Mathematics 9709/52 February/March 2025: question paper and mark scheme

Mark scheme

1(a) 1 1 4 1 3 1 2 1 1 9 3 B1 Order of coins must be consistent with question if not × × + × × + × × = = AG 3 4 5 3 4 5 3 4 5 60 20 identified. 1 1(b) x 0 1 2 3 B1 Table with correct values of x and at least two correct non- zero probabilities. P(X = x) 24 8 2 26 13 9 3 1 , , , , , , , 60 20 5 60 30 60 20 60 B1 One more correct non-zero probability linked with correct 0.4 0.433 0.15 0.0167 x value, need not be in table if clearly identified, accept unsimplified (total of 3 correct probabilities). B1 4 correct probabilities linked with the correct outcomes, may not be in table. Decimals correct to at least 3SF. SC1 for 4 or more probabilities summing to 1 placed in a probability distribution table. 3 1(c) (02 × 24+) 12 × 26 + 22 × 9 + 32 ×1 47 2 M1 Appropriate variance formula using (E(X))2 value. FT [Var(X) =] − their table with 3 or more probabilities (0 < p < 1) which 60 60 need not sum to 1, with an expression no more evaluated 2 1× 26 + 4 × 9 + 9 ×1 47 than in bold. FT acceptable at the bold partially evaluated = − 60 60 stage with their probabilities. 2051 A1 0.5695 < Var(X) ⩽ 0.570. = , 0.570 3600 2051 If M0 scored, SC1 for , 0.570 WWW 3600 Note: 0.57 without more accurate previous value penalised as 2SF. 2

Mark scheme

2(a) 297 B1 148500 [0.55 × 0.3 × 0.15 × 3! =]0.1485, Accept , condone 0.149. 2000 1000000 1 2(b) Method 1 [1 – P(10, 11, 12) = ] M1 One term of the form 12 Cx ( p )x (1 − p )12− x , 1 – {12 C10 0.5510 0.452 + 12 C11 0.5511 0.45 + 0.5512 } = [1 – (0.0338529 + 0.0075229 + 0.0007662) =] 0 < p < 1, x ≠ 0 or 12. A1 Correct unsimplified expression, no terms omitted leading to final answer. Condone omission of last bracket ‘}’ only. = 0.958 B1 0.9575 < p ≤ 0.958. Method 2 [P(0,1,2,3,4,5,6,7,8,9) =] M1 One term of the form 12 Cx ( p )x (1 − p )12− x , 0 < p < 1, x ≠ 0 0.4512 +12 C1 0.551 0.4511 + … + 12 C9 0.559 0.453 or 12. A1 Correct unsimplified expression, no more than 7 ‘middle’ terms omitted leading to final answer. = 0.958 B1 0.9575 < p ⩽ 0.958. 3 2(c) [Mean = 140 ×0.15 =] 21 B1 17 21 and 17.85 (or 17 ) seen, allow unsimplified. [Variance = 14 0 × 0.15 × 0.85 =] 17.85 20 May be in standardisation formula. ( σ = 17.85, 4.224926to at least 4SF implies correct variance). Withhold mark if variance clearly identified as standard deviation, condone N(21, 17.85 ) if standardisation formula correct or variance/standard deviation correctly stated as well. 24.5 − 21 M1 Substituting their µ and their σ into the ± standardisation P(X > 24) = P Z > formula (any number for 24.5), allow σ2 or √σ. 17.85 M1 Use continuity correction 23.5 or 24.5 in their standardisation formula. 3.5 3.5 Note: If no working ± or ± 4.225 seen gains 17.85 M2 BOD. [P( Z > 0.8284 ) = 1 − Φ ( 0.8284 ) ] M1 Appropriate area Φ, from final process, must be a 1 – 0.7961 probability. May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5. = 0.204 A1 Final answer AWRT. 5

