Cambridge A Level Mathematics 9709 — 2019 May/June Paper 6 · Variant 2

1 Two ordinary fair dice are thrown and the numbers obtained are noted. Event S is ‘The sum of the numbers is even’. Event T is ‘The sum of the numbers is either less than 6 or a multiple of 4 or both’. Showing your working, determine whether the events S and T are independent. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 P(S) = 1 2 B1 P(T) = 16 36 4 9       B1 P(S ∩ T) = 10 36 5 18       M1 P(S ∩ T) found by multiplication scores M0 M1 awarded if their value is identifiable in their sample space diagram or Venn diagram or list of terms or probability distribution table (oe) P(S) P(T) ≠ P(S ∩ T) so not independent A1 8/36, 10/36 P(S) × P(T) and P(S ∩ T) seen in workings and correct conclusion stated, www Alternative method for question 1 P(S) = 1 2 B1 P(T) = 16 36 4 9       B1 P(S ∩ T) = 10 36 5 18       M1 P(S ∩ T) found by multiplication scores M0 M1 awarded if their value is identifiable in their sample space diagram or Venn diagram or list of terms or probability distribution table (oe) P(S | T) = 10 16 or P(T | S) = 10 18 P(S|T) ≠ P(S) or P(T | S) ≠ P(T) so not independent A1 Either 18/36, 10/16,P(S) and P(S |T) seen in workings and correct conclusion stated, www Or 16/36, 10/18, P(T) and P(T | S) seen in workings and correct conclusion stated, www 4

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2 The volume of ink in a certain type of ink cartridge has a normal distribution with mean 30 ml and standard deviation 1.5 ml. People in an office use a total of 8 cartridges of this ink per month. Find the expected number of cartridges per month that contain less than 28.9 ml of this ink. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 P( < 28.9) = P 28.9 30 1.5 −   <     z B1 = P(z < –0.733) = 1 – 0.7682 M1 Appropriate area Φ from standardisation formula P(z <….) in final probability solution, Must be a probability, e.g. 1 – 0.622 is M0 = 0.2318 A1 Correct final probability rounding to 0.232. (Only requires M1 not B1 to be awarded Number of cartridges is their 0.2318 × 8 = 1.85, so 2 (Also accept 1 but not both) B1 FT using their 4 SF (or better) value, ans. rounded or truncated to integer, no approximation indicated. 4

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3 The probability that Janice will buy an item online in any week is 0.35. Janice does not buy more than one item online in any week. (i) Find the probability that, in a 10-week period, Janice buys at most 7 items online. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The probability that Janice buys at least one item online in a period of n weeks is greater than 0.99. Find the smallest possible value of n. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) P(at most 7) = 1 – P(8, 9, 10) = 1 – 10C8(0.35)8(0.65)2 – 10C9(0.35)9(0.65)1 – (0.35)10 M1 Binomial term of form 10Cxpx(1 – p)10 – x 0 < p < 1 any p, x ≠ 10,0 [= 1 – 0.004281 – 0.0005123 – 0.00002759] A1 Correct unsimplified (or individual terms evaluated) answer seen Condone 1 – A + B + C leading to correct solution = 0.995 B1 B1 not dependent on previous marks. Alternative method for question 3(i) P(at most 7) = P(0,1,2,3,4,5,6,7) M1 Binomial term of form 10Cxpx(1 – p)10 – x 0 < p < 1 any p, x ≠ 10,0 = (0.65)10 + 10C1(0.35)1(0.65)9+…+ 10C7(0.35)7(0.65)3 A1 Correct unsimplified answer or individual terms evaluated seen = 0.995 B1 3 3(ii) 1 – (0.65)n > 0.99 0.01 > (0.65)n M1 Equation or inequality with (0.65)n and 0.01 or (0.35)n and 0.99 only (Note 1 – 0.99 is equivalent to 0.01 etc.) n > 10.69 M1 Solving their a n = c, 0 < a,c < 1 using logs or Trial and Error If answer inappropriate, at least 2 trials are required for Trial and Error M mark smallest n = 11 A1 CAO 3

