Cambridge A Level Mathematics 9709 — 2020 May/June Paper 6 · Variant 3

1 A random sample of 100 values of a variable X is taken. These values are summarised below. n = 100 Σx = 1556 Σx2 = 29 004 Calculate unbiased estimates of the population mean and variance of X. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Est σ2 = 2 100 29004 "15.56" 99 100   −     or = 2 1 1556 29004 99 100   −     M1 48.4105 = 48.4 (3 sf) A1 3

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5 Sunita has a six-sided die with faces marked 1, 2, 3, 4, 5, 6. The probability that the die shows a six on any throw is p. Sunita throws the die 500 times and finds that it shows a six 70 times. (a) Calculate an approximate 99% confidence interval for p. 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(b) Sunita believes that the die is fair. Use your answer to part (a) to comment on her belief. 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(c) Sunita uses the result of her 500 throws to calculate an !% confidence interval for p. This interval has width 0.04. Find the value of !. 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Mark scheme: 5(a) p = 70 500 or 0.14 B1 z = 2.576 B1 "0.14" ± z × ( ) "0.14" 1 "0.14" 500 − M1 0.100 to 0.180 A1 4 Question Answer Marks 5(b) 0.1666... is within confidence interval Belief supported or justified B1 1 5(c) z × ( ) "0.14" 1 "0.14" 500 − = 0.02 M1 z = 1.289 A1 Φ(‘1.289’) = 0.9013 M1 α = ‘0.9013’ – (1 – ‘0.9013’) M1 80.3% (3 sf) A1 5

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7 A market researcher is investigating the length of time that customers spend at an information desk. He plans to choose a sample of 50 customers on a particular day. (a) He considers choosing the first 50 customers who visit the information desk. Explain why this method is unsuitable. 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The actual lengths of time, in minutes, that customers spend at the information desk may be assumed to have mean - and variance 4.8. The researcher knows that in the past the value of - was 6.0. He wishes to test, at the 2% significance level, whether this is still true. He chooses a random sample of 50 customers and notes how long they each spend at the information desk. (b) State the probability of making a Type I error and explain what is meant by a Type I error in this context. 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(c) Given that the mean time spent at the information desk by the 50 customers is 6.8 minutes, carry out the test. 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(d) Give a reason why it was necessary to use the Central Limit theorem in your answer to part (c). 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Mark scheme: 7(a) Later customers might spend times different from first ones B1 1 7(b) 0.02 B1 Concluding that μ ≠ 6.0, when actually μ = 6.0 B1 2 Question Answer Marks 7(c) H0: μ = 6.0 H1: μ ≠ 6.0 B1 6.8 6.0 4.8 50 − M1 2.582 A1 comp 2.326 M1 Evidence that μ ≠ 6.0 A1 5 7(d) Population distribution unknown B1 1

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