Cambridge A Level Mathematics 9709 — 2015 May/June Paper 7 · Variant 2

1 The independent random variables X and Y have standard deviations 3 and 6 respectively. Calculate the standard deviation of 4X −5Y. [3]

Mark scheme: 1 Var = 16×9 + 25×36 (= 1044) B1 M1 for 16 (or 4²) & 25 (or 5²) used M1 M1 for add any multiples of 9 and 36 only sd = 32.3 or 6√29 or √1044 A1 Total 3

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2 Cloth made at a certain factory has been found to have an average of 0.1 faults per square metre. Suki claims that the cloth made by her machine contains, on average, more than 0.1 faults per square metre. In a random sample of 5 m2 of cloth from Suki’s machine, it was found that there were 2 faults. Assuming that the number of faults per square metre has a Poisson distribution, (i) state null and alternative hypotheses for a test of Suki’s claim, [1] (ii) test at the 10% significance level whether Suki’s claim is justified. [4]

Mark scheme: 2 (i) H0: λ = 0.5 or Pop mean = 0.5, not just Mean = 0.5 or Pop mean (per m2 )= 0.1 H1: λ >0.5 B1 1 Accept µ instead of λ (ii) 1 – e–0.5(1 + 0.5) M1 1 – P(X = 0,1) attempted, any λ. Allow 1 end error = 0.0902 (3 sf) A1 Allow 0.09 comp 0.1 M1 Valid comparison NB 0.9098>0.9 recovers M1A1 M1 Claim justified or there is evidence to oe Accept ‘Reject H0’ if correctly defined support claim A1 4 No contradictions. Total 5

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3 In a golf tournament, the number of times in a day that a ‘hole-in-one’ is scored is denoted by the variable X, which has a Poisson distribution with mean 0.15. Mr Crump offers to pay $200 each time that a hole-in-one is scored during 5 days of play. Find the expectation and variance of the amount that Mr Crump pays. [5]

Mark scheme: 3 λ = 5×0.15 (= 0.75) M1 E(amount) = 200×0.75 = 150 A1 Var(weekly no of hole–in–ones) = 0.75 B1 Var(amount) = 2002 × 0.75 M1 Allow 2002 × their variance (with nothing = 30,000 A1 5 added/subtracted at any stage) (SR probability table can score M1A0 srB1 if var rounds to 30,000 (2sf) ) Total 5

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4 In the past, the flight time, in hours, for a particular flight has had mean 6.20 and standard deviation 0.80. Some new regulations are introduced. In order to test whether these new regulations have had any effect upon flight times, the mean flight time for a random sample of 40 of these flights is found. (i) State what is meant by a Type I error in this context. [2] (ii) The mean time for the sample of 40 flights is found to be 5.98 hours. Assuming that the standard deviation of flight times is still 0.80 hours, test at the 5% significance level whether the population mean flight time has changed. [4] (iii) State, with a reason, which of the errors, Type I or Type II, might have been made in your answer to part (ii). [2]

Mark scheme: 4 (i) Conclude flight times affected B1 Or accept pop mean changed from 6.2 when in fact they have not been. B1 2 although pop mean has not changed from 6.2 (ii) H0: Pop mean (or µ) = 6.2 H0: Pop mean (or µ) ≠ 6.2 B1 .5 98 − 2.6 M1 Allow with 40 instead of √40 Allow SD/Var mix 8.0 A1 (CV method 5.952 or 6.2279 M1 A1) 40 B1 For valid comparison = –1.739 (±) Accept (±)1.74 4 or P(z < –1.739) = 0.041 > 0.025 or 5.98 > 5.952 or comp z = 1.96 6.2 < 6.228 and correct conclusion No evidence that flight times affected (iii) H0 was not rejected oe B1* If in (ii) H0 was rejected, then: Type II B1*dep H0 rejected B1; Type I B1dep 2 Total 8

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5 The volumes, v millilitres, of juice in a random sample of 50 bottles of Cooljoos are measured and summarised as follows. n = 50 Σv = 14 800 Σv2 = 4 390 000 (i) Find unbiased estimates of the population mean and variance. [3] (ii) An !% confidence interval for the population mean, based on this sample, is found to have a width of 5.45 millilitres. Find !. [4] Four random samples of size 10 are taken and a 96% confidence interval for the population mean is found from each sample. (iii) Find the probability that these 4 confidence intervals all include the true value of the population mean. [2]

Mark scheme: 5 (i) 14800/50 or 296 B1 50  4390000 2  M1 Oe − '296' (= 187.755)   49  50  A1 3 = 188 (3 sf) (ii) '187.755' M1 '187.755' 2 × z × = 5.45 oe If ‘2 ×’ omitted: z × = 5.45 M1 50 A1 50 z = 1.406 or 1.405 z = 2.812 or 2.810 A0 Φ(‘1.406’) (= 0.92 or 0.9199) Φ(‘2.812’) (= 0.9975) M1 α = 99.5 or 99 or 100 M1 A0 α = 84 (2 sf) allow 83.98 A1 4 For complete method to find α SR use of biased var(184) scores M1A1(1.4205) Α=84.5 M1A1 (iii) 0.964 M1 = 0.849 (3 sf) A1 2 Total 9 15 ∫ 2

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6 The waiting time, T minutes, for patients at a doctor’s surgery has probability density function given by T k 225 −t2 0 ≤t ≤15, f t = 0 otherwise, where k is a constant. (i) Show that k = 2250.1 [3] (ii) Find the probability that a patient has to wait for more than 10 minutes. [3] (iii) Find the mean waiting time. [4]

Mark scheme: 6 (i) k ∫ ( 225 − t 2 ) dt = 1 M1 Attempt integ f(x) and = 1. Ignore limits 0  3  15 A1 Correct integration and limits k  225t −t  = 1 3 0   A1 3 No errors seen k×[3375 – 1125} = 1 or k×2250 = 1 1 (k = AG) 2250 1 15 (ii) M1 Attempt integ, ignore limits 2250 ∫ (225 − t 2 )d t 10 1  3  15 10 A1 Or 1– ∫o (=  225t −t  ) 2250  3  10 Correct integration and limits. 1 1000 Condone missing k = [2250 – (2250 – )] 2250 3 A1 3 4 = or 0.148 (3 sf) 27 1 15 (iii) M1* Attempt integ xf(x) , ignore limits 2250 ∫ (225t − t 3 )d t 0 1  225t 2 4  15 A1 Correct integration and limits. Condone missing k =  −t  2250 2 4 0   1 50625 50625 M1*dep Sub correct limits into their integral = [ − ] 2250 2 4 A1 4 Accept 5 mins 37 or 38 secs 45 = or 5.625 or 5.63 (3 sf) 8 Total 10

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