Mathematics 9709/51 May/June 2025: question paper and mark scheme

Mark scheme

1 1.48 − μ B1 0.771 < z < 0.773 or −0.773 < z < −0.771 seen. P (Z < ) = 0.22 0.14 1.48 − μ M1 Use of the ±standardisation formula with µ, 1.48 and 0.14 and = − 0.772 equating to a z-value (not 0.78, 0.22, 0.5871, 0.7823, 0.228). 0.14 Condone σ2, σ and continuity correction ±0.005. μ = 1.59 A1 1.585 < µ ⩽ 1.59. If M0 scored SC B1 for correct answer WWW. 3

Mark scheme

2(a) Method 1 total arrangements with As together – arrangements with As together and Os together 7! M1 7! − 6! – k , where k is an integer ⩾ 1. 2! 2 n – 6! M1 n – 6!, where n is an integer > 720. 1800 A1 3 Method 2 arrangements of other 6 letters with the As together and then the Os placed 6×5 M1 5! × p, where p is an integer > 1. 5!× 2 M1 6×5 q× , q × 6C2, q × 6P2, q × 6 × 5, q an integer > 1. 2 1800 A1 3 2(b) Method 1 3 5 M1 3 a [P(X < 6) = ] 1 − ] 1 − a = 5, 6. 4 4 781 A1 0.7626 < p ⩽ 0.763. = 0.763, 1024 2 Method 2 2 3 1 1 3 1 3 1 3 1 3 4 M1 1 1 3 1 3 1 3 1 3 1 3 2 3 4 5 [ P(X < 6) = ] + + + + + + + + + . 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 781 A1 0.7626 < p ⩽ 0.763. = 0.763, 1024 2 2(c) M1 p 2 (1 − p ) × 5 , 0 < p < 1, p ≠ 1 – p. 2 4 4 1 3 405 × 5 = 4096 4 4 = 0.0989 A1 AWRT 2

Mark scheme

3(a) B1 Correct stem, ignore extra values (not in reverse, not split). Gulls Herons If a split stem-and-leaf plot is used (i.e. stem values are repeated), the remaining B marks are available. 9 7 B1 Correct Gulls labelled on left, leaves in order from right to left and 8 6 6 3 2 8 1 3 5 7 lined up vertically (less than halfway to next column), 8 7 2 9 2 3 5 7 9 9 no commas or other punctuation. 4 0 10 8 B1 Correct Herons labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas or other punctuation. Key: 7|9|5 means 9.7 seconds for Gulls and 9.5 seconds for Herons Penalise each error only once in question. E.g. commas in both sets of data If the correct data for Gulls & Herons is transposed, treat as a single error in Gulls and condone in Herons. 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. SC: If 2 separate diagrams drawn, max marks: B1 if both stems correct, B1 if Gulls is correct to the left of the stem B0 B1 if both keys correct including ‘sec’ or ‘s’ 4 3(b) Median = 8.8 (seconds) B1 Clearly identified, e.g. Q2, med, m.. [LQ 8.3, UQ 9.8] M1 9.7 ⩽ UQ ⩽ 10.0 – 8.2 ⩽ LQ ⩽ 8.6. 9.8 – 8.3 Implied if both quartile values are stated and the appropriate IQR is calculated accurately. 1.5 A1 WWW If M0 scored SCB1 for 1.5 WWW. 3 3(c) 175.0 + 30 × 8.4 175.0 + 252.0 M1 Mean = or 50 50 = 8.54 A1 If M0 scored, SCB1 for 8.54 WWW. 2 3(d) 1.38 =2(1823.0 + Σ y ) − their 427.0 2 2 M1 Substitute values into correct variance formula, any form. 50 50 Solve to find Σ y 2 DM1 Rearrange equation to obtain Σ y 2 . 2 Σ y2 427.0 1823.0 = 1.382 + their − . 50 50 50 Σ y 2 = 1918.8 A1 Mark final answer. If one or both M marks not awarded, SCB1 for 1918.8 seen as final answer WWW mark final answer. 3

