Mathematics 9709/55 May/June 2025: question paper and mark scheme

Mark scheme

1(a) B1 Table with correct X values and at least one X 0 1 2 3 4 5 probability correct. Values need not be in order, lines may not be drawn, may be vertical, X and P(X=x) P(X=x) 6 10 8 6 4 2 may be omitted. 36 36 36 36 36 36 Condone any additional X values if probability stated as 0. 1 5 2 1 1 1 6 18 9 6 9 18 B1 Total of four correct probabilities linked with correct outcomes, may not be in table. 0.167 0.278 0.222 0.167 0.111 0.056 B1 All six probabilities are correct and linked with correct outcomes, may not be in table. If decimals are used, condone correct rounding (which will not sum to 1) or one value rounded inaccurately to sum to 1. 3 SCB1 for six probs linked to X-values 0 – 5 summing to 1 with no more than 3 correct. 1(b) [0 × 6] + 10 × 1 + 8 × 2 + 6 × 3 + 4 × 4 + 2 × 5 = 10 + 16 + 18 + 16 + 10 M1 May be implied by use in Variance, accept un- E ( X ) = simplified expression. 36 36 FT their table if their 4 or more non-zero probabilities sum to 1 or 0∙999. 10 ×1 + 8 × 22 + 6 × 32 + 4 × 42 + 2 × 52 their35 2 M1 Appropriate variance formula using their (E(X))2 [Var =] − value. FT their table even if their 4 or more non-zero 36 18 probabilities not summing to 1 210 1225 665 210 − ( their E ( X ) ) is M1. 2 = 36 − 324 = 324 = 2.05 Note: If table is correct, 36 35 665 17 A1 OE. E(X) = , 1.94 , Var(X) = ,2 , 2.05 Answers for E(X) and Var(X) must be identified. 18 324 324 Accept Var = 2.052469… to 3sf or better. 3

Mark scheme

2(a) 14 − 20 M1 14, 20 and 5 substituted into ± Standardisation [P(X < 14) =] P( Z < ) formula, no continuity correction, condone σ2 or √σ. 5 [= P( Z < −1.2) ] = 1 – 0.8849 M1 Appropriate area Φ, from final process, must be probability. (Expect final ans < 0∙5 ). Note: the correct final answer may imply M1 from use of calculator. = 0.115 A1 0.1150 ⩽ z ⩽ 0.1151. [Number =] 150 × 0.1151 = 17.265 so 17 or 18 B1FT FT their probability - final answer must be positive integer. 4 2(b) h − 20 B1 ±0.674 seen CAO – critical value. [P( X > h) = 0.25, so P( Z < ) = 0.75 ] 5 M1 20 and 5 substituted in ±standardisation formula, no h − 20 continuity correction, not σ2, √ σ, equated to a z- = 0.674 5 value. Note: 0.75; 0.25; 0.5987; 0.7734, 0.326 are NOT z- values. h = 23.4 A1 AWRT. Only dependent on M mark. 3

Mark scheme

3(a) Mean = 80 × 0.4 = 32 Var = 80 × 0.4 × 0.6 = 19.2 B1 Correct mean and variance, allow un-simplified. 4 30 (4.381 < σ ⩽ 4.382 or imply correct 5 variance). 26.5 − 32 M1 Substituting their mean and variance into P( X < 27) = P( Z < ) = P( Z < −1.255) ±standardisation formula (any number for 26∙5), not 19.2 σ2, √ σ. M1 Using continuity correction 26∙5 or 27∙5 in their standardisation formula. = 1 − 0.8953 M1 Appropriate area Φ, from final process, must be probability. (Expect final ans < 0∙5). Note: the correct final answer may imply M1 from use of calculator. 0.105 A1 0.1045 < p ⩽ 0.105. 5 3(b) Method 1 [P(0, 1, 2) = ] 10C0 0.40 0.610 + 10C1 0.41 0.69 + 10C2 0.42 0.68 M1 One term 10Cx p x (1 − p )10− x , 0 < p < 1, x ≠ 0 . = 0.0060466 + 0.0403107 + 0.120932 A1 Correct expression, accept un-simplified. = 0.167 [2…] B1 Method 2 [1 – P(3, 4, 5, 6, 7, 8, 9, 10) = ]1 – (10C3 0.47 0.63 +10C4 0.46 0.64 +10C5 M1 One term 10Cx p x (1 − p )10− x , 0 < p < 1, x ≠ 0 . 0.45 0.65 +10C6 0.44 0.66 +10C7 0.43 0.67 +10C8 0.42 0.68 +10C9 0.41 0.69 +10C10 0.40 0.610 ) A1 Correct expression, accept un-simplified. = 0.167 [2893…] B1 3

