Cambridge A Level Mathematics 9709 — 2024 May/June Paper 5 · Variant 1

1 A summary of 20 values of x gives / ( x - 30) = 439 , / ( x - 30) 2 = 12 405 . A summary of another 25 values of x gives / ( x - 30) = 470 , / ( x - 30 ) 2 = 11346 . (a) Find the mean of all 45 values of x. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the standard deviation of all 45 values of x. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) [For all 45 values Mean =] 439 470 30 45  439 470 909 or 45 45 seen. 50.2  A1 If M0 awarded, SC B1 50.2 WWW. Alternative Method for Question 1(a) [For all 45 values Mean =] 25 30 470 20 30 439 45      (M1) 1220 1039 2259 or 45 45  seen. 50.2  (A1) If M0 awarded, SC B1 50.2 WWW. 2 1(b) For all 45 values Sd2 = 2 12405 11346 909 45 45        M1   2 12405 11346 or 23751 909 45 45 their their        sd [= 119.76 ] 10.9  A1 If M0 awarded, SC B1 10.9 WWW. 2

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2 The lengths of the tails of adult raccoons of a certain species are normally distributed with mean 28 cm and standard deviation 3.3 cm. (a) Find the probability that a randomly chosen adult raccoon of this species has a tail length between 23 cm and 35 cm. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The masses of adult raccoons of this species are normally distributed with mean 8.5 kg and standard deviation v kg. 75% of adult raccoons of this species have mass greater than 7.6 kg. (b) Find the value of v. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) P(23 < X < 35) = P( 23 28 35 28 3.3 3.3   Z ) [= P( 1.515 2.121)    Z ] allow 2,  allow ,  no continuity correction. A1 One fully correct ± standardisation formula. [=     Φ 2.121 Φ 1.5151 1  ] 0.9830 0.9351 – 1   M1 Appropriate area Φ, from final process, must be a probability. = 0.918 A1 AWRT 4 2(b) [P(X > 7.6) = P( 7.6 8.5)    Z = 0.75] 7.6 8.5 0.674    B1 0.674 or – 0.674 seen. CAO as critical value. M1 Use of the ± standardisation formula with 7.6, 8.5, σ and a z-value (not 0.75, 0.25, 0.7734, 0.2266, 0.5987 nor 1 – z-value: 0.326, 0.5987). Condone use of 0.9.   1.34  A1 1.33 ⩽ σ ⩽ 1.34 3

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3 The heights, in cm, of 200 adults in Barimba are summarised in the following table. Height (h cm) 130 G h 1 150 150 G h 1 16 0 160 G h 1 170 170 G h 1 175 175 G h 1 195 Frequency 16 32 76 64 12 (a) Draw a histogram to represent this information. [4] (b) The interquartile range is R cm. Show that R is not greater than 15. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) cw 20 10 10 5 20 fd 0.8 3.2 7.6 12.8 0.6 M1 At least four frequency densities calculated f cw 16 e.g. . 20       Condone f cw 0.5  if unsimplified. Accept unsimplified, may be read from graph using their scale no lower than 1 cm = fd 2. A1 All bar heights correct on graph, not FT. Using their suitable linear scale with at least three values indicated, no lower than 1 cm = fd 2. B1 Bar ends at 150, 160, 170, 175, 195. Five bars drawn with a horizontal linear scale no lower than 1 cm = 10 cm, with at least three values indicated, 130 ⩽ horizontal scale ⩽ 195. B1 Axes labelled frequency density (fd) height (h) and cm, OE, or an appropriate title. (Axes may be reversed) 4 3(b) [LQ:] 160 170  h  [UQ:] 170 175  h  175 – 160 = 15 M1 170 175 160 170    h h   UQ and LQ classes seen. A1 175 – 160 = 15 If M0 scored, SC B1 for 175 – 160 = 15. 2

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4 A game for two players is played using a fair 4-sided dice with sides numbered 1, 2, 3 and 4. One turn consists of throwing the dice repeatedly up to a maximum of three times. When a 4 is obtained, no further throws are made during that turn. A player who obtains a 4 in their turn scores 1 point. (a) Show that the probability that a player obtains a 4 in one turn is 37.64 [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Xeno and Yao play this game. (b) Find the probability that neither Xeno nor Yao score any points in their first two turns. 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(c) Xeno and Yao each have three turns. Find the probability that Xeno scores 2 more points than Yao. 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Mark scheme: 4(a) Method 1 [Probability of 4 in 3 throws is] 3 3 37 1 4 64         M1  3 3 1 1 , or . 4 4 s s   A1 AG Method 2 [Probability of 4 in 3 throws is] 2 1 1 3 1 3 37 4 4 4 4 4 64            (M1)     2 1 3 1 1 , or . 4 4 t t t t t t      (A1) AG Method 3 3C1 1 4  2 3 4         3C2 2 1 3 4 4          3C3 3 1 37 4 64         (M1) (A1) AG 2 Question Answer Marks Guidance 4(b) Method 1 4 37 1 0.0317 64         B1FT   4 1 their  (a) , accept unsimplified. 1 Method 2 [Probability no 4s is] 6 3 4       6 3 0.0317 4       (B1FT) Accept unsimplified. 1 4(c) X3 Y1 3 2 37 37 27 3 64 64 64                [= 0.059645] X2 Y0 2 3 37 27 27 3 64 64 64                [= 0.03176] B1 Correct probability for 1 identified scenario. Accept unsimplified. M1 Add values of 2 correct scenarios. Identification may be implied by correct unsimplified expressions (condone omission of × 3). Values may not be probabilities. Probability = 0.0914 A1 If A0 scored, SC B1 for 0.0914 WWW. 3

