Cambridge A Level Mathematics 9709 — 2018 Feb/March Paper 6 · Variant 2

1 There are 900 students in a certain year-group. An identical puzzle is given to each student and the time taken, t minutes, to complete the puzzle is recorded. These times are summarised in the following frequency table. Time taken, t ≤3 3 < t ≤4 4 < t ≤5 5 < t ≤6 6 < t ≤8 8 < t ≤10 10 < t ≤14 t minutes Frequency 120 180 200 160 110 80 50 On the grid, draw a cumulative frequency graph to represent the data. Use your graph to estimate the median time taken by these students to complete the puzzle. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 t cf Med CUMULATIVE FREQUENCY 0 3 0 120 dian value: 4.8 (m 0 200 400 600 800 1000 0 CUMULATIVE FREQUENCY 4 5 6 300 500 6 minutes) 5 TIME, IN M 6 8 10 660 770 85 10 MINUTES 0 14 50 900 M A 15 M1 A1 joined b Linear s mins, all M1 450 seen (indepen A1 FT Correct graph at 4 t to plot cumulativ etween (3,y1) and cales starting at (0 l points correct; (a n in median attem ndent); (4.7 ⩽ m < 4.9) o t cf = 450 ve frequencies at u d (14,y2). Cf table 0,0) and axes labe allow straight line mpt on increasing C or FT from reading ucb and all points not required. elled cf and time i es or curves) CF graph g their increasing in

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2 A selection of 3 letters from the 8 letters of the word COLLIDER is made. (i) How many different selections of 3 letters can be made if there is exactly one L? [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) How many different selections of 3 letters can be made if there are no restrictions? [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) B1 1 2(ii) No L: 6C3 = 20 (1 L: 6C2 = 15) M1 Either 0L or 2L correct unsimplified 2 L: 6C1 = 6 M1 Summing the 3 correct scenarios Total = 41 A1 3

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3 Last Saturday, Sarah recorded the colour and type of 160 cars in a car park. All the cars that were not red or silver in colour were grouped together as ‘other’. Her results are shown in the following table. Type of car Saloon Hatchback Estate Red 20 40 12 Colour of car Silver 14 26 10 Other 6 24 8 (i) Find the probability that a randomly chosen car in the car park is a silver estate car. [1] ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that a randomly chosen car in the car park is a hatchback car. [1] ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the probability that a randomly chosen car in the car park is red, given that it is a hatchback car. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) One of the cars in the car park is chosen at random. Determine whether the events ‘the car is a hatchback car’ and ‘the car is red’ are independent, justifying your answer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) (10/160 =) 1/16, 0.0625 B1 OE 1 3(ii) (90/160) = 9/16, 0.5625 B1 OE 1 3(iii) P(red/hatchback) = P(red hatchback) / P(hatchback) = 40/160 / 90/160 M1 Appropriate probabilities in a fraction = 4/9 A1 OE Altn method: Direct from table M1 for 40/a or b/90, a ≠ 160 A1 for 40/90 oe 2 Question Answer Marks Guidance 3(iv) EITHER: P(red) × P(hatchback) = 72 90 160 160 × ≠ 40 160 (M1 Use correct approach with appropriate probabilities substituted Not independent A1) Numerical comparison and conclusion stated OR: P(red/hatchback) = 40/90 and 40 72 90 160 ≠ (M1 Use correct approach with appropriate probabilities substituted Not independent A1) Numerical comparison and conclusion stated 2

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5 A summary of n values of x gave the following information: Σ x −20 = 136, Σ x −20 2 = 2888. The mean of the n values of x is 24.25. (i) Find the value of n. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find Σx2. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 24.25n – 20n = 136 Or 136 20 24.25 n + = M1 Unsimplified correct equation n = 32 A1 2 5(ii) Using coded information: Variance = 2 2888 136 32 32   −    M1 unsimplified expression for variance = 72.1875 = 72.19 A1 accept answers 72.2 SOI Using uncoded information: Variance = 2 2 24.25 32 x ∑ − Equate with 72.1875 to give M1 Equate two expressions for variance and solve 2 21128 x ∑ = A1 4

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6 The digits 1, 3, 5, 6, 6, 6, 8 can be arranged to form many different 7-digit numbers. (i) How many of the 7-digit numbers have all the even digits together and all the odd digits together? [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) How many of the 7-digit numbers are even? [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) 3! × 4! 3! × 2 M1 4!/3! oe seen multiplied by integer > 1, no addition = 48 A1 3 6(ii) EITHER: Even = Total number of arrangements – Odd numbers = 7!/3! – 3× 6 5 4 3 2 1 3! × × × × × = (7!/3! – 6!/2!) = 840 −360 B1 7!/3! – B1 6!/2! OE = 480 B1 OR: No of arrangements ending in 8: 6! 3! B1 No. ending in 8 or no. ending in 6 correct unsimplified No ending in 6: 6!/2! B1 Both correct and added unsimplified Total: 6! 6!/ 2 120 360 480 3! + = + = B1 3

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7 The weights of packets of a certain type of biscuit are normally distributed with mean 400 grams and standard deviation 3 grams. (i) In a random sample of 6000 packets of this type of biscuit, 225 packets weighed more than 410 grams. Find the value of 3. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) In a random sample of 500 packets of this type of biscuit, how many packets would you expect to find with weights that are more than 1.5 standard deviations from the mean? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) P(X > 410) = 225/6000 = 0.0375 P 410 400 σ −   >     Z = 0.0375: 0.9625 z value = ±1.78 A1 z value: ± 1.78 10 1.78 σ = M1 (410-400)/σ = their z (must be a z value) 5.62 σ = A1 4 7(ii) We need P(Z < −1.5) and P(Z > 1.5) M1 Attempt at P(Z < −1.5) or P(Z > 1.5) 1 – Φ(1.5) seen ( ) ( ) Φ 1.5 1 Φ 1.5 − + − = ( ) 2 2Φ 1.5 − M1 Or equivalent expression with values =2 – 2 × 0.9332 = 0.1336 (0.134) A1 Correct to 3sf Number expected = 500 × 0.1336 = 66.8: 66 or 67 packets B1ft 0.1336 used or FT their 4sf probability times 500, (not 0.9625 or 0.0375) rounded or truncated 4

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8 The results of a survey at a certain large college show that the proportion of students who own a car is 14. (i) Five students at the college are chosen at random. Find the probability that at least four of these students own a car. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) For a random sample of n students at the college, the probability that at least one of the students owns a car is greater than 0.995. Find the least possible value of n. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) For a random sample of 160 students at the college, use a suitable approximate distribution to find the probability that fewer than 50 own a car. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(i) P(4) + P(5) = 4 1 5 0 5 5 4 5 1 3 1 3 C C 4 4 4 4       +             = 0.014648.. + 0.00097656.. M1 Add 2 correct unsimplified binomial terms = 0.0156 or 1 64 A1 3 8(ii) 1 −P(0) > 0.995: 0.75 0.005 n < M1 Equation or inequality involving 0.75n and 0.005 or 0.25n and 0.995 log0.75 log0.005 n < n > 18.4: M1 Attempt to solve their exponential equation using logs, or trial and error May be implied by their answer n = 19 A1 3 8(iii) p = 0.25, n = 160: mean = 160 x 0.25 (= 40) variance = 160 x 0.25 x 0.75 (=30) B1 Correct unsimplified mean and variance P(X < 50) = P 49.5 40 30 −   <     Z M1 Use standardisation formulae must include square root. M1 Use continuity correction ±0.5 (49.5 or 50.5) = P(Z < 1.734) = 0.959 A1 Correct final answer 4

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