Mathematics 9709/52 October/November 2022: question paper and mark scheme

Mark scheme

1(a) 0.2 × x + 0.1× 2x + 0.7 × 0.25 = 0.235 M1 0.2 × x + 0.1× 2x + 0.7 × 0.25 or 0.2x + 0.2x + 0.175 seen. M1 Equating their 3 term expression (2 terms involving x) to 0.235 x = 0.15 A1 3 1(b) P ( car and not late ) M1 0.1 × (1 − 2 × their x ) or 0.1 × 0.7 as numerator [ P(car |not late ) = P ( not late ) and 0.2 × (1 – their x) + 0.1 × (1 – 2 × their x ) + 0.7 × 0.75 with values 0.1 × (1 − 0.3) substituted or 1 – 0.235 or 0.765 as denominator of fraction. 1 − 0.235 Condone 0.2 × (1 – their x) + 0.1 × (1 – × their x) + 0.7 × 0.75 as denominator consistent with 1(a). 0.07 70 14 A1 0.091503267 to at least 3SF. 0.765 = 0.0915, 765 , 153 If M0 scored SC B1 for 0.091503267 to at least 3SF. 2

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2(a) 54.8 − 55.6 M1 Use of ± standardisation formula, with 54.8, 55.6 and 1.2 [P(X<54.8)] = P( Z < ) 1.2 substituted. condone 1.22 , 1.2 or continuity correction of 54.75 or 54.85 [= P( Z < −0.6667)] = 1 − 0.7477 M1 Appropriate area Φ, from final process, must be probability. = 0.2523 A1 0.252 ⩽ p ⩽ 0.2525 If A0 scored S CB1 for 0.252 ⩽ p ⩽ 0.2525 [Expected number =] 400 × 0.2523 = 100.92 B1 FT FT their 4SF (or better) probability from a normal calculation. 100 or 101 Must be a single integer answer. 4 2(b) 1 1 M1 {Both ½ and –½ seen as z-values [P( − < Z < ) = Φ(½ ) – Φ(−½) =] or appropriate use of +½ or −½} 2 2 and {no other z-values in part}. 1 2Φ − 1 56.2 − 55.6 55[.0] − 55.6 2 Condone and seen as z-values. 1.2 1.2 = 2 × their 0.6915 −1 or their 0.6915 − (1 − their 0.6915 ) M1 Calculating the appropriate area from stated phis of z-values which or 2 × ( 0.6915 − 0.5 ) must be ± the same number. 0.383 A1 0.3829 ⩽ z ⩽ 0.383 If A0 scored SC B1 for 0.3829 ⩽ z ⩽ 0.383 3

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3(a) 4 1 B1 May be seen used in calculation. [P(17 or 18) =] = ,0.0185(185…) 216 54 5 53 1 M1 p(1 – p)5, 0 < p < 1 P(X = 6) = . 54 54 0.0169 A1 0.01686 < p ⩽ 0.0169 If A0 scored SC B1 for 0.01686 < p ⩽ 0.0169 3 3(b) 53 7 M1 r [P(X < 8) =] 1 − 53 1 − their or 0.98148 or correct , 54 54 r = 7,8 0 < their p < 1 0.123 A1 0.1225 ⩽ p ⩽ 0.123 Alternative method for Question 3(b) [P(X < 8) =] M1 q + pq + p 2 q + p3q + p 4 q + p5 q + p 6 q , p + q = 1, 0 < p, q < 1, q 2 3 4 1 53 1 53 1 53 1 53 1 + + + + + = their 53 54 54 54 54 54 54 54 54 54 54 5 6 53 1 53 1 + 54 54 54 54 0.123 A1 0.1225 ⩽ p ⩽ 0.123 2

