Mathematics 9709/53 October/November 2024: question paper and mark scheme

Mark scheme

1(a) 0.33 + 0.73 M1 p 3 + q 3 , p + q = 1, p, q > 0 or or 1 – (3 × 0.3 × 0.7 + 3 × 0.3 × 0.7 ) = [1 − 0.63] 2 2 1 − (3 × p 2 × q + 3 × p × q 2 ), p + q = 1, p, q > 0 . 0.37[0] A1 37 . 100 2 1(b) [Mean = 125 × 0.3 =] 37.5 B1 37.5 or 37½ and 26.25, 26¼ seen, allow [Variance = 125 × 0.3 × 0.7 =] 26.25 unsimplified. May be seen in standardisation formula. 105 ( [σ =] 5.12, implies correct 2 variance). 45.5 − 37.5 M1 Substituting their mean and their positive P(X > 45) = P( Z > ) standard deviation into the ±standardising 26.25 formula (any number for 45.5), not their σ2, not their σ . M1 Use continuity corrections 44.5 or 45.5 in their standardisation formula ±8 ±8 Note: or seen gains M2 26.25 5.123 BOD [1 − Φ ( their 1.5614 ) ] = 1 – their 0.9407 M1 Appropriate area Φ, from final process, must be a probability. Expect final answer < 0.5. Note: appropriate final answer implies this M1. 0.0593 A1 0.0592 ⩽ p ⩽ 0.0593. 5

Mark scheme

2(a) B1 Table with correct x values and at least one correct probability linked with the correct x- x 1 2 3 4 6 value. Values need not be in order, lines may not 6 12 6 6 6 be drawn, may be vertical, P(X = x) 36 36 36 36 36 x and P(X) may be omitted. Condone any additional x values if 1 1 1 1 1 probability stated as 0. 6 3 6 6 6 B1 4 correct probabilities linked with the correct x-values, need not be in table, 0.167 0.333 0.167 0.167 0.167 accept unsimplified. B1 5 correct probabilities linked with correct x- values, may not be in table. Decimals correct to at least 3 SF. SC B1 4 or 5 probabilities summing to 1 placed in a probability distribution table with 4 or 5 x-values between 1 and 6 inclusive. 3 2(b) 1 B1 FT FT their table with 4 or 5 probabilities (0 < [E(X) = ( 6 + 24 + 18 + 24 + 36) =] 3 p < 1) summing to 1. 36 1

Mark scheme

3(a) 170 − 176 M1 Using ± standardisation formula with 170, [P(X > 170) = P( Z >] ) 176 and 4.8 substituted appropriately. 4.8 Condone σ 2 and σ . No continuity correction. Φ (1.25) = 0.894 M1 Appropriate area Φ, from final process, must be a probability. Expect final answer > 0.5. A1 0.894 or 0.89435 ⩽ p ⩽ 0.8944. If A0 scored, SC B1 for 0.894 or 0.89435 ⩽ p ⩽ 0.8944, WWW. 3 3(b) P(h < 170) = 1 – 0.8944 = 0.1056 M1 1 – their 3(a) seen or implied by 0.7056 or 0.2944 k − 176 B1 0.540 < z ⩽ 0.541 or −0.541 ⩽ z < 0.540 = Φ−1 ( 0.1056 + 0.6 ) = 0.541 4.8 seen. M1 Use of ±standardisation formula with k, 176, 4.8 equated to a z-value (not 1.25, 0.7601, 0.2399, 0.7056, 0.7257, 0.8313, 0.253 ±0.894, 0.6, 0.4), not 4.82, not 4.8 , no continuity correction. k = 178.6 A1 CAO (answer required to 1 dp). 4