Mark scheme

3(a) B1 At least 4 of 46, 106, 178, 226, 250 cumulative < 9.5 < 14.5 < 19.5 < 24.5 < 29.5 < 39.5 frequencies correct. May be implied by accurate plotting if scale suitable. CF 18 46 106 178 226 250 May be by data table. B1 Linearly scaled axes correctly labelled cumulative frequency (cf) (from 0 to 250), length (oe) and cm (from 5 to 39.5) – or a suitable title, with at least 3 values identified on each. Axes can be the other way round. Axes must be more than 50% of grid. M1 At least 4 points correctly plotted at class upper end points (9.5, 14.5, 19.5, 24.5, 29.5, 39.5) on scaled axes. A1 All points plotted correct, curve drawn (within tolerance) and joined to (4.5,0) and not going above 250 vertically within range. A0 if straight line segments used. 4 3(b) [250 × 0.62 = 155] M1 Clear indication of use of graph at cf 155 is required. Line drawn from 155 on cf axis to meet graph at l = 23 A1FT Must be an increasing cf graph. Expect an answer in the range 22.5 ⩽ l ⩽ 23.5 from correct graph. 2 3(c) Midpoints 7, 12, 17, 22, 27, 34.5 B1 At least 5 correct midpoints seen, may be unsimplified, may be in calculation, may be by data table. 7 ×18 + 12 × 28 + 17 × 60 + 22 × 72 + 27 × 48 + 34.5 × 24 M1 Correct unsimplified mean formula using their 6 Mean = midpoints (not upper bound, lower bound, upper limits, 250 lower limits, cw, fd, f or cf and must be within class) condone 1 error. 5190 If midpoints correct accept or 250 126 + 336 + 1020 + 1584 + 1296 + 828 . 250 = 20.76 A1 19 38 76 Accept 20 , 20 or 20 , not improper fraction 25 50 100 If M1 withheld, SC1 for 20.76 oe WWW. 3

Mark scheme

4(a) 8 7 6 336 1 8 C M1 1 outcome seen as the product of 3 fractions with P(WWW) = × × = , or 16 3 16 15 14 3360 10 C3 8, 7, 6 or 5, 4, 3 or 3, 2, 1 as numerators and 16, 15, 14 or 16, 16, 16 as denominators 5 4 3 60 1 5 C Or P(BBB) = × × = , or 16 3 16 15 14 3360 56 C3 1 outcome correct in terms of combinations. 3 2 1 6 1 3 C3 P(SSS) = × × = , or M1 Sum of 3 correct identified scenarios (may be identified by 16 15 14 3360 560 16 C3 correct unsimplified numerator values). 402 67 A1 0.1195 < p ⩽ 0.120. , ,0.120 3360 560 3 4(b) 3 cases to consider: WBW, WBB, WBS M1 One outcome seen as the product of 3 fractions with 8 5 7 840 1 correct numerators and n, (n - 1), (n - 2) only as P(WBW) = × × ×3 = , denominator, where 8 ⩽ n ⩽ 16 (condone omission of ×3 16 15 14 3360 4 or ×6) 8 5 4 480 5 P(WBB) = × × ×3 = , Or 16 15 14 3360 42 1 outcome correct in terms of combinations with m C 3 as 8 5 3 720 3 denominator where 8 ⩽ m ⩽ 16. P(WBS) = × × ×6 = , 16 15 14 3360 14 Or Must be a probability, no additional ‘divisions’ leading to final answer. 8 C2 × 5 C1 140 P(WBW) = = 16 C3 560 A1 1 identified outcome fully correct (accept unsimplified). 8 C1 × C2 80 5 P(WBB) = = M1 Sum of 3 correctly identified scenarios (may be identified 16 C3 560 by correct unsimplified numerator values). 8 C1 × 5 C1 × 3 C1 120 P(WBS) = = 560 16 C3 2040 17 A1 2040 17 , ,0.607 If 1 or more M mark not scored, SC1 for , ,0.607 3360 28 3360 28 WWW. Method 2 1 – {P(WSS)+P(WWS)+P(WWW)+P(BSS)+P(BBS)+P(BBB)+P(SSS)} 8 3 2 144 3 M1 Two outcomes seen as the product of 3 fractions with P(WSS) = × × ×3 = , correct numerators and n, (n - 1), (n - 2) only as 16 15 14 3360 70 denominator, where 8 ⩽ n ⩽ 16 (condone omission of ×3). 8 7 3 504 3 Attempt at 1 – p must be present. P(WWS) = × × ×3 = , 16 15 14 3360 20 Must be a probability, no additional ‘divisions’ leading to 8 7 6 336 1 final answer. P(WWW) = × × = , 16 15 14 3360 10 A1 2 identified outcomes fully correct (accept unsimplified). 4(b) 5 3 2 90 3 M1 1 – sum of 7 correctly identified scenarios (may be P(BSS) = × × ×3 = , identified by correct unsimplified numerator values). 16 15 14 3360 112 5 4 3 180 3 P(BBS) = × × ×3 = , 16 15 14 3360 56 5 4 3 60 1 P(BBB) = × × = , 16 15 14 3360 56 3 2 1 6 1 P(SSS) = × × = , 16 15 14 3360 560 2040 17 A1 2040 17 , If 1 or more M mark not scored, SC1 for , ,0.607 3360 28 3360 28 WWW. 4