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4(a) Cw 20 20 10 10 40 M1 At least 4 frequency densities calculated Fd 1.6 2.3 9.6 5.2 0.6 f 32 f eg condone if unsimplified , accept unsimplified, cw 20 cw ± 0.5 may be read from graph using their scale no lower than 1 cm = fd 1 A1 All bar heights correct on graph, using their suitable linear scale with at least 3 values indicated, no lower than 1 cm = fd 2. B1 Bar ends at [0,] 20, 40, 50, 60, 100 (at axis), 5 bars drawn 0 ⩽ time axis ⩽ 100, linear scale with at least 3 values indicated. B1 Axes labelled frequency density (fd), time (t) and minutes (mins, m) or appropriate title. (Axes may be reversed). 4 4(b) Midpoints 10 30 45 55 80 B1 At least 4 correct midpoints seen (check data table). [Mean = 43.2 given] M1 Appropriate variance formula with their 5 midpoints (not upper 32 × 102 + 46 × 302 + 96 × 452 + 52 × 552 + 24 × 802 bound, lower bound, class width, frequency density, frequency or [Var =] − 43.22 cumulative frequency). 250 Or Condone 1 frequency error. If correct midpoints seen accept 32 (10 − 43.2 ) + 46 ( 30 − 43.2 ) + 96 ( 45 − 43.2 ) 2 2 2 3200 + 41400 + 194400 + 157300 + 153600 549900 or +52 ( 55 − 43.2 ) + 24 (80 − 43.2 ) 2 2 250 250 250 −{43.22 or 1866.24} . 549900 A1 www, final answer 18.25814887 to at least 3SF. = − 43.22 = 333.36 If M0 earned SC B1 for final answer 18.25814887 to at least 3SF. 250 Sd = 18.3 3

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5(a) Method 1: Scenarios identified ignoring unbiased coin 1 3 3 M1 All 3 different calculations seen unsimplified. P(BH1 BT2) = × = 4 4 16 3 1 3 P(BT1 BH2) = × = 4 4 16 1 1 1 P(BH1 BH2) = × = 4 4 16 3 3 1 7 A1 Clear identification of all scenarios, linked probabilities and sum. + + = AG 16 16 16 16 5(a) Method 2: Scenarios identified with all 3 coins 1 1 3 3 M1 All 6 different calculations seen unsimplified. P(H BH1 BT2) = × × = 2 4 4 32 1 1 3 3 P(T BH1 BT2) = × × = 2 4 4 32 1 3 1 3 P(H BT1 BH2) = × × = 2 4 4 32 1 3 1 3 P(T BT1 BH2) = × × = 2 4 4 32 1 1 1 1 P(H BH1 BH2) = × × = 2 4 4 32 1 1 1 1 P(T BH1 BH2) = × × = 2 4 4 32 1 + 3 + 3 + 1 + 3 + 3 14 7 A1 Clear identification of all scenarios, linked probabilities and sum. P(B) = = = AG 32 32 16 Method 3: 1- P(BT1 BT2) ignoring unbiased coin 3 2 M1 Calculation seen unsimplified 1 – P(BT1 BT2) = 1 − and 1 – probability seen. 4 7 A1 Clear identification of scenario used, linked probability and = calculation. AG 16 5(a) Method 4: 1- P(BT1 BT2) with all 3 coins 1 3 3 1 3 3 M1 Both calculations seen unsimplified 1 – P(H BT1 BT2) – P(T BT1 BT2) = 1 − × × − × × and 1 – 2 probabilities seen. 2 4 4 2 4 4 9 9 7 A1 Clear identification of all scenarios used, linked probabilities and = 1− − = calculation. AG 32 32 16 2 5(b) 1 1 1 1 M1 Their identified P(HHH) or correct as numerator P ( A B ) 2 × 4 × 4 32 and their identified P(B) or correct as denominator. P ( A|B ) = = = P ( B) 7 7 Either numerical expression acceptable. 16 16 1 A1 Accept 0.071428… rounded to at least 3SF. = , 0.0714 14 2 5(c) 1 3 1 1 1 3 1 3 3 15 B1 Table with correct X values and at least one probability. P(1H) = × × + × × + × × = Condone any additional X values if probability stated as 0. 2 4 4 2 4 4 2 4 4 32 1 1 1 1 3 1 1 1 3 7 P(2H) = × × + × × + × × = B1 P(1) or P(2) correct, need not be in table, accept unsimplified. 2 4 4 2 4 4 2 4 4 32 B1 4 correct probabilities linked with correct outcomes, may not be in X 0 1 2 3 table. p(X) 9 15 7 1 Decimals correct to at least 3 SF. 32 32 32 32 0.28125 0.46875 0.21875 0.03125 SC B1 for 4 probabilities (0 < p < 1) sum to 1 ± 0.005 with P(1) and P(2) incorrect. 3

