Cambridge A Level Mathematics 9709 — 2018 May/June Paper 7 · Variant 1

1 A random sample of 75 values of a variable X gave the following results. n = 75 Σ x = 153.2 Σx2 = 340.24 Find unbiased estimates for the population mean and variance of X. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 est(µ) (= 153.2 ÷ 75) = 2.04 (3 sf) B1 est(σ2) = 7475 ( 340.2475 − "2.04267"2 ) oe M1 = 0.369 (3 sf) A1 Accept 0.368 3

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2 A six-sided die is suspected of bias. The die is thrown 100 times and it is found that the score is 2 on 20 throws. It is given that the probability of obtaining a score of 2 on any throw is p. (i) Find an approximate 94% confidence interval for p. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use your answer to part (i) to comment on whether the die may be biased. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 20 0.2×(1− 0.2) M1 Any z ± z × 100 100 z = 1.881 or 1.882 B1 = 0.125 to 0.275 A1 3 2(ii) 1 6 is within this range B1ft Both statements needed No evidence of bias concerning 2 1

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4 The volume, in millilitres, of a small cup of coffee has the distribution N 103.4, 10.2 . The volume of a large cup of coffee is 1.5 times the volume of a small cup of coffee. (i) Find the mean and standard deviation of the volume of a large cup of coffee. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that the total volume of a randomly chosen small cup of coffee and a randomly chosen large cup of coffee is greater than 250 ml. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) mean= 155.1 B1 var = 1.52 × 10.2 ( = 22.95) M1 or 1.5 × √10.2 sd = √"22.95" = 4.79 A1 3 4(ii) mean = 103.4 + “155.1” (= 258.5) B1ft Both. ft their 155.1 and 22.95. Accept var = 10.2 + “22.95” (=33.15) sd. 250 −"258.5" (= –1.476) M1 Standardising – no sd/var mix. Their "33.15" mean/sd must be from an attempt at combination 1– ɸ(–1.476) = ɸ(1.476) M1 For area consistent with their working = 0.930 (3 sf) A1 Allow 0.93 4

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5 The mass, in kilograms, of rocks in a certain area has mean 14.2 and standard deviation 3.1. (i) Find the probability that the mean mass of a random sample of 50 of these rocks is less than 14.0 kg. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Explain whether it was necessary to assume that the population of the masses of these rocks is normally distributed. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) A geologist suspects that rocks in another area have a mean mass which is less than 14.2 kg. A random sample of 100 rocks in this area has sample mean 13.5 kg. Assuming that the standard deviation for rocks in this area is also 3.1 kg, test at the 2% significance level whether the geologist is correct. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 14 − 14.2 M1 For stand'n; must have √50 (= – 0.456) 3.1 50 1 – Φ(“0.456”) M1 for area consistent with their working = 0.324 (3 sfs) A1 3 5(ii) No because n large B1 Accept n > 30 1 5(iii) H0: µ = 14.2 B1 H1: µ < 14.2 or ‘pop mean’, but not just ‘mean’ 13.5 − 14.2 M1 For stand'n; must have √100 3.1 100 = –2.258 A1 comp –2.054 (or –2.055) M1 Valid comparison of z values or areas (0.0119 < 0.02) There is evidence (at 2% level) that mean A1ft Ft their z. Correct conclusion no mass in this area < 14.2 contradictions 5

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7 The number of absences by girls from a certain class on any day is modelled by a random variable with distribution Po 0.2 . The number of absences by boys from the same class on any day is modelled by an independent random variable with distribution Po 0.3 . (i) Find the probability that, during a randomly chosen 2-day period, the total number of absences is less than 3. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that, during a randomly chosen 5-day period, the number of absences by boys is more than 3. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) The teacher claims that, during the football season, there are more absences by boys than usual. In order to test this claim at the 5% significance level, he notes the number of absences by boys during a randomly chosen 5-day period during the football season. (a) State what is meant by a Type I error in this context. [1] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) State appropriate null and alternative hypotheses and find the probability of a Type I error. [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (c) In fact there were 4 absences by boys during this period. Test the teacher’s claim at the 5% significance level. [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................

Mark scheme: 7(i) Po(1.0) B1 Seen or implied e–1 (1 + 1 + 122 ) M1 Allow any λ. Allow one end error. = 0.920 (3 sfs) A1 3 7(ii) P(X > 3) = 1 – e–1.5(1+1.5+ 1.52 2 + 1.53!3 ) M1 Allow any λ. Allow one end error = 0.0656 A1 2 7(iii)(a) Incorrectly concluding that more absences B1 In context than usual when there are not oe 1 7(iii)(b) H0: λ = 1.5 (or 0.3) B1 Or µ H1: λ > 1.5 (or 0.3) Both P(X > 4) = “0.0656” – e–1.5 × 1.54!4 M1 or 1 – e–1.5(1+1.5+ 1.52 2 + 1.53!3 + 1.54!4 ) = 0.0186 (3 sf) P(Type I) = 0.0186 or 0.0185 A1ft Ft their P(X > 4) if less than 0.05 3 7(iii)(c) P(X > 3) = "0.0656" B1ft Ft their (ii) 0.0656 > 0.05 M1 No evidence of more than usual male A1ft Ft their P(X>3). Correct conclusion. absences No contradictions. 3

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