Mathematics 9709/52 May/June 2021: question paper and mark scheme

Mark scheme

1(a) 6 B1 WWW 1 1(b) 3 4 5 1 5 1 5 1 5 1 5 6 M1 p3(1 – p) + p4(1 – p) + p5(1 – p) + p6(1 – p), 0 < p < 1 + + + 6 6 6 6 6 6 6 6 0.300 (0.2996…) A1 At least 3s.f. Award at most accurate value. Alternative method for Question 1(b) 3 5 5 7 M1 p3 – p7, 0 < p < 1 − 6 6 0.300 (0.2996…) A1 At least 3s.f. Award at most accurate value. 2 1(c) 5 9 M1 1 – pn, 0 < p < 1, n = 9, 10 1− 6 0.806 A1 Alternative method for Question 1(c) 1 15 15 15 2 8 M1 p + p(1 – p) + p(1 – p)2 + p(1 – p)3 + p(1 – p)4 + p(1 – p)5 + p(1 + + +Λ + – p)6 + p(1 – p)7 + p(1 – p)8 (+ p(1 – p)9), 0 < p < 1 6 66 66 66 As per answer for minimum terms shown 0.806 A1 2

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2 72 B1 1.79 < z ⩽ 1.80, –1.80 ⩽ z < –1.79 seen P ( X > 1.1) = 2000 ( = 0.036 ) z = ±1.798 1.1 − 1.04 B1 1.1 and 1.04 substituted in ±standardisation formula, allow = 1.798 continuity correction, not σ2 or √ σ σ 0.06 M1 Equate their ±standardisation formula to a z-value and to solve σ = 1.798 for the appropriate area leading to final answer (expect σ < 0.5). 0.06 Accept ± = z − value σ σ = 0.0334 A1 0.03335 ≤ σ ≤ 0.0334. At least 3 3s.f. 4

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3(a) P(not late) = 0.4 × 0.45 + 0.35 × 0.3 + 0.25 × (1 – x) M1 0.4 × p + 0.35 × q + 0.25 × r , or p = 0.45, 0.55, q = 0.3, 0.7 and r = (1 – x), x P(late) = 0.4 × 0.55 + 0.35 × 0.7 + 0.25x 0.18 + 0.105 + 0.25 (1 – x) = 0.48 A1 Linear equation formed using sum of 3 probabilities and 0.48 or or 0.52 as appropriate. 0.22 + 0.245 + 0.25x = 0.52 Accept unsimplified. x = 0.22 A1 Final answer 3 3(b) P ( train ∩ late ) B1 0.35 × 0.7 or 0.245 seen as numerator of fraction P ( train late ) = P ( late ) M1 P(late) seen as a denominator with their probability as numerator 0.35 × 0.7 0.35 × 0.7 their p their p = or (Accept or ) 1 − 0.48 0.4 × 0.55 + 0.35 × 0.7 + 0.25 × their 0.22 0.52 0.22 + 0.245 + 0.25 × their 0.22 49 A1 = 0.471 or 104 3

Mark scheme

4(a) B1 Table with correct X values and at least one probability X –1 0 1 2 3 Condone any additional X values if probability stated as 0. 1 2 1 3 2 P(X) B1 2 correct probabilities linked with correct outcomes, may not be 9 9 9 9 9 in table. B1 3 further correct probabilities linked with correct outcomes, may not be in table. SC if less than 2 correct probabilities seen, award SCB1 for sum of their 4 or 5 probabilities in table = 1 3 4(b) −1 × 1 + ( 0 × 2 ) + 1 × 1 + 2 × 3 + 3 × 2 M1 May be implied by use in variance, accept unsimplified E ( X ) = = expression. 9 FT their table if their 3 or more probabilities sum to 1 or 0.999 −1 + 1 + 6 + 6 9 [Var (X) =] M1 Appropriate variance formula using their (E(X))2 value. ( ) −12 × 1 + 02 × 2 + 12 × 1 + 22 × 3 + 32 × 2 − ( their E ( X ) ) 2 FT their table even if their 3 or more probabilities not summing to 1. 9 1 + 0 + 1 + 12 + 18 − ( their E ( X ) ) 2 9 4 16 A1 Answers for E(X) and Var(X) must be identified E(X ) = or 1.33 and Var ( X ) = or 1.78 3 9 3 N.B. If method FT for M marks from their incorrect (b), expressions for E(X) and Var(X) must be seen unsimplified with all probabilities <1

Mark scheme

5(a) [( 0.7 ) =] 0.343 3 B1 Evaluated WWW Alternative method for Question 5(a) [(0.15)3 + 3C1(0.15)2(0.55) + 3C2(0.15)(0.55)2 + (0.55)3 =] 0.343 B1 Evaluated WWW 1 5(b) 1 – (0.859 + 9C1 0.151 0.858 + 9C2 0.152 0.857) M1 One term: 9Cx px (1 – p)9-x for 0 < x < 9, any 0 < p < 1 [1 – (0.231617 + 0.367862 + 0.259667)] A1 Correct expression, accept unsimplified. 0.141 A1 0.1408 ⩽ ans ⩽ 0.141, award at most accurate value. Alternative method for Question 5(b) 9 C3 0.153 0.856 + 9C4 0.154 0.855 + 9C5 0.155 0.854 + 9C6 0.156 0.853 + M1 One term: 9Cx px (1 – p)9-x for 0 < x < 9, any 0 < p < 1 9 C7 0.157 0.852 + 9C8 0.158 0.85 + 0.159 A1 Correct expression, accept unsimplified. 0.141 A1 0.1408 ⩽ ans ⩽ 0.141, award at most accurate value. 3 5(c) Mean = [ 60 × 0.15 =] 9 B1 Correct mean and variance, allow unsimplified. (2.765 ≤ σ ≤ 2.77 imply correct variance) Variance = [ 60 × 0.15 × 0.85 = ] 7.65 11.5 − 9 M1 Substituting their mean and variance into ±standardisation ( X ≥ 12 ) = P Z > formula (any number for 11.5), not σ2 or √ σ 7.65 M1 Using continuity correction 11.5 or 12.5 in their standardisation formula. 1 − Φ ( 0.9039 ) = 1 − 0.8169 M1 Appropriate area Φ, from final process, must be probability. 0.183 A1 Final AWRT 5

