Mathematics 9709/52 May/June 2025: question paper and mark scheme

Mark scheme

1 B1 Table with correct values of x and at least one non- x 0 1 2 3 zero probability correct. Condone extra x values if probability stated as 0. P(X = x) 1 6 11 6 . 24 24 24 24 B1 Two more correct non-zero probabilities linked with correct outcome (3 correct probabilities 0.0416 0.25 0.458 0.25 present). 7 3 Accept probabilities not in table if clearly identified. B1 Four correct probabilities linked with the correct outcomes. Accept probabilities not in table if clearly identified. Non-exact decimals correct to at least 3SF. SCB1 for 4 non-zero probabilities (not all ¼) in table with correct x values adding to 1 if B1 max scored. 3

Mark scheme

2 Mean = 160 ×0.7 = 112 B1 112 and 33.6 (CAO) seen, allow un-simplified. Variance = 160 × 0.7 × 0.3 = 33.6 May be in standardisation formula. ( σ = 33.6, 5.79655 to at least 4SF implies correct variance). Withold mark if variance clearly identified as standard deviation. ( ) Condone N 112, 33.6 if standardisation formula correct or variance/standard deviation stated correctly as well. 120.5 − 112 M1 Substituting their 112 and their 33.6 into the P(X >120 ) = P( Z > ) ±standardising formula (any number for 120.5), 33.6 allow σ2 or √σ. M1 Use continuity correction 119.5 or 120.5 in their standardisation formula. Note: If no standardisation formula seen 8.5 8.5 ± or ± 5.797 scores M2 BOD. 33.6 [P( Z > 1.466 ) = 1 − Φ (1.466 )] M1 Appropriate area Φ , from final process, must be a probability. = 1 – 0.9287 May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5. = 0.0713 final answer A1 Final answer AWRT. 5

Mark scheme

3(a) 4 4 M1 OE. [P(both same colour) =] × + 12 × 11 4 4 12 11 12 11 16 16 16 15 × and either × or × seen. 16 16 16 15 16 16 11 Decimals to 4SF but condone = 0.733 . 15 No additional terms. 49 A1 ISW. = , 0.6125 80 49 CAO, (0.6125, must be seen). 80 If M mark not scored, 49 SCB1 for , 0.6125 WWW. 80 2 3(b) P (1B 2R ) M1 4 12 × P (1B|2 R ) = seen as a numerator of a single fraction. P ( 2R ) 16 16 4 12 3 × If 0.1875 or is seen as the numerator, the = 16 16 16 4 12 12 11 × + × calculation must be seen in the working for the 16 16 16 15 denominator – or in 3(a), including by the tree diagram. The question will need to be linked if the work is in 3(a). M1 4 12 their × or correct 16 16 12 11 their × from part ( a ) 59 + 16 15 or 0.7375 or 80 or correct seen as the denominator of a single fraction. 15 A1 0.254237… to at least 3 SF. , 0.254 If one or more M not scored 59 15 SCB1 for , 0.254237… to at least 4 SF WWW. 59 3

Mark scheme

4(a) 0.3 × 0.25 × 0.45 × 6 M1 OE. 0.3 × 0.25 × 0.45 × k , k an integer > 1. E.g. 6 = 3! = 3P3 etc. 81 A1 CAO exact answer. 0.2025, 400 2 4(b) Method 1 1−0.78 M1 1−0.7d , d = 8, 9, 0.75, 0.3 are not misreads. = 0.942 A1 0.942 ⩽ p < 0.9425. Method 2 0.3 + 0.3( 0.7 ) + 0.3( 0.7 ) + 0.3( 0.7 ) + 0.3( 0.7 ) + 0.3 (0.7 ) + 0.3 (0.7 ) + 0.3( 0.7 ) M1 0.3 + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 2 3 4 5 6 7 2 3 4 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) +0.3 ( 0.7 ) 5 6 7 8 = 0.942 A1 0.942 ⩽ p < 0.9425. 2 4(c) (0.3)2 (0.7)5 ×6 M1 (p)2 (1 – p)5 ×k , 0 < p < 1, k = 5 or 6 . = 0.0908 A1 AWRT. 2