Mark scheme

4(a) B1 Fully correct labelled tree diagram for each trio of branches clearly labelled ‘accepted’,’rejected’ and ‘test again’ for each intersection (no additional branches). B1 All correct probabilities on 8 required branches in correct positions. Ignore additional branches. 2 4(b) [P(A) + P(T A) + P(T T A) =] 0.3 + 0.5 × 0.3 + 0.5 × 0.5 × 0.25 M1 0.3 + k + j, [= 0.3 + 0.15 + 0.0625 ] where either k = 0.5 × 0.3 and 0 ⩽ j < 1 or 0 ⩽ k < 1 and j = 0.5 × 0.5 × 0.25. 41 A1 CAO. 0.5125, 80 2 4(c) P ( T A) + P ( T T A) B1 0.5 × 0.3 + 0.5 × 0.5 × 0.25 or their (b) – 0.3 seen as P ( 2nd test taken | A) = = = a numerator of a fraction (accept evaluated 0.2125). P ( accepted ) 0.5 × 0.5 × 0.25 + 0.5 × 0.3 M1 Conditional probability formula used with their (b) or 0.5125 or ( 0.3 + 0.5 × 0.3 + 0.5 × 0.5 × 0.25 ) correct in denominator. 17 A1 0.41463… to 3SF or better. 0.2125 80 17 = = 0.415, 0.5125 41 41 80 3 4(d) M1 (1 – m)3, m = their (b) or correct. ( ) 39 3 3 (1 − 0.5125) or 0.2 + 0.5 × 0.2 + 0.52 × 0.75 or 0.48753 or ( 3 ) 80 0.116 A1 2

Mark scheme

5(a) Smarts Teasers B1 Correct stem cannot be upside down, ignore extra 9 0 values. 8 8 3 1 1 2 5 8 9 8 2 2 1 5 5 B1 Correct Smarts labelled on left, leaves in order from 8 0 3 2 6 7 9 9 right to left and lined up vertically (less than halfway 1 4 to next column), no commas or other punctuation. B1 Correct Teasers labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas or other punctuation. Condone misalignment, commas, reverse order or omission of label if error penalised already in Smarts. If the correct data for Smarts & Teasers is transposed, treat as a single error in Smarts and condone in Teasers. If 2 errors on Teasers and 1 error is repeated in Smarts, B0B0. Key 8|2|5 means 28 minutes for Smarts and 25 minutes for Teasers B1 Correct single key for their diagram, need both teams identified and ‘minutes’ stated at least once here or in leaf headings or title. SC If 2 separate diagrams drawn, SCB1 if both keys meet these criteria (Max B1, B0, B0, B1) 4 5(b) Smarts: LQ 13 M 22 UQ 30 B1 All correct seen or plotted, even if Smarts Box Plot is not labelled. Box-and-whisker diagram B1 All five key values for Teasers plotted accurately using their linear scale and labelled. B1 All five key values for Smarts, FT their values, plotted accurately using their linear scale and labelled B1 There must be two Box Plots. Whiskers not through either box or drawn at corners of boxes; single linear scale with at least three values stated and labelled time and minutes. SCB1 if there is no scaled and correctly labelled line but the two Box Plots are correct relative to each other, may or may not be labelled. lowest Q1 Q2 Q3 highest Smarts 9 13 22 30 41 Teaser 12 18 25 37 39 4 5(c) Smarts are quicker B1 Comment in context about central tendency. Smarts’ times are more consistent B1 Comment in context about spread. 2

