Mathematics 9709/62 May/June 2019: question paper and mark scheme

Mark scheme

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

Mark scheme

2  28.9 − 30  B1 Using ± standardising formula, no continuity correction, not σ 2 or √σ, P( < 28.9) = P  z <   1.5  = P(z < –0.733) M1 Appropriate area Φ from standardisation formula P(z <….) in final = 1 – 0.7682 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 B1 FT using their 4 SF (or better) value, ans. rounded or truncated to integer, = 1.85, so 2 (Also accept 1 but not both) no approximation indicated. 4

Mark scheme

3(i) P(at most 7) = 1 – P(8, 9, 10) M1 Use of normal approximation M0 = 1 – 10C8(0.35)8(0.65)2 – 10C9(0.35)9(0.65)1 – (0.35)10 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 M1 Equation or inequality with (0.65)n and 0.01 or (0.35)n and 0.99 only 0.01 > (0.65)n (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

Mark scheme

4  121 − µ  B1 ± 0.842 seen but B0 if 1 ± 0.842 oe seen z = 0.842 =   so 0.842σ = 121 – µ  σ  M1 One appropriate standardisation equation with a z-value, µ, σ and 121 or 102, condone continuity correction. Not 0.158, 0.42,…  102 − µ  B1 ± 0.58(0) seen but B0 if 1 ± 0.58 oe seen z = –0.58 =   so –0.58σ = 102 – µ  σ  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

Mark scheme

5(i) T B1 First pair of branches labels and probs correct (6/7 and 1/7 or rounding to 5/9 0.857 and 0.143) T (Labelling must be logically…e.g. (T and T) or (T and Not T) would be acceptable) 6/7 4/9 C B1 Either of second top pair or bottom of branches labels and probs correct T 1/7 6/9 C 3/9 B1 Both second pairs of branches labels and probs correct. No additional / C further branches. 3 5(ii) B1 P(1) correct No of toffees 0 1 2 taken (T) B1 P(0) or P(2) correct 3 30 30 B1 FT Correct values in table, any additional values of T have stated , , , prob 63 63 63 probability of zero. For FT Σp = 1, 0.0476(2) 0.476(2) 0.476(2) 3 5(iii) 90 10 B1 Not FT E(X) = ( ) (1.43) 63 7 1 5(iv) 1 6 6 B1 P(C ∩ T) attempt seen as numerator of a fraction, consistent with their tree P (C ∩ T ) × diagram or correct P(1st C | 2nd T) = = 7 9 = 63 P(T ) 1 6 6 5 36 × + × M1 Summing 2 appropriate two-factor probabilities, consistent with their tree 7 9 7 9 63 diagram or correct seen anywhere A1 36 oe or correct unsimplifed expression seen as numerator or 63 denominator of a fraction 1 A1 Final answer oe 6 4

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 B1 Comments referring to no individual data, no information about the provided 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 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 B1 Median, UQ and LQ values seen, may not be identified or identified Min 190 max 375 correctly. (Not read from box plot unless value stated) B1 FT Median and quartiles plotted in box on graph, linear scale 150 200 250 300 350 400 B1 Correct end points, whiskers from ends of box but not through box, not at time minutes 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 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

Mark scheme

7(a) 6 C3 × 3C2 × 1C1 M1 6 Ca × 6–aCb × 6–a–bC6–a–b seen oe 6–a–bC6–a–b can be implied by 1 or omission, 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 P6 6! M1 P6 / (nPn x k) with 3 ⩾ n > 1and 6 ⩾ k an integer ⩾ 1, not 6!/1 = 3 2 1 P3 × P2 × P1 3!× 2! A1 Correct method with no additional terms = 60 A1 Correct answer 3 7(b)(i) 4! 3! M1 A single expression with either 4!/3! × k or 3!/2! × k, k a positive integer × ×2 3! 2! 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 7(b)(ii) 7! B1 Accept unsimplified Total no of arrangements = = 420 (A) 2!3! 6! B1 Accept unsimplified No with 2s together = = 120 (B) 3! 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 B1 k x 5! in numerator, k a positive integer × 3! 2 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