Cambridge A Level Mathematics 9709 — 2016 May/June Paper 7 · Variant 3

1 The time taken for a particular type of paint to dry was measured for a sample of 150 randomly chosen points on a wall. The sample mean was 192.4 minutes and an unbiased estimate of the population variance was 43.6 minutes2. Find a 98% confidence interval for the mean drying time. [3]

Mark scheme: Qu Answer Marks Notes 1 192.4 ± z 43.6150 M1 Allow 43.6 Allow one side for M1 150 B1 z = 2.326 to 2.329 Condone √(43.6/149 ) oe 191 to 194 (3 sf) A1 [3] CWO

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2 In the past, the mean annual crop yield from a particular field has been 8.2 tonnes. During the last 16 years, a new fertiliser has been used on the field. The mean yield for these 16 years is 8.7 tonnes. Assume that yields are normally distributed with standard deviation 1.2 tonnes. Carry out a test at the 5% significance level of whether the mean yield has increased. [5]

Mark scheme: 2 Ho: Pop mean yield = 8.2 or µ = 8.2(not just “mean”) H1: Pop mean yield > 8.2 B1 µ > 8.2 8.7 −8.2 (±) M1 Allow without √ sign (Allow cc) 1.2/ 16 A1 = (±)1.667 M1 Or comp 1 - Φ('1.667') with 0.05 Comp z = 1.645 Or Area comparison 0.0475-0.0478) Valid Comparison z-values (same sign) or areas Reject H0 No Contradictions A1 [5] No follow through for 2 tail test Evidence that mean yield has increased

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3 1% of adults in a certain country own a yellow car. (i) Use a suitable approximating distribution to find the probability that a random sample of 240 adults includes more than 2 who own a yellow car. [4] (ii) Justify your approximation. [2]

Mark scheme: 3 (i) Use of Poisson B1 Mean = 2.4 B1 1 – e-2.4(1 + 2.4 + 2.42 2 ) M1 Allow any λ (Allow one end error) Final answer A1 [4] = 0.43(0) (3 sf) SR Use of binomial: B1 for ans 0.431 (3 sf) (ii) 240 > 50 or n>50 B1 240 × 0.01 = 2.4 < 5 or np<5 or p<0.1 B1 [2] SR n large, p small: B1

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4 The number of sightings of a golden eagle at a certain location has a Poisson distribution with mean 2.5 per week. Drilling for oil is started nearby. A naturalist wishes to test at the 5% significance level whether there are fewer sightings since the drilling began. He notes that during the following 3 weeks there are 2 sightings. (i) Find the critical region for the test and carry out the test. [5] (ii) State the probability of a Type I error. [1] (iii) State why the naturalist could not have made a Type II error. [1]

Mark scheme: 4 (i) H0: Pop mean = 2.5 (or 7.5) or λ = 2.5(Not just “mean”) Allow µ H0: Pop mean < 2.5 (or 7.5) B1 or λ < 2.5 λ = 7.5 P(X ⩽ 2) = e–7.5(1+7.5+ 7.52 2 ) = 0.0203 3 M1 Either P(X⩽2) or P(X⩽3) , allow any λ P(X⩽3)=0.0203 + e–7.5× 7.53! = 0.0591 A1 Both Correct CR is X ⩽ 2 A1 Clear statement Reject H0 A1 [5] Follow through their CR/their P(X⩽2) Evidence that no of sightings fewer (ii) P(Type I) = 0.0203 (3 sf) B1 [1] ft their P(X ⩽ 2) (iii) H0 was rejected oe B1 [1] or Type II is P(not reject H0)oe

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7 Bags of sugar are packed in boxes, each box containing 20 bags. The masses of the boxes, when empty, are normally distributed with mean 0.4 kg and standard deviation 0.01 kg. The masses of the bags are normally distributed with mean 1.02 kg and standard deviation 0.03 kg. (i) Find the probability that the total mass of a full box of 20 bags is less than 20.6 kg. [5] (ii) Two full boxes are chosen at random. Find the probability that they differ in mass by less than 0.02 kg. [5]

Mark scheme: 7 (i) E(T) =20.8 B1 Var(T)= 20 × 0.032 + 0.012(= 0.0181) B1 or √(20 × 0.032 + 0.012 ) = 0.135 (3sf) 20.6 − 20.8 (= –1.487) "0.0181" M1 For standardising (σ must come from combination) 1 – Φ(“1.487”) M1 Area consistent with their working = 0.0684 to 0.686 A1 [5] Any answer within range (ii) E(D) = 0 Var(D) = 2 × 0.0181(= 0.0362) B1 Both (Seen or implied) 0.02 − 0 ( = 0.105) M1 Allow without √ "0.0362) A1 Allow to 3sf Φ(“0.105”) = 0.5418 or 1-Φ(0.015) =0.4582 M1 or 1 – 2(1 – Φ(“0.105”)) Φ(“0.105”) – (1 – Φ(“0.105”) (= 1 – 2 × 0.4582) (= 0.5418 – 0.4582) A1 [5] = 0.0836/0.0837

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