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4 It is known that 20% of male giant pandas in a certain area weigh more than 121 kg and 71.9% weigh more than 102 kg. Weights of male giant pandas in this area have a normal distribution. Find the mean and standard deviation of the weights of male giant pandas in this area. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 z = 0.842 = 121 µ σ −       so 0.842σ = 121 – µ B1 M1 One appropriate standardisation equation with a z-value, µ, σ and 121 or 102, condone continuity correction. Not 0.158, 0.42,… z = –0.58 = 102 µ σ −       so –0.58σ = 102 – µ B1 ± 0.58(0) seen but B0 if 1 ± 0.58 oe seen Solving M1 Correct algebraic elimination of µ or σ from their two simultaneous equations to form an equation in one variable, condone 1 numerical slip σ = 13.4 µ = 110 A1 If M0A0 scored (i.e. no algebraic elimination seen), SC B1 can be awarded for both answers correct Consistent use of σ 2 or √σ throughout apply MR penalty to A mark or SC B mark. 5

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5 Maryam has 7 sweets in a tin; 6 are toffees and 1 is a chocolate. She chooses one sweet at random and takes it out. Her friend adds 3 chocolates to the tin. Then Maryam takes another sweet at random out of the tin. (i) Draw a fully labelled tree diagram to illustrate this situation. [3] (ii) Draw up the probability distribution table for the number of toffees taken. 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(iii) Find the mean number of toffees taken. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) Find the probability that the first sweet taken is a chocolate, given that the second sweet taken is a toffee. 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Mark scheme: 5(i) T 5/9 T 6/7 4/9 C T 1/7 6/9 C 3/9 C B1 0.857 and 0.143) (Labelling must be logically…e.g. (T and T) or (T and Not T) would be acceptable) B1 Either of second top pair or bottom of branches labels and probs correct B1 Both second pairs of branches labels and probs correct. No additional / further branches. 3 5(ii) No of toffees taken (T) 0 1 2 prob 3 63 , 0.0476(2) 30 63 , 0.476(2) 30 63 , 0.476(2) B1 P(1) correct B1 P(0) or P(2) correct B1 FT Correct values in table, any additional values of T have stated probability of zero. For FT Σp = 1, 3 5(iii) E(X) = 90 63 (10 7 ) (1.43) B1 Not FT 1 Question Answer Marks Guidance 5(iv) P(1st C | 2nd T) = ( ) ( ) ∩ P C T P T = 1 6 6 7 9 63 1 6 6 5 36 7 9 7 9 63 × = × + × B1 P(C ∩ T) attempt seen as numerator of a fraction, consistent with their tree diagram or correct M1 Summing 2 appropriate two-factor probabilities, consistent with their tree diagram or correct seen anywhere A1 36 63 oe or correct unsimplifed expression seen as numerator or denominator of a fraction 1 6 oe A1 Final answer 4