Mark scheme

5(a) 56 − 65.8 75 − 65.8 M1 Use of ±standardisation formula once with 65.8, 9.6 and [P(56 < X < 75) =] P( <Z < ) 9.6 9.6 either 56 or 75. No continuity correction, not σ 2 , not σ . P ( −1.0208 < Z < 0.9583) = [ Φ ( 0.9583) + Φ (1.0208) − 1 ] M1 Appropriate probability area Φ, from final process. Must be a probability. (expect > 0.5). = 0.8309 + 0.8463 – 1 or 0.8309 – (1 – 0.8463) or 0.8309 – 0.1537 or (0.8309 – 0.5) + (0.8463 – 0.5) or 0.3309 + 0.3463 = 0.677 A1 AWRT. If 1 or more M mark not awarded, SC1 for final answer 0.677 AWRT WWW. 3 5(b) 59.1 − 72.4 B1 1.282 or – 1.282 seen cao (critical value). P Z < σ = 0.10 M1 ±standardisation formula with 59.1, 72.4, σ equating to a z-value (not 0.1, 0.9, 0.5398, 0.4602, 0.8159, 0.1841). 59.1 − 72.4 Condone continuity correction of ±0.05, not σ2 and not √σ. = −1.282 σ 13.3 Condone ± = −1.282 . σ σ = 10.4 A1 AWRT. Signs must be consistent throughout. If M1 not awarded, SC1 σ = 10.4 WWW. 3 5(c) ( 0.9 )4 ( 0.1) = 0.0656 [1] B1 1 5(d) Method 1 [P(X < 5) =] 1 − ( 0.9 ) M1 1 − ( 0.9)d d = 4, 5. 4 = 0.344 A1 0.3439. Method 2 [P(X < 5) =] 0.1 + ( 0.1)( 0.9) + ( 0.1)( 0.9) + ( 0.1)( 0.9) M1 0.1 + ( 0.1)( 0.9 ) + ( 0.1)( 0.9 )2 + ( 0.1)( 0.9 )3 + ( 0.1)( 0.9 )4 2 3 or or 1 – ( ( 0.1)( 0.9) + ( 0.1)( 0.9) + ( 0.1)( 0.9) + ( 0.9) ) 4 5 6 7 1 – ( ( 0.1)( 0.9 ) + ( 0.1)( 0.9 ) + ( 0.1)( 0.9 ) + ( 0.9 ) ) . 4 5 6 7 = 0.344 A1 0.3439. 2

Mark scheme

6(a) 10! B1 CAO. 2!3!2! = 151200 1 6(b) Method 1 Total – Ys together 8! 7! M1 8! 7! − = − , a = 1, 2 b = 1, 2 c = 1, 3. 2!2! 2! 2!a!c! a!b!c! [10080 – 2520 =] 7560 A1 2 Method 2 3 Bs treated as a single unit ^ ^ ^ ^ BBB ^ and Ys inserted 6! 7 M1 6! 7 6! 6! 7 P × C2 × C2 or × 7 P2 or × 2 2! d! d! d! 2 or 6! 7 × 6 6! 6! 7 × 6 or × or ( 6 + 5 + 4 + 3 + 2 + 1) × d! e d! 2! 2 d = 1, 2, 3 e = 1, 2. or 6! ( 6 + 5 + 4 + 3 + 2 + 1) 2! [360 × 21 =] 7560 A1 2 6(c) Method 1 8! *M1 8! ×5 × d , b = 1,2, c = 1,3 and b ≠ c, d ≥ 1. 2!3! b!c ! DM1 Multiply by 5. 16 800 A1 Method 2 8 P4 × 5! *M1 8 P4 × d , b = 1,2, c = 1,3 and b ≠ c, d ≥ 1. 2!3! b !c ! DM1 Multiply by 5!. 16 800 A1 3 6(d) Method 1 YRB_ 2 ×2 × 3 C1 × 3 C1 = 36 M1 One correct identified unsimplified expression YRRB 2 ×1× 3 C1 =6 (3 C1 ≠ 3 C2 ). YR__ 2 ×2 × C2 3 = 12 YRR_ 2 ×1× 3 C1 =6 B1 Correct outcome/value for 2 clearly identified scenarios, accept unsimplified WWW. M1 Sum of 4 correct identified scenarios. 60 A1 Method 2 YR__ 5C × 3C × 2C M1 5 C seen with YR^^ identified. 3 1 1 3 M1 5 C × a, a = 2, 3, 6. 3 B1 5 C × 3 C × 2 C or 5 C × 3 × 2. 3 1 1 3 60 A1 4