Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 6 · Variant 2

1 Twelve tourists were asked to estimate the height, in metres, of a new building. Their estimates were as follows. 50 45 62 30 40 55 110 38 52 60 55 40 (i) Find the median and the interquartile range for the data. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Give a disadvantage of using the mean as a measure of the central tendency in this case. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) UQ = 57.5, LQ = 40 B1 IQR = UQ – LQ M1 55 ⩽ UQ ⩽ 62 – 38 ⩽ LQ ⩽ 45 17.5 A1 NFWW 3 1(ii) Result will be disproportionately affected by 110 B1 Affected by an extreme/large value There is a large outlier …contains outliers such as 110… Not ‘mean affected by extreme values’ 1

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2 Benju cycles to work each morning and he has two possible routes. He chooses the hilly route with probability 0.4 and the busy route with probability 0.6. If he chooses the hilly route, the probability that he will be late for work is x and if he chooses the busy route the probability that he will be late for work is 2x. The probability that Benju is late for work on any day is 0.36. (i) Show that x = 0.225. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given that Benju is not late for work, find the probability that he chooses the hilly route. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 1.6x = 0.36 x = 0.225 A1 Fully justified by algebra AG 2 Question Answer Marks Guidance 2(ii) P( H L') = ( ) ( ) 0.4 1 0.4 1 0.225 0.4 0.775 1 0.36 0.64 0.4 0.775 0.6 0.55 x − × − × = = − × + × M1 Correct numerical numerator of a fraction. Allow unsimplified. M1 Denominator 0.36 or 0.64. Allow unsimplified. 31 or 0.484 64 A1 3

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3 The speeds, in km h−1, of 90 cars as they passed a certain marker on a road were recorded, correct to the nearest km h−1. The results are summarised in the following table. Speed (km h−1) 10 −29 30 −39 40 −49 50 −59 60 −89 Frequency 10 24 30 14 12 (i) On the grid, draw a histogram to illustrate the data in the table. [4] (ii) Calculate an estimate for the mean speed of these 90 cars as they pass the marker. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 0.5 2.4 3 1.4 0.4 M1 At least 3 frequency densities calculated (frequency ÷ class width) e.g. 10 10 10 , or 20 19 19.5       may be read from graph using their scale, 3SF or exact All heights correct on graph. A1 Bar ends of 9.5, 29.5, 39.5, 59.5, 89.5 B1 Axes labelled: Frequency density (fd) and speed/km h-1 (or appropriate title). Linear scales 9.5 ⩽ horizontal axis ⩽ 89.5, 0 ⩽ vertical axis ⩽ 3, 5 bars with no gaps B1 4 Question Answer Marks Guidance 3(ii) 19.5 10 34.5 24 44.5 30 54.5 14 74.5 12 90 their × + × + × + × + × =195 828 1335 763 894 90 + + + + = 4015 803 or 90 18 M1 Uses at least 4 midpoint attempts (e.g. 19.5 ± 0.5). Allow unsimplified expression. 1 11 44 or 44.6 (km h ) 18 − A1 Final answer not an improper fraction NFWW 2

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4 In Quarendon, 66% of households are satisfied with the speed of their wificonnection. (i) Find the probability that, out of 10 households chosen at random in Quarendon, at least 8 are satisfied with the speed of their wificonnection. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) A random sample of 150 households in Quarendon is chosen. Use a suitable approximation to find the probability that more than 84 are satisfied with the speed of their wificonnection. 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Mark scheme: 4(i) a+b = 10, 0 < a,b < 10 A1 Correct unsimplified expression 0.284 B1 CAO 3 Question Answer Marks Guidance 4(ii) 0.66 150 99 np = × = ( ) 0.66 1 0.66 150 33.66 npq = × − × = B1 Accept evaluated or unsimplified µ, σ2 numerical expressions, condone 33.66 5.8017 or 5.802 σ = = CAO P(X > 84) = P 84.5 99 33.66 Z −   >     M1 ± Standardise, 99 33.66 x their their − , condone σ2, x a value M1 84.5 or 83.5 used in their standardisation formula (= P( ) 2.499 Z > − ) M1 Correct final area 0.994 A1 Final answer (accept 0.9938) SC if no standardisation formula seen, B2 P(Z > -2.499) = 0.994 5

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6 The heights, in metres, of fir trees in a large forest have a normal distribution with mean 40 and standard deviation 8. (i) Find the probability that a fir tree chosen at random in this forest has a height less than 45 metres. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that a fir tree chosen at random in this forest has a height within 5 metres of the mean. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ In another forest, the heights of another type of fir tree are modelled by a normal distribution. A scientist measures the heights of 500 randomly chosen trees of this type. He finds that 48 trees are less than 10 m high and 76 trees are more than 24 m high. (iii) Find the mean and standard deviation of the heights of trees of this type. 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Mark scheme: 6(i) P(X < 45) = P 45 40 8 Z −   <     = P(Z < 0.625) M1 ± Standardise, no continuity correction, σ2 or σ , formula must be seen 0.734(0) A1 CAO 2 6(ii) ( ) ( ) ( ) 1 2 1 2 1 i i − − = − = 2((i) – 0.5) M1 Use result of part (i) or recalculated to find area OE 0.468 A1ft 0 < FT from (i) < 1 or correct. 2 6(iii) P(X < 10) = 48/500 = 0.096 z = –1.305 B1 z = ± 1.305 P(X > 24) = 76/500 = 0.152 z = 1.028 B1 z = ± 1.028 10 1.305 µ σ − = − 24 1.028 µ σ − = M1 Form 1 equation using 10 or 24 with ߤ, ߪ, ݖ–value. Allow continuity correction, not ߪଶ, √ߪ 14 = 2.333σ M1 OE Solve two equations in σ and µ to form equation in one variable [ ] [ ] 6. 00 , 17.8 3 σ µ = = A1 CAO, WWW 5

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7 (i) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that all three Os are together and both Ts are together. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that the Ts are not together. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the probability that a randomly chosen arrangement of the 9 letters of the word TOADSTOOL has a T at the beginning and a T at the end. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) Five letters are selected from the 9 letters of the word TOADSTOOL. Find the number of different selections if the five letters include at least 2 Os and at least 1 T. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) B1 Evaluated 1 7(ii) Total no of arrangements: 9! 2!3! 30240 = B1 Accept unevaluated No with Ts together = 8! 3! 6720 = B1 Accept unevaluated With Ts not together: 30 240 – 6720 M1 correct or 9! 8!, , integers 1 m n m n − > or their identified total – their identified Ts together 23 520 A1 CAO Alternative method for question 7(ii) 7! 8 7 3! 2 × × B1 7! × (k > 0) in numerator, cannot be implied by 7P2, etc. B1 3! × (k > 0) in denominator M1 7! 3! their their × 8C2 or 8P2 23 520 A1 CAO 4 Question Answer Marks Guidance 7(iii) Number of arrangements = 7! 3! Probability = 7! 840 3! 9! 30240 3!2! their their = M1 identified number of arrangements with T at ends identified total number of arrangements their their 7! , integers 1 9! m or m n n > 1 36 or 0.0278 A1 Final answer 2 7(iv) OOT_ _ 4C2 = 6 OOTT_ 4C1 = 4 OOOT_ 4C1 = 4 OOOTT = 1 M1 4Cx seen alone or 4Cx x k ≥1, k an integer, 0< x <4 A1 4C2 x k, k = 1 oe or 4C1 x m, m = 1 oe alone M1 Add 3 or 4 identified correct scenarios only, accept unsimplified (Total) = 15 A1 CAO, WWW Only dependent on 2nd M mark 4

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