Mathematics 9709/53 May/June 2024: question paper and mark scheme

Mark scheme

1(a) B1 Table with correct X values and at least one correct probability x 2 3 4 5 6 associated with the correct X value. Values need not be in order, lines may not be drawn, may be 1 4 10 12 9 P(X = x) vertical, X and P(X) may be omitted. 36 36 36 36 36 Condone any additional X values if probability stated as 0. 1 1 5 1 1 B1 Three other probabilities associated with correct x values, need not 36 9 18 3 4 be in table, accept unsimplified. B1 Five correct probabilities linked with correct outcomes, may not be Decimal equivalent 3sf: in table. 0.0278, 0.111, 0.278, 0.333, 0.25 Decimals correct to at least 3sf. SC B1 for five probabilities summing to 1 placed in a probability distribution table with the correct x values. 3 1(b) 1 × 2 + 4 × 3 + 10 × 4 + 12 × 5 + 9 × 6 M1 Accept unsimplified expression or sum of fractions seen. May be E(X) = = calculated in variance. 36 FT their table with five probabilities summing to 2 12 40 60 54 14 0.999 ⩽ total ⩽ 1 (0 < p < 1). + + + + = or 4.67 36 36 36 36 36 3 1 × 22 + 4 × 32 + 10 × 42 + 12 × 52 + 9 × 62 14 2 M1 Appropriate variance formula using their (E(X))2 value. Var(X) = − FT their table with 4 or more probabilities. 36 3 (0 < p < 1) which need not sum to 1. Note: If table is correct, then 196 14 2 824 206 or or 22.89 − or 21.78 or implies M1. 36 9 9 3 824 196 10 A1 1 [= − = 22.89 − 21.78] = 1 , 1.11[1] , 1.1 36 9 9 9 3 1(c) 10 9 M1 their P ( 4 ) + their P ( 6 ) + , all probabilities (0 < p < 1). P(X even | X > 3) = 36 36 their P ( 4 ) + P ( 5 ) + P ( 6 ) 31 their 19 36 If sample space seen in any part of the question, then M1 . their 31 19 A1 0.613 = 31 2

Mark scheme

2(a) 1.93 − 1.64 M1 Using ± standardisation formula with 1.93, 1.64 and 0.25 PZ > = P(Z > 1.16) 0.25 substituted, not σ 2 , not σ , no continuity correction. 1 – 0.8770 M1 Appropriate area Φ resulting from a standardisation, from final process, must be probability. Note: the appropriate probability answer implies this M1. 0.123 A1 0.123 ⩽ p < 0.12303 If M0 M0, SC B1 if no standardisation shown. 3 2(b) 1.56 − μ B1 −0.441 < z1 < −0.439 or = −0.44 0.439 < z1 < 0.441 seen. σ 1.86 − μ B1 z2 = 0.674 or z2 = – 0.674 seen, CAO, critical value. = 0.674 σ M1 Use of the ±standardisation formula once with μ, σ equating to a z- value (not 0.33, 0.67. 0.25, 0.75, 0.6293, 0.5987, 0.7486, 0.7734, (1 – 0.44), (1 – 0.674)). Condone continuity correct ± 0.005, not σ 2, σ . Solve, obtaining values for μ and σ M1 Solve two equations in μ and σ using the elimination method, substitution method or other appropriate approach to obtain values μ = 1.68, σ = 0.269 for both μ and σ. A1 AWRT μ = 1.68, σ = 0.269. If one or both of the M marks have not been awarded, SC B1 for both correct. 5

Mark scheme

3(a) B1 Correct structure and probabilities for Box A branches. Box A Box B B1 Completely correct structure and one correct probability for a Box B branch including label for G or Y. B1 Completely correct structure and second correct probability on a Box B branch including label G or Y. B1 Completely correct structure and remaining two probabilities correct on Box B branches, including labels for G or Y. SC B1 if correct shape diagram but only four correct algebraic probs for GG, GY, YG and YY. 4 3(b) 6 5 3 1+ x M1 P(GG) + P(YY) = P(same colour) = × + × 9 5+ x 9 5+ x 6 3 5 3 6 1+ x or × their + or × their 9 9 5+ x 9 9 5+ x 6 5 3 1+ x 8 M1 15 ( x + 11) = 24 ( x + 5 ) OE × + × = and arrange as a linear equation 9 5 + x 9 5 + x 15 8 Accept sum of their products equated to and rearranged to form 15 a linear equation. Solve: x = 5 A1 3

Mark scheme

4(a) B1 Correct stem, ignore extra values (not in reverse, not split). Penguins Dolphins If a split stem-and-leaf plot is used (i.e. stem values are repeated), the remaining B marks are available. 9 5 3 6 8 5 5 4 2 4 1 3 8 9 9 B1 Correct Penguins labelled on left, leaves in order from right to left and lined up vertically (less than halfway to next column), no 9 8 6 0 5 0 1 4 6 6 commas or other punctuation. 8 6 1 6 0 1 4 B1 Correct Dolphins labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas 2 7 1 or other punctuation. If the correct data for Penguins & Dolphins is transposed, treat as a Key: 2 | 4 | 1 means 42 seconds for Penguins and 41 seconds for single error in Penguins and condone in Dolphins. Dolphins B1 Correct key for their diagram, need both clubs labelled and ‘sec’ or ‘s’ stated at least once here, or in leaf headings or title. If two separate diagrams drawn: SC B1 if both keys meet these criteria. 4 4(b) For Dolphins, median is 51 B1 Plotted on box. LQ = 48, UQ = 60 B1 Plotted on box. Correct end points for whiskers and diagram labelled Dolphins B1 Correct end points of whiskers (36 and 71). Whiskers not through box, not drawn at corners of boxes, diagram labelled. 3 4(c) Dolphins have more consistent times than Penguins B1 Reason given in context. or Can be reference to either the central tendency or spread. Penguins are faster (have faster times) than Dolphins 1

