Cambridge A Level Mathematics 9709 — 2009 May/June Paper 7 · Variant 1

1 In Europe the diameters of women’s rings have mean 18.5 mm. Researchers claim that women in Jakarta have smaller fingers than women in Europe. The researchers took a random sample of 20 women in Jakarta and measured the diameters of their rings. The mean diameter was found to be 18.1 mm. Assuming that the diameters of women’s rings in Jakarta have a normal distribution with standard deviation 1.1 mm, carry out a hypothesis test at the 212% level to determine whether the researchers’ claim is justified. [5]

Mark scheme: 1 H0 : µ = 18.5 B1 Both hypotheses correct H1 : µ < 18.5 181. − 185. Test statistic z = M1 Standardising, must have 20 1.1( / 20) = –1.626 A1 For correct z CV z = ±1.96 M1 Correct comparison with correct CV or finding area on LHS of –1.626 and comparing with 2.5 % (OR comparison with 2.241 oe if one-tail test set up) Not enough evidence to support the claim A1ft Correct conclusion must ft their CV and their that fingers are smaller. z. No contradictions [5]

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2 The weights in grams of oranges grown in a certain area are normally distributed with mean µ and standard deviation σ. A random sample of 50 of these oranges was taken, and a 97% confidence interval for µ based on this sample was (222.1, 232.1). (i) Calculate unbiased estimates of µ and σ2. [4] (ii) Estimate the sample size that would be required in order for a 97% confidence interval for µ to have width 8. [3]

Mark scheme: 2 (i) µˆ = 227.(1) B1 Correct mean B1 2.17 seen σˆ 2 5 = 2.17 × M1 Solving an equation with 5 or 10 on the LHS 50 σˆ and some z value × on the RHS n ˆσ 2 = 265 or 266 A1 Correct answer [4] 163. (ii) 4 = 2.17 × B1ft Correct equation ft their wrong z if the same n as in part (i) and their σ M1 Solving an equation with their z and σ, and width 4 or 8 n = 78 A1 Correct answer (whole number) [3]

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3 Major avalanches can be regarded as randomly occurring events. They occur at a uniform average rate of 8 per year. (i) Find the probability that more than 3 major avalanches occur in a 3-month period. [3] (ii) Find the probability that any two separate 4-month periods have a total of 7 major avalanches. [3] (iii) Find the probability that a total of fewer than 137 major avalanches occur in a 20-year period. [4]

Mark scheme: 3 (i) λ = 2 B1 Correct mean (used) P(X > 3) =1 – P(0, 1, 2, 3) 2  3  2 2 − 2  M1 Poisson 1 – P(0,1,2,3) or P(0,1,2) or P(1,2,3)  1 + 2 + + = 1 – e  2 !3    = 1 – 0.857 = 0.143 A1 Correct answer [3] (ii) λ = 16/3 B1 Correct new mean  7  (16 / 3) −16 / 3   M1 P(7) using a different mean from (i) P(7) = e  !7    = 0.118 A1 Correct final answer [3] (iii) X ~ N(160, 160) B1 Correct mean and variance M1 Standardising attempt with or without cc must have sq rt  1365. − 160  P(X < 137) = P  z <  M1 Cc of 136.5 or 137.5 and area < 0.5  160  = P(z < –1.858) = 1 – 0.9684 = 0.0316 A1 Correct answer [4] GCE A/AS LEVEL – May/June 2009 9709 71

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4 In a certain city it is necessary to pass a driving test in order to be allowed to drive a car. The probability of passing the driving test at the first attempt is 0.36 on average. A particular driving instructor claims that the probability of his pupils passing at the first attempt is higher than 0.36. A random sample of 8 of his pupils showed that 7 passed at the first attempt. (i) Carry out an appropriate hypothesis test to test the driving instructor’s claim, using a significance level of 5%. [5] (ii) In fact, most of this random sample happened to be careful and sensible drivers. State which type of error in the hypothesis test (Type I or Type II) could have been made in these circumstances and find the probability of this type of error when a sample of size 8 is used for the test. [4]

Mark scheme: 4 (i) H0 : p = 0.36 B1 Both hypotheses correct H1 : p > 0.36 P(7) = 8C7 × (0.36)7 (0.64)1 = 0.00401 M1 Evaluating P(7) or P(8) P(8) = (0.36)8 = 0.000282 A1 Correct answer for both Σ P = 0.00429 < 0.05 M1 Comparing their prob sum to 0.05 oe Accept driving instructor’s claim B1 Correct conclusion cwo no contradictions [5] (ii) Type I error B1 Correct answer P(6) = 8C6 × (0.36)6 (0.64)2 = 0.02496 M1 Evaluating P(6) P(5) = 8C5 × (0.36)5 (0.64)3 = 0.08876, B1 Correct P(5) and showing this is not in the CR > 0.05 either by Σ P > 0.05 or P(5) > 0.05 P(Type I error) = 0.0292 or 0.0293 A1 Correct answer NB Marks for part (ii) may be awarded in part (i) but not vice versa. [4] 6 ∫ 2

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5 The time in minutes taken by candidates to answer a question in an examination has probability density function given by k(6t −t2) 3 ≤t ≤6, f(t) = 0 otherwise, where k is a constant. (i) Show that k = 1 [3] 18. (ii) Find the mean time. [3] (iii) Find the probability that a candidate, chosen at random, takes longer than 5 minutes to answer the question. [2] (iv) Is the upper quartile of the times greater than 5 minutes, equal to 5 minutes or less than 5 minutes? Give a reason for your answer. [2]

Mark scheme: 5 (i) ∫ 3 k ( 6t − t 2 ) dt = 1 M1 For equating to 1 and a sensible attempt to integrate 2 3 6 k [3t −t / 3] 3 = 1 k([108 – 216/3] – [27 – 9]) = 1 A1 Correct integration and correct limits k = 1/18 AG A1 Given answer correctly obtained [3] 6 (ii) mean = ∫ 3 k (6t 2 − t 3 ) dt M1 Attempt to evaluate the integral of tf(t) (t or x) 6  4  3 =  k ( 2t −t )  A1 Correct integral and correct limits (condone  4  3 loss of k) = k(432 – 324) – k(54 – 81 / 4) 33 = (4.13) 8 A1 Correct answer [3] 6 (iii) ∫ 5 k (6t − t 2) dt M1 Attempt to evaluate the integral between 5 and 6 6 oe  3  100 2 = k 3t −t  = k ( 36 − )  3  5 3 4 = (0.148) A1 Correct answer 27 [2] (iv) the area on the left is > 0.75 M1 sensible reason or (iii) is < 0.25 UQ is less than 5 A1ft ft their (iii) SR B1ft correct but 0.25/0.75 implied [2] GCE A/AS LEVEL – May/June 2009 9709 71

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