Cambridge A Level Mathematics 9709 — 2017 May/June Paper 7 · Variant 1

1 On average, 1 clover plant in 10 000 has four leaves instead of three. (i) Use an approximating distribution to calculate the probability that, in a random sample of 2000 clover plants, more than 2 will have four leaves. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Justify your approximating distribution. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) B1 1−e−0.2 (1 + 0.2 + 2 0.2 2 ) M1 1 – Poisson P(0, 1, 2, 3) attempted, any λ, allow one end error = 0.00115 (3 sf) A1 SR: using Bin, ans 0.00115: B1 Total: 3 1(ii) n large (n > 50) B1 np = 0.2 < 5 or p small B1 Total: 2

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2 Past experience has shown that the heights of a certain variety of plant have mean 64.0 cm and standard deviation 3.8 cm. During a particularly hot summer, it was expected that the heights of plants of this variety would be less than usual. In order to test whether this was the case, a botanist recorded the heights of a random sample of 100 plants and found that the value of the sample mean was 63.3 cm. Stating a necessary assumption, carry out the test at the 2.5% significance level. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Assume sd still = 3.8 B1 or sd unchanged H0: µ = 64.0 H1: µ < 64.0 B1 3.8 100 63.3 64.0 − M1 Standardising with their values (no sd / var mixes) Must have √100 = –1.842 A1 comp "1.842" with z-value "1.842" < 1.96 M1 comp +ve with +ve or –ve with –ve or comp Φ ("1.842") with 0.975 0.9672 < 0.975 OE No evidence that heights are shorter A1FT OE FT their zcalc Total: 6

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3 (a) The waiting time at a certain bus stop has variance 2.6 minutes2. For a random sample of 75 people, the mean waiting time was 7.1 minutes. Calculate a 92% confidence interval for the population mean waiting time. 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(b) A researcher used 3 random samples to calculate 3 independent 92% confidence intervals. Find the probability that all 3 of these confidence intervals contain only values that are greater than the actual population mean. 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(c) Another researcher surveyed the first 75 people who waited at a bus stop on a Monday morning. Give a reason why this sample is unsuitable for use in finding a confidence interval for the mean waiting time. 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Mark scheme: 3(a) 7.1 ± z × 2.6 75 M1 score M1) seen z = 1.751 B1 6.77 to 7.43 (3 sfs) A1 Must be an interval Total: 3 3(b) 0.043 M1 Allow 0.083 for M1 = 0.000064 A1 Total: 2 3(c) e.g. Particular day or time of day B1 Allow "Not random" Total: 1

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5 Large packets of sugar are packed in cartons, each containing 12 packets. The weights of these packets are normally distributed with mean 505 g and standard deviation 3.2 g. The weights of the cartons, when empty, are independently normally distributed with mean 150 g and standard deviation 7 g. (i) Find the probability that the total weight of a full carton is less than 6200 g. 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Small packets of sugar are packed in boxes. The total weight of a full box has a normal distribution with mean 3130 g and standard deviation 12.1 g. (ii) Find the probability that the weight of a randomly chosen full carton is less than double the weight of a randomly chosen full box. 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Mark scheme: 5(i) W ~ N(6210, 171.88) B2 seen or implied. B1 each parameter 6200 "6210" "171.88" − (= – 0.763) M1 Standardising with their values. No sd / var mix 1 – Φ(“0.763”) M1 For area consistent with their mean = 0.223 (3 sfs) A1 Total: 5 Question Answer Marks Guidance 5(ii) E(C – 2B) = ̶ 50 M1 “6210”–2(3130) (or E(2B–C)=50 Var(C – 2B) = "171.88" + 22 × 12.12 (= 757.52) M1 0 ( 50) "757.52" −− (= 1.817) M1 Standardising with their values Φ(“1.817”) M1 For area consistent with their mean = 0.965 (3 sfs) A1 Total: 5

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6 The number of sports injuries per month at a certain college has a Poisson distribution. In the past the mean has been 1.1 injuries per month. The principal recently introduced new safety guidelines and she decides to test, at the 2% significance level, whether the mean number of sports injuries has been reduced. She notes the number of sports injuries during a 6-month period. (i) Find the critical region for the test and state the probability of a Type I error. 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(ii) State what is meant by a Type I error in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) During the 6-month period there are a total of 2 sports injuries. Carry out the test. 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(iv) Assuming that the mean remains 1.1, calculate the probability that there will be fewer than 30 sports injuries during a 36-month period. 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Mark scheme: 6(i) mean = 6.6 B1 B1 for 6.6 (could be scored in iii) P(X ⩽ 1) = e–6.6 (1 + 6.6) = 0.0103 M1 Allow incorrect λ in both probs P(X ⩽ 2) = e–6.6(1 + 6.6 + 2 6.6 2 )= 0.0400 M1A1 A1 for both values CR is X ⩽ 1 DA1 Dep on at least one M P(Type I error) = P(X ⩽ 1) = 0.0103 B1FT FT their P(X ⩽ 1) Total: 6 6(ii) Wrongly concluding that (mean) no of (sports) injuries has decreased B1 Must be in context Total: 1 Question Answer Marks Guidance 6(iii) H0: λ = 6.6 H1: λ < 6.6 B1 Can be scored in (i). Allow µ or λ / 1.1 or 6.6 or P(X ⩽ 2) = 0.0400 > 0.02 2 not in CR M1 No evidence mean no. of injuries has decreased A1FT Total: 3 6(iv) N(39.6, 39.6) B1 May be implied 29.5 39.6 39.6 − (= −1.605) M1 Allow with wrong or no cc Φ(“–1.605”) = 1 – Φ(“1.605”) M1 For area consistent with their mean = 0.0543 (3 sfs) A1 Total: 4

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