Mathematics 9709/55 October/November 2025: question paper and mark scheme

Mark scheme

1(a) 5 6 1 15625 B1 = 0.0558, 6 6 279936 1 1(b) 2 3 5 1 5 1 5 1 4 M1 (1 − p )2 p + (1 − p )3 p + (1 − p )4 p + + 6 6 6 6 6 6 Or Or 2 5 5 5 p 2 − p5 0 < p <1 − 6 6 2275 A1 = , 0.293 7776 2

Mark scheme

2(a) 3 B1 May be seen used in formulae. [ 4 p + 0.4 = 1,] p = 0.15 , 20 [E(X) = ] 1× 0.15 + 2 × 0.4 + 3 × 0.3 + 4 × 0.15 M1 Accept un-simplified numerical expression. May be calculated in 49 variance. [ = 0.15 + 0.8 + 0.9 + 0.6 = 2.45, ] 1× p + 2 × 0.4 + 3 × 2 p + 4 × p, 0 < p < 1 , ∑p = 1. 20 [Var (X) =] 0.15 × 12 + 0.4 × 22 + 0.3 × 32 + 0.15 × 42 − (their 2.45) 2 M1 Appropriate variance formula using their (E(X))2 value FT their p, acceptable at the bold partially evaluated stage. [0.15 × 1 + 0.4 × 4 + 0.3 × 9 + 0.15 × 16 − 6.0025] Must be a numerical expression, [6.85 – 6.0025] condone ∑p ≠ 1 . Accept 0.15 + 1.6 + 2.7 + 2.4 − 2.452 or 6.0025 . 339 A1 If either or both M marks not awarded, SCB1 for correct answer = 0.8475, WWW. 400 4 2(b) Method 1 [P(X < 8) = 1 – (P(8, 9, 10) =] M1 One term of form 10 Cx ( p )x (1 − p )10− x . 1 – {10 C8 ( 0.4) ( 0.6) + 10 C9 ( 0.4) ( 0.6) + 10 C10 ( 0.4) } 8 2 9 1 10 With 0 < p < 1, x ≠ 0 or 10. [= 1 – (0.0106168 + 0.00157286 + 0.00010486)] A1 Correct un-simplified expression, no terms omitted, leading to final answer. Condone omission of final bracket ‘}’ . = 0.988 B1 0.9875 < p ⩽ 0.988. Method 2 [P(X < 8) = P(0, 1, 2, 3, 4, 5, 6, 7) =] M1 One term of form 10 Cx ( p )x (1 − p )10− x . 0 ( 0.6 ) + 10 C1 ( 0.4) ( 0.6) + 10 C2 ( 0.4) ( 0.6) + 10 C3 10 C 10 1 9 2 8 With 0 < p < 1, x ≠ 0 or 10. ( 0.4)3 ( 0.6)7 + 10C4 ( 0.4)4 ( 0.6)6 + 10C5 ( 0.4)5 ( 0.6)5 + 10C6 A1 Correct un-simplified expression, no more than 5 ‘middle’ terms ( 0.4)6 ( 0.6)4 + 10C7 ( 0.4)7 ( 0.6)3 omitted, leading to final answer. = 0.988 B1 0.9875 < p ⩽ 0.988. 3

