Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 7 · Variant 3

1 Test scores, X, have mean 54 and variance 144. The scores are scaled using the formula Y = a + bX, where a and b are constants and b > 0. The scaled scores, Y, have mean 50 and variance 100. Find the values of a and b. [4]

Mark scheme: 1 50 = a + b × 54 B1 100 = b2 × 144 or 10 = b × 12 B1 b = 5 oe M1 Solving two simultaneous equations 6 a = 5 A1 [4] Both correct 0 35 × 0 65 M1 For √(pq/n) in equation

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2 35% of a random sample of n students walk to college. This result is used to construct an approximate 98% confidence interval for the population proportion of students who walk to college. Given that the width of this confidence interval is 0.157, correct to 3 significant figures, find n. [5]

Mark scheme: .0 35 × .0 65 (pq ) q 2 2 × z × = 0.157 M1 For equation of the form n 2 × z × f(n) = 0.157 z = 2.326 B1 n = 4 × 2.3262 × 0.35 × 0.65 ÷ 0.1572 M1 Rearrange to form n = ... from a correct (=199.738…) equation in n, but allow any z and/or factor of “2” errors n = 200 A1 [5] cao

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3 Jack has to choose a random sample of 8 people from the 750 members of a sports club. (i) Explain fully how he can use random numbers to choose the sample. [3] Jack asks each person in the sample how much they spent last week in the club caf´e. The results, in dollars, were as follows. 15 25 30 8 12 18 27 25 (ii) Find unbiased estimates of the population mean and variance. [3] (iii) Explain briefly what is meant by ‘population’ in this question. [1]

Mark scheme: 3 (i) Number all members B1 Explain the selection of 3-digit random B1 numbers Omit repeats OR omit nos. over 750 (until B1 have 8 nos.) [3] (ii) Est (µ) = 20 B1 8  3636 2  M1 1/7 × (3636 – 1602/8) Est (σ2) =  − 20  7  8  436 A1 (7.89...)2 M1A1, but 7.89... only M1A0 = or 62.3 (3 sfs) 7 [3] (iii) Amounts spent last week in café by all club B1 members [1] 1 M1 Int = 1 ignore limits ∫

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4 The random variable X has probability density function given by ke−x 0 ≤x ≤1, f(x) = 0 otherwise. e (i) Show that k = [3] e −1. (ii) Find E(X) in terms of e. [4]

Mark scheme: − M1 Int 1, ignore limits 4 (i) ∫0 ke xd x = 1 = 1 [− ke − x ]01 A1 Correct integral & limits, & = 1 (= –ke–1 – (–ke0) ) e − 1 = k × = 1 or k(e – 1) = e e e A1 Correctly obtained, no errors seen k = AG e − 1 [3] GCE AS/A LEVEL – October/November 2011 9709 73 e 1 − M1 Attempt ∫xf(x)dx, ignore limits (ii) e − 1 ∫0 xe xd x e − x 1 1 − x M1* Attempt integration by parts the correct = – ∫0 ( − e )dx ) way round, ignore limits ([x ( − e ) ]0 e − 1 e x 1 − x 1 M1dep* Attempt second integral of the form = ([− xe− – [e ) ±∫e–xdx, ignore limits e − 1 ]0 ]0 e (= (–e–1 – 0 – (e–1 – 1) ) ) e − 1 e 2 e − 2 A1 e = (1 – ) or oe Accept k instead of throughout e − 1 e e − 1 e − 1 [4] except ans

More questions on Continuous random variables

5 Records show that the distance driven by a bus driver in a week is normally distributed with mean 1150 km and standard deviation 105 km. New driving regulations are introduced and in the next 20 weeks he drives a total of 21 800 km. (i) Stating any assumption(s), test, at the 1% significance level, whether his mean weekly driving distance has decreased. [6] (ii) A similar test at the 1% significance level was carried out using the data from another 20 weeks. State the probability of a Type I error and describe what is meant by a Type I error in this context. [2]