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5 In a certain area in the Arctic the probability that it snows on any given day is 0.7, independent of all other days. (a) Find the probability that in a week (7 days) it snows on at least five days. 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A week in which it snows on at least five days out of seven is called a ‘white’ week. (b) Find the probability that in three randomly chosen weeks at least one is a white week. 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In a different area in the Arctic, the probability that a week is a white week is 0.8 . (c) Use a suitable approximation to find the probability that in 60 randomly chosen weeks fewer than 47 are white weeks. 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Mark scheme: 5(a) Method 1 [P(5, 6, 7) =] 7C5 0.75 0.32 + 7C6 0.76 0.31 + 0.77 [ = 0.31765 + 0.24706 + 0.08235] M1 One term 7Cx   7 , 1 x x p p   with 0 1, 0 p x    or 7. A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. = 0.647 B1 0.647 ⩽ p < 0.6475 Method 2 [P(5, 6, 7) = 1 – P(0, 1, 2, 3, 4) =] 1 – {0.37 +7C1 0.71 0.36 +7C2 0.72 0.35 +7C3 0.73 0.34 +7C4 0.74 0.33} (M1) One term 7Cx   7 , 1 x x p p   with 0 1, 0 p x    or 7. (A1) Correct expression, accept unsimplified, no terms omitted leading to final answer. Condone omission of final bracket ‘}’. If other brackets omitted, allow recovery if 1 – 0.35294 seen. = 0.647 (B1) 0.647 ⩽ p < 0.6475 3 Question Answer Marks Guidance 5(b) Method 1 [1 – P(0 white weeks) =] 1 – (1 – 0.647)3 M1 1 – p3, 0 < p < 1, p = 1 – their (a), or correct. 0.956 A1 Method 2 [P(1, 2, 3 white weeks) = ] 2 2 3 3 0.647 0.353 3 0.647 0.353 0.647      (M1)     2 2 3, 3 1 3 1 q q q q q       q = their (a), or correct. 0.956 (A1) 2 Question Answer Marks Guidance 5(c) [Mean = 60 0.8 ] 48   [Variance = 6 0 0.8 0.2 ] 9.6    B1 48 and 9.6, 3 48 9 , 5 5 seen, allow unsimplified. May be seen in the standardisation formula ([σ =] 3.098 ⩽ σ ⩽ 3.1[0] implies correct variance). Incorrect notation penalised but values can be used as anticipated in remainder of question. P(X < 47) = P 46.5 48 9.6 Z         M1 Substituting their µ and σ into ± standardising formula (any number for 46.5), not their σ2 or . their M1 Use continuity correction 46.5 or 47.5 in their standardised formula. Note: 1.5 1.5 or 3.098 9.6   seen gains M2 BOD. [P( 0.4841)  Z =   1 Φ 0.4841 ]  1 – 0.6858 M1 Appropriate area Φ, from final process, must be a probability. Expect final answer < 0.5. Note: appropriate final answer implies this M1. = 0.314 A1 0.314 ⩽ p < 0.3145 5

More questions on Probability

6 Harry has three coins: • One coin is biased so that the probability of obtaining a head when it is thrown is 1.3 • The second coin is biased so that the probability of obtaining a head when it is thrown is 1.4 • The third coin is biased so that the probability of obtaining a head when it is thrown is 1.5 Harry throws the three coins. The random variable X is the number of heads that he obtains. (a) Draw up the probability distribution table for X. 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Harry has two other coins, each of which is biased so that the probability of obtaining a head when it is thrown is p. He throws all five coins at the same time. The random variable Y is the number of heads that he obtains. (b) Given that P ( Y = 0 ) = 6P ( Y = 5 ) , find the value of p. 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Mark scheme: 6(a) x 0 1 2 3 P(X = x) 24 60 26 60 9 60 1 60 2 5 13 30 3 20 1 60 0.4 0.433 0.15 0.0167 B1 Table with correct X values and at least one probability. Values need not be in order, lines may not be drawn, may be vertical, X and P(X) may be omitted. Condone any additional X values if probability stated as 0. B1 P(X = 1) or P(X = 2) correct and identified, need not be in table, accept unsimplified. B1 Two more correct and identified probabilities, need not be in table, accept unsimplified. B1 4 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. SC B1 for four probabilities summing to 1 placed in a probability distribution table with the correct x values. 4 Question Answer Marks Guidance 6(b) [P(Y = 0) =]   2 2 3 4 1 ; 3 4 5    p [P(Y = 5) =] 2 1 1 1 3 4 5   p B1 Either   2 , 2 3 4 1 3 4 5 p     not   2 2 1 5 ; p   or 2, 1 1 1 3 4 5 p    not 2 1 60 . p    2 2 2 3 4 1 1 1 1 6 3 4 5 3 4 5          p p   2 2 24 1 6        p p 2 3 8 4 0    p p M1 Equating and forming a 3 term quadratic equation. Their P(Y = 0) = 6 × their P(Y = 5). 2 3  p A1 Not dependent on B1. A0 if p = 2 seen and not clearly rejected. SC B1 if 2 3  p obtained from a correct quadratic with more than three terms. If p = 2 seen and not clearly rejected, SC B0. 3