Mark scheme

4(a) −1.2 1.2 M1 OE P( <Z < ) 2.5 2.5 Using ± standardisation formula, not σ 2 , not σ, no continuity correction Use of ±standardisation formula once with 32.4, 2.5 and either 31.2 or 33.6. No continuity correction, not σ 2 , not σ 1.2 1.2 Implied by either − or seen 2.5 2.5 [= Φ ( 0.48) + Φ ( −0.48) = 2Φ ( 0.48 ) − 1 ] M1 Calculating the correct probability area (leading to their final 2 × 0.6844 – 1 or 2 × (0.6844 – 0.5) or 0.6844 – 0.3156 probability). This may be implied by the correct or appropriate probability area. 0.3688 A1 Expected number = 0.3688 × 600 = 221.28 so 221 or 222 B1 FT FT their 4-figure probability to obtain a single integer answer. No approximation indicated, condone use of 3sf probability here if more accurate answer seen earlier. 4 4(b) Method 1 [P(2 ⩽ X < 8) = 1 – P(0, 1, 8, 9) = ] M1 One term of the form 9Cx ( p ) x (1 − p )9− x , 0 < p < 1, x ≠ 0 or 9 1 – (0.49 + 9C1 0.48 0.6 + 9C8 0.410.68 + 0.69) = A1 Correct un-simplified expression. Condone omission of last bracket only. If both brackets omitted in [1 – 0.000262144 – 0.00353894 – 0.060466176 – 0.010077696 =] un-simplified expression, allow recovery for final stated calculation of 1 – 0.07434… or better. 0.926 B1 0.925 < p ⩽ 0.926 from correct working. Method 2 [P(2 ⩽ X < 8) = P(2, 3, 4, 5, 6, 7) = ] M1 One term of the form 9Cx ( p ) x (1 − p )9− x , 0 < p < 1, x ≠ 0 or 9. 9 C2 0.47 0.62 +9C3 0.46 0.63 +9C4 0.45 0.64 +9C5 0.44 0.65 +9C6 0.43 A1 Correct un-simplified expression. 0.66 +9C7 0.42 0.67 0.926 B1 0.925 < p ⩽ 0.926 from correct working. 3 4(c) Mean = 80 ×0.6 = 48 B1 48 and 19.2 (CAO) seen, allow un-simplified. Variance = 80 × 0.6 × 0.4 = 19.2 May be in standardisation formula. (4.38178… to at least 4SF identified as σ implies correct variance). Do not condone clear incorrect identification. 50.5 − 48 M1 Substituting their 48 and their 19.2 into the ±standardising [P(X > 50 ) =] P( Z > ) formula (any number for 50.5), allow σ2 or √σ. 19.2 M1 Use continuity correction 49.5 or 50.5 in their standardisation formula. 2.5 2.5 Note: If no working ± or ± 4.382 seen gains B2. 19.2 [P( Z > 0.571 ) = 1 − Φ ( 0.571) = ] M1 Appropriate probability area, from final process, must be a 1 – 0.7160 = probability. May be implied by a sketch of the required probability area. Expect final answer < 0.5. 0.284 A1 AWRT 5

Mark scheme

5(a) 3G 2P 1D: 9C3 × 6C2 × 5C1 = 6300 B1 Correct outcome/value for either the 1st or 2nd scenario clearly identified, accept un-simplified, WWW. 4G 1P 1D: 9C4 × 6C1 × 5C1 = 3780 M1 2 correct outcomes/values obtained, accept un-simplified 5G 0P 1D: C5 [× C0] × C1 = 630 9 6 5 M1 Sum of 3 correct scenarios, may be identified by un-simplified Condone 5C1 =5 and 6C1 = 6 expression. Total: 10710 A1 CAO If one or both M marks not awarded, SCB1 for 10710 WWW. 4 5(b) Ways of selecting 1st band: 9C3 × 6C1 × 5C1 = 2520 M1 9 C3 × 6C1 × 5C1 or 9C3 × 6 × 5 seen, Ways of selecting 2nd band: 6C3 × 5C1 × 4C1 = 400 condone × 3 or ×3! . Ways of selecting 3rd band: 1 × 4C1 × 3C1 = 12 M1 their 2520 × their 400 × their 12 seen, accept un-simplified, [Total number of ways =] 2520×400 ×12 = condone × 3 or ×3! . 12096000 A1 Condone 12100000. If one or both M marks not awarded, SCB1 for 12096000 (CAO) WWW. 3

Mark scheme

6(a) Method 1 6 5 4 3 4! 3 M1 AG. P(X = 2) = × × × × = 10 9 8 7 2!2! 7 6 5 4 3 4 3 6 5 × × × × k or × × × × k for k an integer, k > 1. 10 9 8 7 10 9 8 7 A1 4 4! C2 may be seen for . 2!2! Method 2 6 C2 × 4 C2 15 × 6 3 M1 AG. = = 6 10C4 210 7 C2 × 4 C2 seen as the numerator of a fraction. Condone use of permutations if used consistently. A1 2 6(b) B1 Table with correct values of x and at least one further non-zero x 0 1 2 3 4 probability correct. Condone extra x values if probability stated as 0. P(X = x) 1 4 3 8 1 , , , , B1 Third probability correct. 210 35 7 21 14 Accept probabilities not in table if clearly identified. 0.00476 0.114 0.381 0.0714 B1 Fourth probability correct. Accept probabilities not in table if clearly identified. B1 Fifth probability correct. Accept probabilities not in table if clearly identified. 4 SCB1 for 4 further non-zero probabilities adding to , 0.5712 if 7 B2 max scored. 4 6(c) 3 8 M1 3 8 + + their seen as the numerator of a fraction. [P(2B, 3B | 3B1R or 2B2R or 1B 3R) =] 7 21 7 21 4 3 8 + + 35 7 21 B1 FT 4 3 8 their + + their seen as the denominator of a fraction. 35 7 21 17 A1 Accept 0.87628…to at least 3SF. 170 85 = 21 = , , 0.876 97 194 97 105 3