Cambridge A Level Mathematics 9709 — 2023 May/June Paper 5 · Variant 3

2 Anil is a candidate in an election. He received 40% of the votes. A random sample of 120 voters is chosen. Use an approximation to find the probability that, of the 120 voters, between 36 and 54 inclusive voted for Anil. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Mean =120 0.4 48 Var =120 0.4 0.6 28.8    B1 48 and 28 4 5 , 28.8 seen, allow unsimplified. (5.366 ⩽ σ ⩽ 5.367 or 12 5 5 implies correct variance). P(36 54 X   ) = P( 35.5 48 54.5 48) 28.8 28.8 Z     M1 Substituting their µ and σ into one ±standardisation formula (any number for 35.5 or 54.5), condone σ2 and √σ. M1 Using continuity correction 35.5, 36.5 or 53.5, 54.5 once in their standardisation formula. Note: 12.5 28.8  or 6.5 28.8  seen gains M2 BOD. [= P( 2.3292 1.211) Z    =] 0.8871 + 0.9900 – 1 M1 Appropriate area Φ, from final process. Must be a probability. Expect final answer > 0.5 . Note: correct final answer implies this M1. = 0.877 A1 0.877 ≤ p < 0.8772 . 5

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4 The times taken, in minutes, to complete a cycle race by 19 cyclists from each of two clubs, the Cheetahs and the Panthers, are represented in the following back-to-back stem-and-leaf diagram. Cheetahs Panthers 9 8 7 4 8 7 3 2 0 8 6 8 9 8 7 9 1 7 8 9 9 6 5 3 3 1 10 2 3 4 4 5 6 9 8 2 11 1 2 8 4 12 0 6 Key: 7 . 9 . 1 means 97 minutes for Cheetahs and 91 minutes for Panthers (a) Find the median and the interquartile range of the times of the Cheetahs. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The median and interquartile range for the Panthers are 103 minutes and 14 minutes respectively. (b) Make two comparisons between the times taken by the Cheetahs and the times taken by the Panthers. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Another cyclist, Kenny, from the Cheetahs also took part in the race. The mean time taken by the 20 cyclists from the Cheetahs was 99 minutes. (c) Find the time taken by Kenny to complete the race. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Median = 99 [minutes] B1 [IQR =] 106 – 83 M1 105 ⩽ UQ ⩽ 112 – 82 ⩽ LQ ⩽ 87. 23 [minutes] A1 www. If M0 scored SC B1 for 23 www. 3 4(b) The times for the Cheetahs are faster than the times for the Panthers B1 Correct statement comparing central tendency in context. The times for the Cheetahs are more spread than the times for the Panthers B1 Correct statement comparing range/IQR in context. 2 4(c) [Total time including Kenny = 99 × 20 = ]1980 B1 Accept unsimplified. [Kenny’s time =] 1980 – 1862 M1 For their 1980 – their 1862. = 118 [minutes] A1 Accept 1 hour 58 mins. Alternative Method for Question 4(c) 1862 Kenny's time 99 20 their   [Kenny’s time = 99 20 1862   ] B1 1862 Kenny's time 99 20 their   seen. M1 For their 99 ×20 – their 1862. = 118 [minutes] A1 Accept 1 hour 58 mins. 3