Mark scheme

5(a) Cumulative frequency graph drawn B1 At least 5 points plotted accurately at class upper end points: (10, 34), (20, 86), (30, 142), (40, 208), (60, 265), (90, 300). Linear cf scale 0 ⩽ cf ⩽ 300 and linear time scale 0 ⩽ time ⩽ 90 with at least 3 values identified on each. Bar/histograms score B0. Required axes must be over 50% of grid size. Axes can be the other way round. B1 All points plotted correct, curve drawn, no line segments and joined to (0,0) passing within ½ square of points. Axes labelled cumulative frequency (cf), time (t) and minutes (min or m) – or a suitable title. Curve must be <300 for 0 ⩽ t < 90. 2 5(b) Line drawn from 180 on cf axis to meet graph M1 Use of graph must be seen. Must be an increasing graph. Annotate this mark on 5(a) grid. [k =] 35 minutes A1 FT Strict FT, reading at their curve condone use of t. 2 5(c) Frequencies: [34,] 52, 56, 66, 57, 35 B1 May be un-simplified and/or in variance calculation. Midpoints: 5, 15, 25, 35, 50, 75 B1 At least 5 correct midpoints seen, may be un- simplified, may be in calculation, may be by data table. 34 × 5 + 52 × 15 + 56 × 25 + 66 × 35 + 57 × 50 + 35 × 75 M1 Correct un-simplified mean formula with their 6 Mean = midpoints (not upper bound, lower bound, upper 300 limits, lower limits, class width, frequency density, 170 + 780 + 1400 + 2310 + 2850 + 2625 frequency or cumulative frequency and must be 300 within class) and their frequencies (not cw, cf, or 10135 fd), accept un-simplified. = 300 Condone 1 value error on numerator. The ‘table’ approach may be used with multiplications and additions implied by appropriate values. = 33.8 A1 WWW. 2027 47 33.78 ⩽ mean ⩽ 33.8, , 33 . 60 60 2027 47 If M not scored, SCB1 for , 33 , 33.78 . 60 60 10135 (Note: scores A0). 300 5(c) 34 × 52 + 52 ×152 + 56 × 252 + 66 × 352 + 57 × 502 + 35 × 752 10135 2 M1 Correct un-simplified Variance formula with their Var = − their midpoints and their frequencies – their mean2 for 300 300 2 variance or standard deviation. 34 × 25 + 52 × 225 + 56 × 625 + 66 × 1225 + 57 × 2500 + 35 × 5625 10135 Or − their Condone one value error on numerator. 300 300 Condone their Σ f from denominator of mean [= 1559.25 – 1141.31 = 417.936] calculation if not 300. sd = 20.4 A1 20.4 ⩽ σ < 20.45 WWW. At least one of the mean or the standard deviation must be linked to the value, and no incorrect identifications, for this mark to be scored. 6

Mark scheme

6(a) Method 1 M _ _ _ _ _ _ _ _ _M (arranging the remaining letters and inserting the As) 4! × 5C1 M1 4! × m, m , m 1 . M1 n × 5C1 or n × 5C4 or n × 5, 5 P allow n × 5P1 or n × 4 , n , n > 1 . 4! e.g. 4! × 5C1× k (oe), k , k ⩾ 1 scores M1M1. 120 A1 Method 2 Total number of arrangements with Ms at end – arrangements with Ms at end and at least 2 As together 8! M1 8! [Total number of arrangements: = 1680 × r − 5!× m, m , m 3, r = 1, 2 . 4! 4! Arrangements with As together: (AAAA) ^ ^ ^ ^ 5! M1 k − 5!×13 × r , k , k > 1560 r = 1,2 . (AAA^) (A) ^^^ 5! × 4 4 (AA^) (AA) ^^^ 5!× 2! 4×3 (A^) (A^) (2A) ^^ 5!× ] 2! 8! − 5!× 13 4! = 120 A1 3 6(b) Method 1 M _ _ _ M _ _ _ _ _ 8! B1 8! ×6 × b, b , b 1 . 4! 4! M1 c! × 6 , c = 8, 9, 10 d = 1, 2!, 4! or 2! × 4!. d 10080 A1 Method 2 M ^ ^ ^ M _ _ _ _ _ letters between Ms arranged and the treated as a single item 6! 8 B1 6! × P3 × b, b , b 1 . 4! 4! M1 c! 8 × P3 , c = 6,7,8 d = 1, 2!, 4! or 2!×4!. d 10080 A1 3 6(c) Method 1 MMAAA 1 B1 One identified outcome correct (excluding 4 MMAA _ C1 4 MMAAA or MAAAA), accept un-simplified. 4 MAA _ _ C2 6 MAAA _ 4 C1 4 M1 Five correct scenarios added MAAAA 1 (values do not need to be correct). [Total] 16 A1 Method 2 2 Ms cannot be present in the scenarios 5 MAA ^ ^ C2 10 B1 One identified outcome correct (excluding 5 MAAA ^ C1 5 MAAAA), accept un-simplified. 5 MAAAA C0 1 M1 Three correct scenarios added (values do not need to be correct). [Total] 16 A1 3