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6(a) 8! M1 8! k = 1 or 2, m = 1 or 3, not k = m = 1 2!3! k !× m! no additional terms 3360 A1 2 6(b) Method 1 Arrangements Rs at ends – Arrangements Rs at ends and Os together 6! M1 6! 6! [Os not together = ] – 4! – m, 1 ⩽ k ⩽ 3, m an integer, condone 2 × − m . 3! k! k! M1 w – 4! or w – 24, w an integer Condone w – 2 × 4! 96 A1 Method 2 identified scenarios R _ _ _ R, Arrangement No Os together + 2Os and a single O 4 C3 × 3! + 4C2 × 2 × 3! M1 4C3 × 3! + r or 4× 3! + r or 4P3 × 3! + r, r an integer. Condone 2 × 4C3 × 3! + r. 2 × 4× 3! + r or 2 × 4P3 × 3! + r. M1 q + 4C2 × 3! × k or q + 4P2 × 3! × k, k = 1,2, q an integer [24 + 72 =] 96 A1 3 6(c) Method 1 Identified scenarios OORR 3 C 2 × 2 C 2 × 3 C0 = 3 × 1 = 3 B1 Outcomes for 2 identifiable scenarios correct, accept unsimplified. 3 ORR_ C1 × 2 C 2 × 3 C1 = 3 × 1 × 3 = 9 OOR_ 3 C 2 × 2 C1 × 3 C1 = 3 × 2 × 3 = 18 M1 Add 4 or 5 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, 3 OR_ _ C1 × 2 C1 × 3 C 2 = 3 × 2 × 3 = 18 condone use of permutations. OOOR 3 C3 × 2 C1 × 3 C0 = 1 × 2 = 2 Total 50 A1 All correct and added 50 M1 their '50' Probability = 8 8 , accept numerator unevaluated C4 C4 6(c) cont’d 50 A1 or 0.714 70 Method 2 Identified outcomes ORTM 3 C1 × 2 C1 = 6 B1 Outcomes for 5 identifiable scenarios correct, accept 3 unsimplified. ORTW C1 × 2 C1 = 6 ORMW 3 C1 × 2 C1 = 6 M1 Add 9, 10 or 11 identified correct scenarios only values, no 3 2 additional incorrect scenarios, no repeated scenarios, accept ORRM C1 × C2 = 3 unsimplified, condone use of permutations. 3 ORRW C1 × 2 C2 = 3 3 ORRT C1 × 2 C2 = 3 3 OROR C2 × 2 C2 = 3 3 OROT C2 × 2 C1 = 6 3 OROM C2 × 2 C1 = 6 3 OROW C2 × 2 C1 = 6 3 OROO C3 × 2 C1 = 2 Total 50 A1 All correct and added 50 M1 their '50' Probability = 8 8 , accept numerator unevaluated. C4 C4 50 A1 or 0.714 70 5

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7(a) Includes all data B1 Reference to either including all/raw data or further statistical processes are possible that cannot be found using data from box-and-whisker, eg frequency, mean, mode or standard deviation not only median, IQR, range or spread which can be found from both. 1 7(b) B1 Correct stem can be upside down, ignore extra values Amazons Giants B1 Correct Amazons labelled on left, leaves in order from right to 8 17 5 left and lined up vertically (less than halfway to next column), 4 2 1 18 2 4 7 9 no commas or other punctuation. 8 6 0 19 2 3 5 5 5 B1 Correct Giants labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no 5 2 1 20 4 commas or other punctuation. 5 21 B1 Correct single key for their diagram, need both teams identified and ‘cm’ stated at least once here or in leaf headings or title. Key: 1|18|2 means 181 cm for Amazons and 182 cm for Giants SC for if 2 separate diagrams drawn, award SCB1 if both keys meet these criteria (Max B1, B0, B0, B1) 4 7(c) [UQ = 202 (cm), LQ = 182 (cm)] M1 201 ⩽ UQ ⩽ 205 – 181 ⩽ LQ ⩽ 184 [IQR =] 202 – 182 = 20 (cm) A1 WWW 2 7(d) [Σ11 = 2132 B1 Both Σ11 and Σ15 found. Accept unevaluated. Σ15 = 191.2 × 15 = 2868 ] their 2868 = their 2132 + (180 + 185 + 190) + h M1 Forming an equation for the height using their Σ11 and Σ15. 181 (cm) A1 Alternative method for Question 7(d) [Σ15 = 191.2 × 15 = 2868 B1 Σ15 found using the mean and raw data methods. Accept Σ15 = 2687 + h ] unevaluated. their 2868 = their 2687 + h M1 Forming an equation for the height using their Σ15 expressions. 181 (cm) A1 Alternative method for Question 7(d) [Σ15 = 2687 + h B1 Σ15 found using raw data method and statement on calculating Σ15 new mean. Accept unevaluated. = 191.2 ] 15 their 2687 + h M1 Forming an equation for the height using their Σ15 expressions = 191.2 15 181 (cm) A1 3 N.B. All methods can be presented as a logical numerical argument which can be condoned if clear.