Mark scheme

4(a) B1 Cf values 28, 60, 102, 130, 144, 150 seen, H(cm) 10–19 20–29 30–39 40–44 45–49 50–54 55–59 Condone omission of 10. May be implied by accurate plotting (scale UB 19.5 29.5 39.5 44.5 49.5 54.5 59.5 no less than 1cm = 10). May be by data table. cf [10] 28 60 102 130 144 150 B1 Linearly scaled axes correctly labelled cumulative frequency (cf) (from 0 to 150) and height (h) and centimetres (cm) (from 9.5 to 59.5) with at least 3 values identified on each. Axes can be the other way round. M1 At least 4 points plotted at upper boundary ± 0.5, (e.g. allow (19, 19.5 or 20, 10) etc.) on correctly scaled axes. (9.5,0), (19.5,10), (29.5, 28), (39.5, 60), (44.5,102), (49.5, 130), (54.5, 144), (59.5,150). A1 All points plotted correctly, curve drawn (within tolerance), joined to (9.5, 0) and not going beyond above 150 vertically. A0 if straight line segments used. 4 4(b) [150 × 0.3 = 45] M1 Use of graph must be seen. Line drawn from 45 on cf axis to meet graph at h = 36 A1 FT Must be an increasing cf graph. Expect an answer in range 35 ⩽ h ⩽ 37 for a correct graph. 2 4(c) Midpoints 14.5, 24.5, 34.5, 42, 47, 52, 57 B1 At least 6 correct midpoints seen, may be unsimplified, may be in calculation, may be by data table. 10 ×14.5 + 18 × 24.5 + 32 × 34.5 + 42 × 42 + 28 × 47 + 14 × 52 + 6 × 57 M1 Correct unsimplified mean formula with Mean = their midpoints (not ub, lb, upper limits, 150 lower limits, cw, fd, f or cf and must be 145 + 441 + 1104 + 1764 + 1316 + 728 + 342 = within class). 150 If midpoints correct, accept partially evaluated. 5840 584 14 A1 Accept answers wrt 38.9 WWW = , , 38 , 38.9 If M1 withheld, SC B1 for 150 15 15 5840 584 14 , ,38 , 38.9. 150 15 15 sd2 = M1 Correct unsimplified variance formula with 10 ×14.52 + 18 × 24.52 + 32 × 34.52 +42 × 422 + 28 × 472 + 14 × 522 + 6 × 572 5840 2 their midpoint (not ub, lb, upper limits, − their lower limits, cw, fd, f or cf and must be 150 150 within class). 2 244285 5840 If midpoints correct, accept partially [= − their ] evaluated 150 150 [= 112.76] A1 AWRT 10.6 WWW. standard deviation 112.76 = 10.6 If second M1 withheld, SC B1 for 10.6 WWW. 5

Mark scheme

5(a) Method 1 [P(0, 1, 2) =] 8C2(0.75)6 (0.25)2 + 8C1(0.75)7(0.25)1 + (0.75)8 M1 One term 8Cx . [= 0.31146 + 0.26697 + 0.10011] = ( p )x (1 − p )8− x , 0 < p < 1, 0 < x < 8 . A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. = 0.679 B1 AWRT. 3 Method 2 [P(0, 1, 2) = 1 – P(3, 4, 5, 6, 7, 8) = ] 1 – {8C3(0.75)5 (0.25)3 + 8C4(0.75)4 (0.25)4 + M1 One term 8Cx ( p ) x (1 − p )8− x , 8 C5(0.75)3 (0.25)5 + 8C6(0.75)2 (0.25)6 + 8C7(0.75) (0.25)7 + (0.25)8} 0 < p < 1, 0 < x <8. A1 Correct expression, accept unsimplified, condone omission of up to 3 ‘middle’ terms, leading to final answer. = 0.679 B1 AWRT. 3 5(b) Method 1 1 − 0.756 M1 1 − 0.75n , n = 6,7 . 3367 A1 0.82202148… to at least 3SF. = 0.822, 4096 Method 2 0.25 + 0.25 × 0.75 + 0.25 × 0.752 + 0.25 × 0.753 + 0.25 × 0.754 + 0.25 × 0.755 M1 Summing 6 or 7 terms – condone extra term 0.25 × 0.756 . =0.822 A1 Method 3 1 - 0.757 − 0.25 × 0.756 M1 Correct expression. =0.822 A1 2 5(c) Method 1 P(2nd gold 5th is first unwrapped). R G RorG RorG U 0.25 × 0.3 × 0.55 × 0.55 × 0.45[ = 0.01021] M1 a × 0.3× b × c × 0.45 0 < a, b, c < 1. on its own or as a numerator a ≠ 0.3, 0.45 b, c ≠ 0.45. multiplied in that order, or correct. A1 5 correct probabilities multiplied. Method 2 P(2nd gold 5th is first unwrapped).4 possible scenarios M1 a × 0.3× b × c × 0.45 0 < a, b, c < 1. a ≠ 0.3, 0.45 b, c ≠ 0.45. RGRGU 0.25 × 0.3× 0.25 × 0.3× 0.45 [= 0.00253125] 4 terms in this form seen added on their own or as a numerator. RGRRU 0.25 × 0.3× 0.25 × 0.25 × 0.45 [= 0.002109375] A1 All probabilities correct and attempt to sum RGGRU 0.25 × 0.3× 0.3× 0.25 × 0.45 [= 0.00253125] the 4 scenarios. RGGGU 0.25 × 0.3× 0.3× 0.3 × 0.45 [= 0.0030375] [Total 0.010209375] For either approach [P(5th is first unwrapped) =] ( 0.55) ( 0.45)[ = 0.041178] B1 4 0.25 × 0.3 × 0.55 × 0.55 × 0.45 M1 their P ( 2nd gold 5th is first unwrapped ) [P(2nd is first gold | 5th is first unwrapped) =] . their P ( 5th is first unwrapped ) ( 0.55) ( 0.45) 4 Their probabilities must be clearly 0.010209375 = 0.0411778125 identified if incorrect. 30 A1 0.24793… = 0.248, 121 5(c) Method 3 M1 0.25 0.3 First chocolate is red and second gold Either or . 0.55 or 0.45 0.55 or 0.45 0.25 0.3 P(red given that it is wrapped) × P(gold given that it is wrapped) = × 0.55 0.55 A1 0.25 0.3 Either or . 0.55 0.55 B1 Both probs correct, can be unsimplified. M1 Multiplying their identified P(red given wrapped) by their identified P(gold given wrapped) or correct. 30 A1 = 0.248 , 121 5