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7 The eight digits 1, 2, 2, 3, 4, 4, 4, 5 are arranged in a line. (a) How many different arrangements are there of these 8 digits? [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the number of different arrangements of the 8 digits in which there is a 2 at the beginning, a 2 at the end and the three 4s are not all together. 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Three digits are selected at random from the eight digits 1, 2, 2, 3, 4, 4, 4, 5. (c) Find the probability that the three digits are all different. 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Additional page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. ..................................................................................................................................................................

Mark scheme: 7(a) 8! 2!3!        3360 1 7(b) Number of arrangements with 2s at the end – number of arrangements with 2s at the end and the 4s together 2 _ _ _ _ _ _ 2 – 2 _ (444) _ _ 2 6! 3! 4! M1 6! 3!  r s , r = 1, 2 and s a positive integer (including 0). B1 4! Seen either alone or in t – 4!, t an integer value > 24. M1 6! 4! , 1, 2 and 1, 2. 3! r u r u      = 96 A1 4 Question Answer Marks Guidance 7(c) Method 1 2s 4s 1,3,5 0 0 3 3C3 1 0 1 2 3C1  3C2 9 1 0 2 2C1  3C2 6 1 1 1 2C1  3C1  3C1 18 [Total 34 ways] M1 One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. A1 Two correct outcomes evaluated, accept unsimplified. M1 Four correct scenarios added. [Total number of selections = ] 8C3 [= 56] B1 Used as denominator of probability expression. [Probability =] 8 3 34 17 , 0.607 28 C       A1 Question Answer Marks Guidance 7(c) Method 2 Combinations of 3 numbers 1,2,3 1C1  2C1  1C1 2 1,2,4 1C1  2C1  3C1 6 1,2,5 1C1  2C1  1C1 2 1,3,4 1C1  1C1  3C1 3 1,3,5 1C1  1C1  1C1 1 1,4,5 1C1  3C1  1C1 3 2,3,4 2C1  1C1  3C1 6 2,3,5 2C1  1C1  1C1 2 2,4,5 2C1  3C1  1C1 6 3,4,5 1C1  3C1  1C1 3 (M1) One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. (A1) Five correct outcomes evaluated, accept unsimplified. (M1) Ten correct scenarios added. [Total 34 ways] [Total number of selections = ] 8C3 [= 56] (B1) 8C3 or 56 as denominator of probability expression. [Probability =] 8 3 34 17 , 0.607 28 C       (A1) Question Answer Marks Guidance 7(c) Method 3 1,2,3 1 2 1 3! 8 7 6    12 336 1,2,4 1 2 3 3! 8 7 6    36 336 1,2,5 1 2 1 3! 8 7 6    12 336 1,3,4 1 1 3 3! 8 7 6    18 336 1,3,5 1 1 1 3! 8 7 6    6 336 1,4,5 1 3 1 3! 8 7 6    18 336 2,3,4 2 1 3 3! 8 7 6    36 336 2,3,5 2 1 1 3! 8 7 6    12 336 2,4,5 2 3 1 3! 8 7 6    36 336 3,4,5 2 3 1 3! 8 7 6    18 336 (M1) One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. (A1) Five correct outcomes evaluated, accept unsimplified. (M1) Ten correct scenarios added. Question Answer Marks Guidance 7(c) (B1) 336 or 8×7×6 seen as a denominator. [Probability ] 17 , 0.607 28  (A1) Method 4 444 3 2 1 8 7 6   6 336 445 3 2 1 3 8 7 6    18 336 443 2 3 1 3 8 7 6    18 336 442 2 3 1 3 8 7 6    36 336 441 2 3 1 3 8 7 6    18 336 225 2 3 1 3 8 7 6    6 336 224 2 3 1 3 8 7 6    18 336 223 2 3 1 3 8 7 6    6 336 221 2 3 1 3 8 7 6    6 336 (M1) 1-1 correct calculation, unsimplified for an identified scenario not containing three 4s. (A1) Five correct probabilities evaluated, accept unsimplified. 7(c) (M1) Nine correct scenarios subtracted. Question Answer Marks Guidance (B1) 336 or 8  7  6 seen as a denominator. [Probability] 132 204 1 , , 0.607 336 336  (A1) 5

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