Mark scheme: 5 (i) Assume pop sd same (105) B1 H0: Pop mean = 1150 B1 Allow “µ” but not just “mean” H1: Pop mean < 1150 21800 M1 105 − 1150 Allow ÷ . (Accept “totals” method) 20 20 105 20 = ±2.556 or 2.56 A1 Or 0.0053 if prob/area comparison used Compare with z = ±2.326 M1 Correct comparison of z or prob/area (for a clear 2 tail test compare with ±2.576) consistent with their test Evidence that mean distance decreased A1ft In context. Allow mean dist decreased [6] ft their z and/or clear 2 tail test (ii) 0.01 B1 Concluding there has been a decrease when B1 In context there has not. [2] GCE AS/A LEVEL – October/November 2011 9709 73

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6 Ranjit goes to mathematics lectures and physics lectures. The length, in minutes, of a mathematics lecture is modelled by the variable X with distribution N(36, 3.52). The length, in minutes, of a physics lecture is modelled by the independent variable Y with distribution N(55, 5.22). (i) Find the probability that the total length of two mathematics lectures and one physics lecture is less than 140 minutes. [4] (ii) Ranjit calculates how long he will need to spend revising the content of each lecture as follows. Each minute of a mathematics lecture requires 1 minute of revision and each minute of a physics lecture requires 112 minutes of revision. Find the probability that the total revision time required for one mathematics lecture and one physics lecture is more than 100 minutes. [4]

Mark scheme: 6 (i) E(Tot) = 2 × 36 + 55 (= 127) B1 (Or ±13) Var(Tot) = 2 × 3.52 + 5.22 (= 51.54) B1 140 − 127 M1 For standardising with their mean and (= 1.811) var. Allow without √ or with false cc, "51. 54" but their mean and variance must involve parameters from both given distributions Φ(“1.811”) = 0.965 (3 sfs) A1 [4] (ii) E(RM) = 36 + 1.5 × 55 (= 118.5) B1 (Or ±18.5) Var(RM) = 3.52 + 1.52×5.22 (= 73.09) B1 100 − 1185 M1 For standardising with their mean and (= –2.164) var. Allow without √ or with false cc, 73. 09 but their mean and variance must involve parameters from both given distributions 1 – Φ(“–2.164”) = Φ (“2.164”) 0.985 (3 sfs) A1 [4] 1 2

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7 The numbers of men and women who visit a clinic each hour are independent Poisson variables with means 2.4 and 2.8 respectively. (i) Find the probability that, in a half-hour period, (a) 2 or more men and 1 or more women will visit the clinic, [4] (b) a total of 3 or more people will visit the clinic. [3] (ii) Find the probability that, in a 10-hour period, a total of more than 60 people will visit the clinic. [4]

Mark scheme: 7 (i) (a) 1 – e–1.2(1 + 1.2) (= 0.3374) M1 M1 for Poisson either P(0 or 1) or P(0) 1 – e–1.4 (= 0.7534) A1 with λ = 1.2 or 2.4 or 1.4 or 2.8, accept one end error Both expressions fully correct (1 – e–1.2(1 + 1.2)) × (1 – e–1.4) M1 Their Poisson P(0 or 1) × P(0) = 0.254 (3 sfs) A1 [4] (i) (b) λ = 2.6 seen B1 1 – e–2.6(1 + 2.6 + 2.62 ÷ 2) M1 Poisson 1 – P(0, 1, 2), allow 1 – P(0, 1, 2, 3), with attempt at combined λ for M and W. Accept combination method: at least 4 correct terms and “1 –” M1; all terms correct B1 = 0.482 (3 sfs) A1 [3] GCE AS/A LEVEL – October/November 2011 9709 73 (ii) N(52, 52) B1 Seen or implied 605. − 52 M1 Standardising with N(λ, λ) with (= 1.179) 52 λ = 10 × 5.2 or 10 × 2.6 Allow with wrong or no cc or no √ 1 – Φ(“1.179”) M1 Their correct area (= 1 – 0.8808) = 0.119 (3 sfs) A1 [4]

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