Mark scheme

3(a) Tree diagram with all 2, 4, 8 branches B1 Structure, including labelling H, T, R & Bs and 4 or more correct probabilities. 1 1 8 7 6 9 7 7 8 6 6 9 6 9 B1 Remaining probabilities correctly placed. , , , , , , , , , , , , 2 2 15 15 15 15 14 14 14 14 15 15 15 15 2 3(b) 1 7 6 1 9 9 M1 OE. × × + × × FT P(HBB) + P(TBB) from incorrect tree diagram 2 15 14 2 15 15 or 1 a a −1 1 b b × × + × × , a =8, 7 b = 6, 9 2 15 14 2 15 15 7 A1 = , 0.28 25 2 3(c) 7 18 B1 18 [P(1R or 2R) = 1 – P(BB) = 1 − =] , 0.72 FT 1 – their 3(b) or seen. 25 25 25 1 8 1 7 8 6 M1 FT from incorrect tree diagram P(X ∩ 1R or 2R) = × ×1 + × × = or 2 15 2 15 14 15 1 a 1 15 − a a × ×1 + × × 2 15 2 15 14 1 a a − 1 15 − a 1 a − 1 a or × × + + × × 2 15 14 14 2 15 14 a = 8, 7. 6 M1 their P ( X 1R or 2R ) P ( X 1R or 2R ) 15 Identified values substituted into or P(X|1R or 2R) = = their P (1R or 2R ) P (1R or 2R ) 18 correct. 25 5 A1 = , 0.556 9 4

Mark scheme

4(a) Pelicans Swans B1 Correct stem, ignore extra values (not in reverse, not split) 6 15 8 If a split stem-and-leaf plot is used (i.e. stem values are repeated) 7 5 4 0 16 2 3 5 the remaining B marks are available. 8 3 1 0 17 0 0 4 4 4 8 7 6 5 4 2 2 18 0 1 2 3 B1 Correct Pelicans labelled on left, leaves in order from right to left 19 1 and lined up vertically no commas or other punctuation. Key: 1|17|8 represents 171 cm for Pelicans and 178 cm for Swans B1 Correct Swans labelled on same diagram, leaves in order from right to left and lined up vertically, no commas or other punctuation. Penalise each error only once in question. E.g. commas in both sets of data. Missing or incorrect leaves can be penalised in both cases. B1 Correct key, for their diagram, need both team names and ‘cm’ at least once here, or in leaf headings or a suitable title. If 2 separate diagram drawn max marks: B1 if both stems correct, B1 if Pelicans correct to the left of the stem, B1 if both keys correct including ‘cm’ and team name. 4 4(b) Median = 173 (cm) B1 Accept Q2, must be identified. [IQR =] 184 – 165 M1 182 ⩽ UQ ⩽ 185 − 164 ⩽ LQ ⩽ 167 Implied if both quartile values are stated and appropriate IQR calculated accurately. = 19 [cm] A1 WWW. If M0 scored, SC1 for 19 WWW. 3 4(c) Box-and-whisker plot on provided grid B1 All 5 key values for Swans plotted accurately in standard format using a linear scale with 3 identified values. Labelled Swans. Swans 158 165 174 181 191 Vertical lines may be omitted at both limits. Pelican 156 165 173 184 187 Condone freehand lines Check accuracy in the middle of the vertical line using daylight B1FT All 5 key values for Pelicans, FT from 4b or correct, plotted rule accurately in standard format using a linear scale with 3 identified values. Labelled Pelicans. Vertical lines may be omitted at both limits B1 Whiskers not through box for both, not drawn at corners of boxes, single linear scale for the diagram and labelled “heights” (OE) and ‘cm’ . • If only one plot is labelled, assume other plot is as anticipated and mark as in scheme. • If neither of 2 plots is labelled, SCB1 for 1 fully ‘correct’ set of values plotted. • If no scale or incorrect linear scale stated, SCB1 if both correct relative to each other. 3 4(d) Swans are taller or Swans’ heights more consistent. B1 Appropriate comment about heights. If they say Pelicans’ heights are more consistent, they need to refer to the Range 1