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6 (i) Give one advantage and one disadvantage of using a box-and-whisker plot to represent a set of data. 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(ii) The times in minutes taken to run a marathon were recorded for a group of 13 marathon runners and were found to be as follows. 180 275 235 242 311 194 246 229 238 768 332 227 228 State which of the mean, mode or median is most suitable as a measure of central tendency for these times. Explain why the other measures are less suitable. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Another group of 33 people ran the same marathon and their times in minutes were as follows. 190 203 215 246 249 253 255 254 258 260 261 263 267 269 274 276 280 288 283 287 294 300 307 318 327 331 336 345 351 353 360 368 375 (a) On the grid below, draw a box-and-whisker plot to illustrate the times for these 33 people. [4] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Find the interquartile range of these times. 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Mark scheme: 6(i) Advantage: comment referring to spread or range or shape B1 Comments referring to quartiles, IQR, Range, median, shape, skewness, data distribution, spread score B1 Any comments with reference to mean or standard deviation or any other ‘disadvantage’ will score B0 Comments referring to ‘5-value plot’, comparison with another data set, overview or ease of drawing/plotting/reading require an appropriate advantage statement. Disadvantage: comment referring to limited data information provided B1 Comments referring to no individual data, no information about the number of values, unable to calculate mean, standard deviation, variance and mode score B1 Any comments with reference to median, shape or any other ‘advantage’ will score B0 Comments referring to ‘size of data set’ or ‘average’ require an appropriate disadvantage statement. Comments referring to outliers are ignored in all cases (as outliers are not in the syllabus content) unless supported by an appropriate advantage / disadvantage statement. If comments not clearly identified, assume first comment is the advantage. 2 Question Answer Marks Guidance 6(ii) Not mean as data skewed by one large value B1 Comment which identifies 768 (or ‘a very large number’) as the problem. Condone the use of ‘outlier’ Not mode as frequencies all the same B1 Comment which indicates that no mode exists (e.g. all the data is different, there is no repeated number, all the values are different) Median B1 Median identified as choice, dependent upon statements for mean and mode being given, even if incorrect or very general. SC: Mean is identified as most suitable Not mode as frequencies all the same SCB1 Comment which indicates that no mode exists Not median as not all values used SCB1 Comment which indicates limitation of median e.g. median is not in middle of range. 3 6(iii)(a) LQ = 256 or 256.5 Med = 280 UQ = 329 Min 190 max 375 150 200 250 300 350 400 time minutes B1 Median, UQ and LQ values seen, may not be identified or identified correctly. (Not read from box plot unless value stated) B1 FT Median and quartiles plotted in box on graph, linear scale B1 Correct end points, whiskers from ends of box but not through box, not at top or bottom of box B1 Uniform scale from 190 to 375 (need at least 3 linear identified points min) and labelled ‘time’ and ‘minutes’ (can be in title) No time axis or time axis with no scale attempt, Max B1B0B0B0 4 Question Answer Marks Guidance 6(iii)(b) IQR = their 329 – their 256 = 73 or 72.5 B1 FT Must follow through only from their stated values (condone if correct quartiles stated here), not reading from graph. 1

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7 (a) A group of 6 teenagers go boating. There are three boats available. One boat has room for 3 people, one has room for 2 people and one has room for 1 person. Find the number of different ways the group of 6 teenagers can be divided between the three boats. 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(b) Find the number of different 7-digit numbers which can be formed from the seven digits 2, 2, 3, 7, 7, 7, 8 in each of the following cases. (i) The odd digits are together and the even digits are together. 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(ii) The 2s are not together. 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Mark scheme: 7(a) M1 condone use of permutations, = 20 × 3 A1 Any correct method seen no addition/additional scenarios = 60 A1 Correct answer Alternative method for question 7(a) 6 6 3 2 1 3 2 1 P 6! 3! 2! P P P = × × × M1 6P6 / (nPn x k) with 3 ⩾ n > 1and 6 ⩾ k an integer ⩾ 1, not 6!/1 A1 Correct method with no additional terms = 60 A1 Correct answer 3 7(b)(i) 4! 3! 2 3! 2! × × M1 A single expression with either 4!/3! × k or 3!/2! × k, k a positive integer seen oe (condone 2 identical expressions being added) M1 Correctly multiplying their single expression by 2 or 2 identical expressions being added. = 24 A1 Correct answer 3 Question Answer Marks Guidance 7(b)(ii) Total no of arrangements = 7! 2!3! = 420 (A) B1 Accept unsimplified No with 2s together = 6! 3! = 120 (B) B1 Accept unsimplified With 2s not together: their (A) – their (B) M1 Subtraction indicated, possibly by their answer, no additional terms present = 300 ways A1 Exact value www Alternative method for question 7(b)(ii) 3 _ 7 _ 7 _ 7 _ 8 _ 5! 6 5 3! 2 × × B1 k x 5! in numerator, k a positive integer B1 m x 3! In denominator, m a positive integer M1 Their 5!/3! multiplied by 6C2 only (no additional terms) = 300 ways A1 Exact value www 4

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