Mark scheme

6(a) Method 1 7 5 4 3 M1 7 5 4 3 × × × ×4 × × × ×n , n a positive integer. 12 11 10 9 12 11 10 9 M1 a c e g × × × × 4 , a ,b ,c ,d ,e ,f ,g ,h positive b d f h integers. 14 A1 , 0.141 [4…] 99 Method 2 7 C1 × 5C3 M1 7 C1 × 5C3 given as the numerator of a fraction. 12 C4 M1 12 C4 given as the denominator of a fraction. 14 A1 , 0.141[4…] 99 3 12 j 6(b) C4 × 8C4 [× 4C4] M1 C4 × k , j = 12, 8 k a positive integer > 1. 34650 A1 SCM1 for 12C4 × 8C4[ × 4C4] ÷3! SCA1 for 5775. 2 6(c) Method 1 – summing no of ways with at least one man and one woman in each team 3M 1W + 3M 1W + 1M 3W M1 (7Cm × 5C4-m ) 1 ⩽ m ⩽ 3 seen multiplied by at least 3M 1W + 2M 2W + 2M 2W one other Combination in form nCr. 3! (7C3 × 5C1 ) × ( 4C3 × 4C1 ) [× (1C1 × 3C3)] × M1 No of ways for two correctly identified scenarios, or 2! correct, added, no incorrect. = 175 ×16 × 3= 8400 3! (7C3 × 5C1 ) × ( 4C2 × 4C2 ) [×(2C2 × 2C2 )] × 2! = 175 × 36 ×3 = 18900 27300 A1 SC A1 for 4550. SC B1 for 9100 if only one M1 has been awarded. Method 2 – subtracting ways with only men/women in a team from total p 4M 0W + 3M 1W + 0M 4W M1 C4 where p = 7, 6, 5 or 4 4M 0W + 2M 2W + 1M 3W seen multiplied by at least 1 other Combinations in form nCr, r ≠ 0, n ≠ r . 7 C4 [× 5C0] × 3C3 × 5C1 × 3! M1 No of ways for two correctly identified scenarios = 175 x 6 = 1050 added (or correct) and subtracted from 34650 or their (b). 7 C4 [× 5C0] × 3C2 × 5C2 × 3! = 1050 × 6 = 6300 34650 – (1050 +6300) 27300 A1 SC A1 for 4550. 6(c) Method 3 – subtracting ways with only men/women in a team from total M1 (7C4 × 8C4) or (5C4 × 8C4) seen. 3! all male team 7C4 × 8C4 × = 7350 M1 Subtracting (all male + all female – overlap) correctly 2! identified or correct from 34650 or their (b). 3! all female team 5C4 × 8C4 × = 1050 2! 3! all male AND all female = 7C4 × 5C4 × = 1050 1! 34650 – (7350+1050-1050) 27300 A1 SCA1 for 4550. Method 4 – subtracting ways with only men in a team as this includes the way with only women 3! M1 (7C4 × 8C4) seen. all male team 7C4 × 8C4 × = 7350 2! M1 Subtracting from 34650 or their (b). 34650 – 7350 27300 A1 SCA1 for 4550. 3 6(d) Method 1 G__H___ M1 5! × n , n = 2,4,8. 5! × 2 × 4 960 A1 Method 2 5 P2 × 2! × 4! or 5C2 × 2 × 2! × 4! M1 5 P2 × n or 5C2 × 2 × n where n = 2!, 4! or 2! ×4! 960 A1 2