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Mathematics 9709/53 May/June 2023: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_s23_qp_539709_s23_ms_539709/53/M/J/23

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/53/M/J/23
9709_s23_ms_53.pdf

9709/53 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 14 marks

Question

Two fair coins are thrown at the same time repeatedly until a pair of heads is obtained. The number of throws taken is denoted by the random variable X . (a) State the value of E X . [1] (b) Find the probability that exactly 5 throws are required to obtain a pair of heads. [1] (c) Find the probability that fewer than 7 throws are required to obtain a pair of heads. [2]

Mark scheme

1(a) 1 B1 P ( HH ) = 4 [E(X) =] 4 1 1(b) B1 81 3 1 4 P ( X = 5 ) = = 0.0791 1024 4 4 1 1(c) 3 6 M1 1 – pn, 0 < p < 1, n = 6, 7 [P(X < 7) = ] 1 − 4 or p + p (1 − p ) + p (1 − p ) ++ p (1 − p ) , where n = 4, 5. 2 5 2 n 1 3 1 3 1 3 1 or + × + × + + × 4 4 4 4 4 4 4 3367 A1 Accept 0.82202148… to at least 3SF. = , 0.822 4096 2

Question 25 marks

Question

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 B1 4 Var = 120 × 0.4 × 0.6 = 28.8 48 and 28 , 28.8 seen, allow unsimplified. 5 12 5 (5.366 ⩽ σ ⩽ 5.367 or implies correct variance). 5 35.5 − 48 54.5 − 48 M1 Substituting their µ and σ into one ±standardisation formula (any P( 36 X 54 ) = P( <Z < ) number for 35.5 or 54.5), condone σ2 and √σ. 28.8 28.8 M1 Using continuity correction 35.5, 36.5 or 53.5, 54.5 once in their standardisation formula. ±12.5 ±6.5 Note: or seen gains M2 BOD. 28.8 28.8 [= P( −2.3292 < Z < 1.211) =] 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

Question 37 marks

Question

The random variable X takes the values 1, 2, 3, 4. It is given that P X = x = kx x + a, where k and a are constants. (a) Given that P X = 4 = 3P X = 2, find the value of a and the value of k. [4] (b) Draw up the probability distribution table for X , giving the probabilities as numerical fractions. [1] (c) Given that E X = 3.2, find Var X . [2]

Mark scheme

3(a) [P(X = 4) = 3P(X = 2)] M1 Using P(X = 4) = 3P(X = 2) to form an equation in a and k. 4k ( 4 + a ) = 3 × 2 k ( 2 + a ) 16k + 4ak = 12k + 6ak a=2 A1 If M0 scored, SC B1 for a = 2 www. 3k + 8k + 15k + 24k = 1 M1 Using sum of probabilities = 1 to form an equation in k: k (1 + a) + 2k (2 + a) + 3k (3 + a)+ 4k (4 + a) = 1. 1 A1 1 k= If M0 scored, SC B1 for k = www. 50 50 4 3(b) B1 FT X 1 2 3 4 X 1 2 3 4 P(X) 3 8 15 24 P(X) k (1 + a) 2k (2 + a) 3k (3 + a) 4k (4 + a) ,0.06 , 0.16 , 0.3 , 0.48 50 50 50 50 0 < p < 1 for all outcomes, must be numerical. 1 3(c) 3 8 15 24 M1 Correct formula for variance method from their probability Var(X) = × 1 + × 22 + × 32 + × 42 − 3.22 distribution table, 0 ⩽ their P(x) ⩽ 1. 50 50 50 50 3 + 32 + 135 + 384 256 Accept − . 50 25 21 A1 If M0 score SC B1 for 0.84 www. [= 11.08 – 3.22 =] 0.84[0], 25 2

Question 48 marks

Question

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 B1 Correct statement comparing central tendency in context. Panthers The times for the Cheetahs are more spread than the times for the B1 Correct statement comparing range/IQR in context. Panthers 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 + their Kenny's time B1 1862 + their Kenny's time = 99 = 99 seen. 20 20 [Kenny’s time = 99 × 20 − 1862 ] M1 For their 99 ×20 – their 1862. = 118 [minutes] A1 Accept 1 hour 58 mins. 3

Question 57 marks

Question

Jasmine throws two ordinary fair 6-sided dice at the same time and notes the numbers on the uppermost faces. The events A and B are defined as follows. A: The sum of the two numbers is less than 6. B: The difference between the two numbers is at most 2. (a) Determine whether or not the events A and B are independent. [4] (b) Find P B | A′ . [3]

