, Jenny’s run time has a standard deviation of 4.0 minutes. (a) State suitable null and alternative hypotheses for Jenny’s test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that Jenny will make a Type I error. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Jenny found that the sample mean run time was 27.2 minutes. Explain briefly whether it is possible for her to make a Type I error or a Type II error or both. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) H0: pop mean run time = 28.2 mins H1: pop mean run time < 28.2 mins B1 Allow ‘μ’. Not ‘mean journey time’ 1 7(b) 27 28.2 4/ 40  [= –1.897] M1 For standardising Must have √40 Φ(< ‘–1.897’) = 1 – Φ(‘1.897’) M1 For correct area consistent with these values 0.0289 (3 sf) A1 3 7(c) H0 is not rejected so… M1 Type II error can be made and Type I error cannot be made A1 Both needed (accept ‘only a Type II error could be made’) 2

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