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6 The mass of grapes sold per day by a large shop can be modelled by a normal distribution with mean 28kg. On 10% of days less than 16kg of grapes are sold. (a) Find the standard deviation of the mass of grapes sold per day. 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The mass of grapes sold on any day is independent of the mass sold on any other day. (b) 12 days are chosen at random. Find the probability that less than 16kg of grapes are sold on more than 2 of these 12 days. 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(c) In a random sample of 365 days, on how many days would you expect the mass of grapes sold to be within 1.3 standard deviations of the mean? 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Mark scheme: 6(a)   16 28 P 16 P 0.1 X Z                   16 28 1.282    M1 Use of the ±standardisation formula with 16, 28, σ and a z-value (not 0.1, 0.9, 0.282, 0.5398, 0.8159) equated to a z-value. Condone continuity correct ±0.5, not 2,  . Condone 12 1.282    . 9.36  A1 3 6(b) [1 − P(0, 1, 2) =] 1 – (12C0(0.1)0 (0.9)12 + 12C1 (0.1)1 (0.9)11 + 12C2 (0.1)2 (0.9)10 ) [1 – (0.2824 + 0.3766 + 0.2301)] M1 One term 12Cx   12 1 x x p p   , 0 1 p  . 0,1,2 x  . A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. 0.111 B1 0.1108699… rounded to at least 3SF. Alternative Method for Question 6(b) P(3,4,5,6,7,8,9,10,11,12) = 12C3 (0.1)3 (0.9)9 + 12C4 (0.1)4 (0.9)8+ … + 12C11 (0.1)11 (0.9)1 + 12C12 (0.1)12 (0.9)0 [0.08523 + 0.02131 + … + 1.08×10-10 + 1×10-12] M1 One term 12Cx   12 1 x x p p   , 0 1 p  . 0,1,2 x  . A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. 0.111 B1 0.1108699… rounded to at least 3SF. 3 Question Answer Marks Guidance 6(c) [P( 1.3 1.3 Z    ) = 2 Φ(1.3) – 1 ] = 2 × 0.9032 – 1 B1 Identifying at least one of −1.3 or 1.3 as the appropriate z-values. M1 Calculating the appropriate probability area from 2 symmetrical z-values (leading to their final answer, expect > 0.5). = 0.806, 504 625 A1 0.8064, 0.806 ⩽ p < 0.8065 . [In 365 days 0.8064 365  ] = 294 or 295 B1 FT Strict FT their at least 4-figure probability (not z-value). Final answer must be positive integer, no approximation or rounding stated. 4 Question Answer Marks Guidance 7(a) Method 1: Total number of arrangements – number of arrangements with Cs together 10! 9! 2!4! 4!  [75600-15120] M1 10! , ! ! c a b  a ≠ b, a = 1, 2, b = 1, 4, with c being a positive integer. M1 ! 4! e d  , e = 8, 9, 10, with d being a positive integer. = 60480 A1 Exact value only. SC B1 for final answer 60480 www. Method 2: Arrangements ^ ^ C ^ C ^ ^ ^ ^ ^ 8! 9 8 4! 2   M1 8! 4! f  seen, with f being a positive integer. M1 9 8 g h   , with g being a positive integer, h = 1, 2. g × 9C2 and g × 9P2 are acceptable. = 60480 A1 Exact value only. SC B1 for final answer 60480 www. 3

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7 (a) Find the number of different arrangements of the 10 letters in the word CASABLANCA in which the two Cs are not together. 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(b) Find the number of different arrangements of the 10 letters in the word CASABLANCA which have an A at the beginning, an A at the end and exactly 3 letters between the 2 Cs. 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Five letters are selected from the 10 letters in the word CASABLANCA. (c) Find the number of different selections in which the five letters include at least two As and at most one C. 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Mark scheme: 7(b) AC^^^C^^^A 6! 2! × 4 6! 2! × s, with s being a positive integer. M1 ! ! t r × 4, r = 1, 2, 3 and t = 8, 7, 6. 1440 A1 Alternative Method for Question 7(b) 6 3 4 P 3! 2!   M1 6 3P 2! × k, with k being a positive integer. M1 4 × 3! × 6P ! m n , m = 2, 3 and n = 1, 2, 3. 1440 A1 3 Question Answer Marks Guidance 7(c) Scenarios AA _ _ _ 5C3 = 10 AAA _ _ 5C2 = 10 AAAA _ 5C1 = 5 B1 Correct number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www. M1 Add 3 values for 2, 3 and 4 As, no additional, incorrect or repeated scenarios. Accept unsimplified. 25 A1 Alternative Method 2 for Question 7(c) Scenarios: AAC _ _ 4C2 = 6 AA _ _ _ 4C3 = 4 AAAC _ 4C1 = 4 AAA _ _ 4C2 = 6 AAAAC 1 AAAA _ 4 B1 Correct total number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www (e.g., both values for AAC^^ and AA^^^ shown would be fine for 2As). M1 Add 6 values of appropriate scenarios only, no additional, incorrect or repeated scenarios. Accept unsimplified. 25 A1 3

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