Mark scheme

5(a) 10 B1 10 5 P(A) = Accept P(A) = , , 0.278 36 36 18 24 24 2 P(B) = and P(B) = , , 0.667 . 36 36 3 8 B1 P(A ∩ B) = 36 10 24 M1 Their P(A) × their P(B) seen numerically, × 0 ⩽ their P(A), P(B) ⩽ 1. 36 36 5 8 A1 FT Multiplication evaluated correctly and compared with intersection = , 0.185 ≠ that is not a product of multiplication, conclusion stated, notation 27 36 P(A), P(B) and P(A ∩ B) used. Events are not independent 4 5(b) P ( B A′ ) M1 16 P ( B A′ ) = , 0.4444 P ( B|A′ ) = = P ( A′ ) 36 or their P(B) – their P(A ∩ B) seen as numerator or denominator 16 of conditional probability fraction. 36 10 M1 1 − 10 26 P ( A′ ) = 1 − , , 0.7222 or 1 – their P(A) seen as 36 36 36 denominator of conditional probability fraction. 8 A1 16 8 = Final answer , , 0.6153846 to at least 3SF. 13 26 13 Alternative Method for Question 5(b): Direct from outcome tables Number of outcomes ( B A′ ) M1 Number of outcomes ( B A′ ) = 16 seen as numerator or P ( B|A′ ) = = Number of outcomes ( A′ ) denominator of conditional probability fraction. 16 26 M1 Number of outcomes ( A′ ) = 26 seen as denominator of conditional probability fraction. A1 16 8 Final answer , , 0.6153846 to at least 3SF. 26 13 3

Question 610 marks

Question

The mass of grapes sold per day by a large shop can be modelled by a normal distribution with mean 28 kg. On 10% of days less than 16 kg of grapes are sold. (a) Find the standard deviation of the mass of grapes sold per day. [3] 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 16 kg of grapes are sold on more than 2 of these 12 days. [3] (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? [4]

Mark scheme

6(a) 16 − 28 B1 ±1.282 seen, cao – critical value. P ( X < 16 ) = P Z < σ = 0.1 M1 Use of the ±standardisation formula with 16, 28, σ and a z-value 16 − 28 (not 0.1, 0.9, 0.282, 0.5398, 0.8159) equated to a z-value. = −1.282 σ Condone continuity correct ±0.5, not σ 2 , σ . 12 Condone ± = −1.282 . σ σ = 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 M1 One term 12Cx ( p ) x (1 − p )12− x , 0 < p < 1. x ≠ 0,1,2 . (0.1)2 (0.9)10 ) [1 – (0.2824 + 0.3766 + 0.2301)] 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+ M1 One term 12Cx ( p ) x (1 − p )12− x , 0 < p < 1. x ≠ 0,1,2 . … + 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] A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. 0.111 B1 0.1108699… rounded to at least 3SF. 3 6(c) [P( −1.3 < Z < 1.3 ) B1 Identifying at least one of −1.3 or 1.3 as the appropriate z-values. = 2 Φ(1.3) – 1 ] = 2 × 0.9032 – 1 M1 Calculating the appropriate probability area from 2 symmetrical z-values (leading to their final answer, expect > 0.5). 504 A1 0.8064, 0.806 ⩽ p < 0.8065 . = 0.806, 625 [In 365 days 0.8064 ×365 ] B1 FT Strict FT their at least 4-figure probability (not z-value). = 294 or 295 Final answer must be positive integer, no approximation or rounding stated. 4

Question 79 marks

Question

(a) Find the number of different arrangements of the 10 letters in the word CASABLANCA in which the two Cs are not together. [3] (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. [3] 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. [3] 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) Method 1: Total number of arrangements – number of arrangements with Cs together 10! 9! M1 10! − [75600-15120] − c, a ≠ b, a = 1, 2, b = 1, 4, with c being a positive integer. 2!4! 4! a !b ! M1 e! d− , e = 8, 9, 10, with d being a positive integer. 4! = 60480 A1 Exact value only. SC B1 for final answer 60480 www. Method 2: Arrangements ^ ^ C ^ C ^ ^ ^ ^ ^ 8! 9 × 8 M1 8! × × f seen, with f being a positive integer. 4! 2 4! M1 9×8 g× , with g being a positive integer, h = 1, 2. h g × 9C2 and g × 9P2 are acceptable. = 60480 A1 Exact value only. SC B1 for final answer 60480 www. 3 7(b) AC^^^C^^^A M1 6! × s, with s being a positive integer. 6! 2! ×4 2! M1 t! × 4, r = 1, 2, 3 and t = 8, 7, 6. r! 1440 A1 Alternative Method for Question 7(b) 4 × 6 P3 × 3! M1 6 P3 × k, with k being a positive integer. 2! 2! M1 6 Pm 4 × 3! × , m = 2, 3 and n = 1, 2, 3. n! 1440 A1 3 7(c) Scenarios B1 Correct number of ways for identified scenarios of 2 or 3 As, 5 AA _ _ _ C3 = 10 accept unsimplified, www. 5 AAA _ _ C2 = 10 5 AAAA _ C1 = 5 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: B1 Correct total number of ways for identified scenarios of 2 or 3 As, 4 AAC _ _ C2 = 6 accept unsimplified, www (e.g., both values for AAC^^ and 4 AA _ _ _ C3 = 4 AA^^^ shown would be fine for 2As). 4 AAAC _ C1 = 4 4 AAA _ _ C2 = 6 M1 Add 6 values of appropriate scenarios only, no additional, AAAAC 1 incorrect or repeated scenarios. Accept unsimplified. AAAA _ 4 25 A1 3

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On this page, free: the mark scheme for AS & A Level Mathematics 9709/53, May/June 2023 (9709_s23_ms_53), and the question paper (9709_s23_qp_53), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/53 for May/June 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.

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