Mark scheme

7(a) 72.4 − 74.8 76.3 − 74.8 M1 Use of ±standardisation formula once with 74.8, 3.2 P <Z< and either 72.4 or 76.3. 3.2 3.2 No continuity correction, not σ2, not σ . [ Φ ( 0.75) + Φ ( 0.4688 ) − 1] M1 Calculating the appropriate probability area, = 0.7734 + 0.6804 – 1 must be a probability (assume a 4SF value in the calculation it is a or ( 0.7734 − 0.5) + ( 0.6804 − 0.5) probability, assume a 3SF value in the calculation it or 0.7734 – 0.3196 is not a probability). or 0.6804 – 0.2266 (leading to their final answer, expect < 0.5). or 0.2734 + 0.1804 Unless there is a misread, the probability area calculation must be structured as shown. = 0.4538 A1 AWRT 0.454 WWW. Expected number = 0.4538 × 120 = 54.46 B1 FT Strict FT their 4-figure probability × 120 (check so 54 with calculator). (or 55) One integer answer. (One integer answer stated) No indication of approximation, e.g. ≈, , about, 2 SF . 4 7(b) 0.202 − μ B1 0.841 < z1 < 0.843 or −0.843 < z1 < −0.841 seen. = 0.842 σ B1 0.582 < z2 < 0.584 or −0.584 < z2 < −0.582 seen. 0.185 − μ = − 0.583 M1 Use of the ±standardisation formula once with µ, σ σ equating to a z-value (not 0.20, 0.80, 0.28, 0.72, 0.5793, 0.4207, 0.6103, 0.3897, 0.7881, 0.2119, 0.7642, 0.2358, –0.417, 0.417, -0.158 0.158, etc). No continuity correction, not σ2, not σ . Solve, obtaining values for µ and σ M1 Solve 2 equations in µ and σ with an attempt at the elimination method, substitution method or other μ = 0.192, σ = 0.0119 appropriate approach to obtain values for both µ and σ . A1 AWRT μ = 0.192, σ = 0.0119 . There must be consistency with signs in the 0.017 solution, e.g. σ = = 0.0119 is not −1.425 acceptable, A0. If one or both the M marks have been withheld, SCB1 for both correct WWW. 5 7(c) Method 1 [P(X < 3) = P(0, 1, 2) = ] M1 One term of the form 10Cx ( p ) x (1 − p )10− x , 0.810 + 10C1 0.89 0.2 + 10C2 0.88 0.22 0 < p < 1 , x ≠ 0 or 10. [= 0.107374 + 0.268435 + 0.301989 ] A1 Correct un-simplified expression. 0.678 B1 0.677 < p ⩽ 0.678. Method 2 [P(X < 3) = 1 – P(3, 4, 5, 6, 7, 8, 9, 10) = ] M1 One term of the form 10Cx ( p ) x (1 − p )10− x , 1 – {10C3 0.87 0.23 + 10C4 0.86 0.24 + … + 0.210} 0 < p < 1 , x ≠ 0 or 10. A1 Correct un-simplified expression. 0.678 B1 0.677 < p ⩽ 0.678 3