Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 7 · Variant 3

1 The standard deviation of the heights of adult males is 7.2 cm. The mean height of a sample of 200 adult males is found to be 176 cm. (i) Calculate a 97.5% confidence interval for the mean height of adult males. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State a necessary condition for the calculation in part (i) to be valid. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(i) 176 ± z × 7.2 M1 need correct form must be z 200 z = 2.24 B1 allow 2.241 and 2.242 175 to 177 A1 cwo 3 1(ii) Sample random B1 oe. both words essential 1

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4 Small drops of two liquids, A and B, are randomly and independently distributed in the air. The average numbers of drops of A and B per cubic centimetre of air are 0.25 and 0.36 respectively. (i) A sample of 10 cm3 of air is taken at random. Find the probability that the total number of drops of A and B in this sample is at least 4. 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(ii) A sample of 100 cm3 of air is taken at random. Use an approximating distribution to find the probability that the total number of drops of A and B in this sample is less than 60. 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Mark scheme: 4(i) λ = 10×0.25 + 10×0.36 ( = 6.1 ) B1 1 – e-6.1 (1 + 6.1 + 6.12 2 + 6.13!3 ) M1 1 – P(X ⩽ 3), any λ Allow one end error = 0.857 A1 Allow 0.858 3 4(ii) λ = 61 B1 ft Ft from (i) N(‘61’, ‘61’) M1 N with µ = λ, any λ. May be implied 59.5 − 61 (= –0.192) M1 Standardise with their mean and variance '61' Allow no or wrong cc. not 61/100 Φ(‘–0.192’) = 1 – Φ(‘0.192’) M1 Correct area consistent with their working = 0.424 A1 5

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5 The times, in months, taken by a builder to build two types of house, P and Q, are represented by the independent variables T1 ∼N 2.2, 0.42 and T2 ∼N 2.8, 0.52 respectively. (i) Find the probability that the total time taken to build one house of each type is less than 6 months. 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(ii) Find the probability that the time taken to build a type Q house is more than 1.2 times the time taken to build a type P house. 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Mark scheme: 5(i) T1 + T2 ~ N( 5, 0.42 + 0.52 ) B1 or N( 5, 0.41 ) 6 − 5 (= 1.562) M1 Allow cc '0.41' Φ(‘1.562’) M1 Correct area consistent with their working = 0.941 A1 4 5(ii) Var(T2 - 1.2T1) = 0.52 + 1.22 × 0.42 B1 Or similar using 1.2T1 – T2 (= 0.4804) T2 – 1.2T1 – N(0.16, 0.4804) B1 ft Only ft attempt at combination. no ft for neg var. 0 − '0.16' M1 Standardise with their mean and variance. (= -0.231) Allow cc '0.4804' P(T2 – 1.2T1) > 0 = Φ(‘0.231’) M1 Correct area consistent with their working = 0.591 (3 sfs) A1 5

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7 A mill owner claims that the mean mass of sacks of flour produced at his mill is 51 kg. A quality control officer suspects that the mean mass is actually less than 51 kg. In order to test the owner’s claim she finds the mass, x kg, of each of a random sample of 150 sacks and her results are summarised as follows. n = 150 Σx = 7480 Σx2 = 380 000 (i) Carry out the test at the 2.5% significance level. 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You may now assume that the population standard deviation of the masses of sacks of flour is 6.856 kg. The quality control officer weighs another random sample of 150 sacks and carries out another test at the 2.5% significance level. (ii) Given that the population mean mass is 49 kg, find the probability of a Type II error. 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Mark scheme: 7(i) H0: µ = 51 H1: µ < 51 B1 Or popn mean … x = 7480150 = 49.8667 = 49.9 B1 s2 = 150149 ( 380000150 − ( 74815 ) 2 ) M1 Correct subst in s2 or 2s formula = 46.9620 = 47.0 or s = 6.85 Biased var scores M0 49.8667 '46.962' − 51 allow 49.9'47'− 51 M1 Allow 49.8667 to 49.9 in numerator 150 150 Need sqrt 150 = ( – ) 2.025 = ( – ) 1.965 A1 Accept 2.02 or 2.03 Accept –2.0264 –1.9651 provided correct working comp z = 1.96 M1 or comp 1 – ɸ(2.025) with 0.025 There is evidence that µ < 51 A1 ft no contradictions biased var B1B1M0M1A0M1A1ft (max 5/7) accept cv method xcrit = 49.9028 M1A1 49 867 < 49.9… M1A1 7 7(ii) x6.856− 51 = –1.96 M1 Need 51 and sqrt 150 and correct form 150 x = 51 – 1.097 = 49.9 A1 This may have been found in part (i) Rejection region is x < 49.9 49.9 − 49 (= 1.608 to 1.614) M1 Need 49 and sqrt 150 and correct form 6.856 150 P( x > 49.9 | µ = 49) = 1 – Φ(‘1.608’) M1 P(Type II error) = 0.0539 A1 Allow 0.0533 to 0.0539 5

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