Mark scheme

5(a) 6S 8C 6 (= 28) M1 One product using 2 or 3 combinations 8 Cx × 5 Cy × 12 Cz with x + 5S 1H 8C × 5C (= 280) y + z = 6 (one of y or z may be 0) and linked to a correct identified 5 1 5S 1T 8C × 12 C (= 672) scenario. 5 1 4S 2H 8C × 5C (= 700) 4 2 4S 1H 1T 8C × 5 C × 12 C (= 4200) B1 2 correct identified outcomes, accept un-simplified. 4 1 1 3S 2H 1T 8C × 5 C × 12 C (= 6720) Identification can be implied by un-simplified expressions. 3 2 1 2nd and 3rd can be combined as 8 C5 × 17 C1 and this counts as 2 M1 Add values of 6 correct scenarios, no incorrect/repeated scenarios, scenarios for the second M1 no extra scenarios. Identification can be implied by un-simplified expressions. Total = 12600 A1 If either or both Ms not awarded, SCB1 for 12600 WWW. 4 5(b) Method 1 Total arrangements – arrangements with 3 hurdlers together [With 3 hurdlers not together ] 8! − (6!×3!) B1 6! × 3! seen as a term added or subtracted. M1 8! − ( 6! × k ) , k ≥ 1 , k is an integer. = 36000 A1 Method 2 Summing no of ways with 2 hurdlers together or all separate 6C3 × 3! × 5! + 6C 2 × 2 × 3! × 5! B1 6Cx × 5! or 6Px × 5! seen, x = 2 or 3 seen. [ = 14400 + 21600] M1 ( 6C3 × 5! + 6C 2 × 2 × 5!) ×k , or 6P3 × 5!+ 6P2 × 5! ×k k ≥ 1, k is an integer. = 36000 A1 3 5(c) Method 1 5! × 3! × 4 B1 5! seen in a single term method or as a common factor in a multi term method, no dividing. M1 3! seen in a single term method or as a common factor in a multi term method, no dividing. 2880 A1 3

Mark scheme

6(a) 55.1 − 46.3 M1 Use of ± standardisation formula with 55.1, 46.3 and 6.4 [P(X > 55.1)] = P( Z > ) [= P(Z > 1.375)] 6.4 substituted appropriately, no continuity correction, allow ( ) σ 2 6.42 or σ 6.4 . ( ) [= 1 − Φ (1.375 ) ] M1 Calculating the appropriate probability area (leading to the final answer, expect <0.5). = 1 − 0.9155 = 0.0845 A1 0.0845 ⩽ z ⩽ 0.0846 If one or both M marks not awarded, SCB1 for 0.0845 ⩽ z ⩽ 0.0846. 3 6(b) 48.6 − 39.8 B1 ±1.282 seen, CAO, critical value. [P(X > 4.86) = P Z > = 0.10, σ 48.6 − 39.8 PZ < = 0.90] σ 48.6 − 39.8 M1 ± standardisation formula with 48.6, 39.8 and σ equating to a z- = 1.282 value (not 0.10, 0.90, 0.5398,0.4602, 0.8159, 0.1841, 1 – their z- σ value), not σ 2 , condone continuity correction of ±0.05. 8.8 A1 AWRT. Not dependent on B mark being awarded. Condone 6.87. σ = 1.282 = 6.86 3 6(c) [ Mean = 0.1 × 150 =]15 B1 15 and 13.5 seen, allow un-simplified. May be seen in standardisation formula. [ Variance = 150 × 0.1 × 0.9 =]13.5 3 6 ( σ = 13.5, 3.674 ≤ σ < 3.675 implies correct variance. 2 Withhold mark if variance clearly identified as standard deviation, ( ) condone N 15, 13.5 if standardisation formula correct or variance/standard deviation correctly stated as well. 19.5 − 15 M1 Using 20 (with or without a cc) and their mean and sd found using [P(X < 20) =] P Z < [= P(Z < 1.2247)] 13.5 150 in ± standardisation formula, not σ 2 . M1 Using continuity correction 19.5 or 20.5 in a standardisation formula. ±4.5 ±4.5 Note: or gains M2. 13.5 3.674 Φ (1.2247 ) = 0.8897 M1 Appropriate area Φ, from final process. Must be a probability. Note: correct final answer implies this M1. = 0.890 A1 AWRT. 5