Mark scheme

5(a) ( 0.35 )4 × 0.65 = 0.00975 B1 AWRT 1 5(b) ( 0.35) × ( 0.65) × 4 M1 ( 0.35) × ( 0.65) × k , where k is an integer. 3 2 3 2 1 ⩽ k ⩽ 5 no + or − = 0.0725 A1 2 5(c) Method 1 [1 – P(5, 6, 7) =] M1 One term 7Cx ( p ) x (1 − p )7 − x , 0 < p < 1, 0 < x < 7. 1 – (7C5 0.655 0.352 + 7C6 0.6560.351+ 0.657) [= 1 – (0.29848 + 0.18478 + 0.049022)] A1 Correct expression, accept unsimplified, no terms omitted leading to final answer = 0.468 B1 AWRT Method 2 [P(0, 1, 2, 3, 4) =] (M1) One term 7Cx ( p ) x (1 − p )7 − x , 0 < p < 1, 0 < x < 7. 0.357 +7C1 0.651 0.356 +7C2 0.652 0.355 +7C3 0.653 0.354 +7C4 0.654 0.353 [0.00064 + 0.00836 + 0.04660 + 0.14424 + 0.26787] (A1) Correct expression, accept unsimplified, no terms omitted leading to final answer. = 0.468 (B1) AWRT 3 5(d) [Mean =] 84 ×0.65 = 54.6 B1 54.6 and 19.11 seen, allow unsimplified. May be seen in the [Var =] 84 ×0.65 × 0.35 = 19.11 standardisation formula 7 39 ([σ = ] 4.371 or implies correct variance). 10 Incorrect notation is penalised, but condone use of values in standardisation formula. 50.5 − 54.6 M1 Substituting their µ and their positive σ into the ± standardising P(X > 50) = P Z > 19.11 formula (any number), not their σ2, or their σ . M1 Use continuity correction 49.5 or 50.5 in their standardisation formula. ±4.1 4.1 Note: or ± seen gains M2 BOD. 19.11 4.371 P ( Z > − 0.9379) = Φ ( 0.9379 ) M1 Appropriate area Φ, from final process, must be a probability. 0.826 A1 0.8255 < p ≤ 0.826 5

Mark scheme

6(a) 9! B1 2!3! = 30240 1 6(b) Method 1: Number of arrangements with E at each end – Number of arrangements with E at each end and the three Rs together 7! B1 7! − 5! = − e , 7P4 – e, e a positive integer. 3! 3! M1 5! f − , f > 120, r = 1, 2 r! 720 A1 If no marks scored SC B1 for 840 – 120 = 720. Method 2: Number of arrangements with E at each end and no Rs together + No of arrangements with E at each end and two Rs together 5 C3 × 4! + 4C1 × 5! (B1) 5 P3 5 One of 5C3 × 4!, × 4! , 4C1 × 5! or 5P2 ×4! seen. P3 3! or × 4! + 5P2 ×4! 3! (M1) a × 4! + b × 5! where a and b are integers between 1 and 10 inclusive, or c × 4! + d × 4! where c and d are integers between 1 and 20 inclusive. 240 + 480 = 720 (A1) 3 6(c) Method 1 Group of 5 B1 Correct no of ways for two correct identified scenarios other than three Rs two Es. 3 Rs 2 Es =1 3 Rs 1 E = 2C1 × 4C1 =8 3 Rs 0 Es = 4C2 =6 Group of 4 3 Rs 1 E = 2C1 =2 3 Rs 0 Es = 4C1 =4 [Total =] 21 M1 No of ways for five correct identified scenarios added or correct. [Number of ways of splitting into the two groups =] 9C5 (= 126) M1 Accept evaluated, accept 9C4. seen as a denominator 21 1 A1 Probability = = (0.167) 126 6 6(c) Method 2 3Rs in Group of 5 = 6C2 = 15 (B1) One correct case evaluated accurately and linked with correct scenario. 3Rs in Group of 4 = 6C1 =6 [Total =] 21 (M1) No of ways for two correct scenarios added or correct. [Number of ways of splitting into the two groups =] 9C5 (= 126) (M1) Accept evaluated, accept 9C4. seen as a denominator 21 1 (A1) Probability = = (0.167) 126 6 Method 3: Considering the possible positions of R within the groups 3Rs in Group of 5 (B1) For one correct product unsimplified and linked with correct 5 4 3 15 scenario. × × = 9 8 7 126 (M1) For second correct product. 3Rs in Group of 4 4 3 2 6 × × = 9 8 7 126 15 6 (M1) For adding probabilities of two correct scenarios or correct. + 126 126 21 1 (A1) Probability = = (0.167) 126 6 6(c) Method 4: Probability method Group of 5 (B1) Two correct probabilities linked with correct scenarios, accept 3 2 1 2 1 5! 1 unsimplified. 2Es × × × × × = 9 8 7 6 5 3!2! 126 (M1) Four probabilities with denominators including a factor of 9 × 8 × 7 × 6 × n, where n is 1 or 5. 4 3 2 1 2 5! 8 1E × × × × × = 9 8 7 6 5 3! 126 3 2 1 4 3 5! 6 0E × × × × × = 9 8 7 6 5 3!2! 126 Group of 4 3 2 1 6 4! 6 × × × × = 9 8 7 6 3! 126 1+ 8 + 6 + 6 (M1) Probabilities of four correct scenarios added or correct. 126 21 1 (A1) Probability = = (0.167) 126 6 4