Mark scheme

6(a) 9! B1 2!2! = 90720 1 6(b) Method 1 Total arrangements – arrangements with repeated letters at ends 9! 7! M1 7! − ×2 a− × b a = their 6(a) or correct, b = 1,2. 2!2! 2! 2! M1 7! a− × 2 a = their 6(a) or correct, c = 1,2. c! 85680 A1 FT ft their 6(a) – 5040. Method 2 Adding no of different ways P and S at ends 2 × 7! = 10080 M1 Finding correct number of ways for one of 7! these correctly identified scenarios. P or S at one end only 4×5× = 50400 2! M1 Adding no of ways for 3 correctly identified 7! scenarios. Neither P nor S at an end 5× 4× = 25200 2!2! Total 85680 A1 Method 3 7! M1 Finding correct number of ways for one of P at beginning 7× = 17640 these correctly identified scenarios. 2! 7! S at beginning 7× = 17640 M1 Adding no of ways for 3 correctly identified 2! scenarios. 8! Neither P nor S at beginning 5 × = 50400 2!2! Total 85680 A1 3 6(c) Method 1 arrangements with PP between Ss { S P P S ^ ^ ^ ^ ^ } add arrangements with PP not between Ss { (S ^ ^ S) ^ P P ^ } 6!+ 5!× 5 × 4 M1 6! + d, d an integer ≥ 1, may be implied. 6!+ 5!× 5 × 4 M1 e + 5!× f , e, f integers ≥ 1, may be implied. M1 ( ) e + g !× 5 × 4 or 5 P2 , e an integer ≥ 1, g = 4,5,6. [Total ]= 3120 A1 Method 2 - considers the 6 positions for S ^^S Positions 1 and 6 there are 5 ×5! ways M1 Identifying no of ways if S^^S is in position 1 or 6. Positions 2, 3, 4 and 5 there are 4 ×5! ways M1 Identifying no of ways if S^^S is in position 2, 3, 4 or 5. 2 ×5 × 5!+ 4 × 4 × 5! M1 Adding no of ways for 6 scenarios ( or 26 × 5!). [Total] = 3120 A1 SC B1 for 3120 if any method marks are withheld. 4 6(d) Method 1 Either PP in the group of 5 or PP in the group of 4 5 C3 5 C2 M1 a× 5 C 2 , a × 5 C 3 , or 5 C2 + 5 C3 seen as a 9 + , C5 9 C5 numerator of one or two fractions where a is 1 or 2, no extra terms. 5 C3 + 5 C2 M1 9 C5 or 9 C 4 seen (no addition, 9 C5 multiplication) as a denominator of one or two fractions. 20 10 A1 Probability = , , 0.159 126 63 Method 2 Considering the positions of P and then S 5 4 4 3 5 4 4 3 M1 a × 5 × 4 × 4 × 3 seen as a numerator of a × × × + × × × fraction. 9 8 7 6 7 6 9 8 where a = 1 or 2. 5 4 4 3 5 4 4 3 M1 9 × 8 × 7 × 6 seen as a denominator of a × × × + × × × fraction. 9 8 7 6 7 6 9 8 10 A1 = 63 3