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6(a) [1 – P(10, 11, 12) =] M1 One term 12Cx p x (1 − p )12− x , for 0 < x < 12, 0 < p < 1 1 − ( 12C10 0.910 0.12 + 12C11 0.911 0.11 + 12C12 0.912 0.10 ) = 1 – (0.230128 + 0.376573 + 0.282430) A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.111 B1 Mark the final answer at the most accurate value, 0.1108 < p ⩽ 0.111 WWW. Alternative method for Question 6(a) [P(0,1,2,3,4,5,6,7,8,9) =] M1 One term 12Cx p x (1 − p )12− x , for 0 < x < 12, 0 < p < 1 12 C0 0.90 0.112 +12C1 0.91 0.111 +12C2 0.92 0.110 +12C3 0.93 0.19 +12C4 0.94 0.18 +12C5 0.95 0.17 +12C6 0.96 0.16 +12C7 0.97 0.15 +12C8 A1 Correct expression, accept unsimplified, no terms omitted, leading 0.98 0.14 +12C9 0.99 0.13 ) to final answer. If answer correct condone omission of any 7 of the 8 middle terms. 0.111 B1 Final answer 0.1108 < p ⩽ 0.111 WWW. 3 6(b) [Mean = 80 × 0.9 =] 72, B1 72 and 7.2 seen, allow unsimplified. [ Variance = 80 × 0.9 × 0.1] = 7.2 May be seen in standardisation formula. (2.683 ⩽ σ < 2.684 imply correct variance). 69.5 − 72 M1 Substituting their mean and their variance into ±standardisation P(X > 69) = P( Z > ) 7.2 formula (any number for 69∙5), not their 7.2, not √their 2.683 M1 Using continuity correction 69∙5 or 68∙5 in their standardisation formula. [= P( Z > −0.9317) =] M1 Appropriate area Φ, from final process, must be probability. Φ ( 0.9317 ) 0.824 A1 0.8239 ⩽ p ⩽ 0.8243 WWW. 5 6(c) np = 72, nq = 8 Both greater than 5, [so approximation is valid] B1 np, nq evaluated accurately. both np & nq referenced correctly. > 5 or greater than 5 seen. 1

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7(a) 7! M1 7! b,c = 1,2 b !× c ! 2! 2! 7! × × oe, no further terms present. 2! 2! 5040 A1 2 7(b) Method 1 for first 3 marks: Arrangements of 6 letters including Ls between As 5! × 5 × 2 M1 5! × d, d integer > 1 M1 e! × f × g, e = 5, 6, 7; f = 1, 5; g = 1, 2; f ≠ g, 1 can be implicit. 1200 A1 Method 2 for first 3 marks: Number of arrangements of LL^^^^^ – number of arrangements with the Ls split by an A 6! × 2 – 5! × 2 M1 6! × 2 – h h an integer 1 < h < 1440 M1 k – 5! × 2 k an integer k > 240 1200 A1 Method 3 for first 3 marks: Alternative approaches to Method 1 ^A ^ ^ ^ ^ ^ A 5 P1 × 1P1 × 5P5 × 1P1 = 600 M1 LL treated as a single unit. M1 1200 A1 7(b) Final 2 marks of Question 7(b) 9! B1 Accept unsimplified. [Total number of arrangements =] = 90720 May be seen as denominator of probability. 2!2! 1200 5 B1 FT their 1200 Probability = , , 0.0132 unsimplified B1 FT if their 1200 and their 90 720 90720 378 their 90720 supported by work in this part. 5 7(c) Method 1: Scenarios identified Both As and Ls removed 5 A____ C4 = 5 B1 1 correct, identified outcome/value 5 AA _ _ _ C3 = 10 for A, AL or AAL scenario, accept unsimplified 5 5 AL _ _ _ C3 = 10 C5–x cannot be used in place of 5Cx 5 AAL _ _ C2 = 10 M1 Add 4 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. [Total =] 35 A1 Value stated WWW. Method 2: 1 A fixed, 1 L removed No other scenarios can be present anywhere in solution 7 7 A^^^^ C4 M1 Ch, 3 ⩽ h ⩽ 5 7 B1 C4 oe, no other terms, scenario identified. [Total =] 35 A1 Value stated. Method 3: 1 A fixed, both Ls removed A ^ ^ ^ ^ = 6C4 = 15 B1 Correct outcome/value for 1 identified scenario, accept A L ^ ^ ^ = 6C3 = 20 unsimplified. WWW M1 Add 2 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. [Total =] 35 A1 Value stated. 3