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Mathematics 9709/51 October/November 2023: question paper and mark scheme

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

9709_w23_qp_519709_w23_ms_519709/51/O/N/23

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/51/O/N/23
9709_w23_ms_51.pdf

9709/51 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

120 100 80 Cumulative frequency 60 40 20 0 0 10 20 30 40 50 Time (seconds) The times taken by 120 children to complete a particular puzzle are represented in the cumulative frequency graph. (a) Use the graph to estimate the interquartile range of the data. [2] 35% of the children took longer than T seconds to complete the puzzle. (b) Use the graph to estimate the value of T . [2]

Mark scheme

1(a) [IQR =] 31 – 23.7 M1 30.5 < UQ < 31.25 – 23.25 < LQ ⩽ 24 Evidence of graph use must be seen at least once. 7.3 A1 7.0 ⩽ IQR ⩽ 7.5 If M0 scored, SC B1 for 7.0 ⩽ IQR ⩽ 7.5 www. 2 1(b) [65% of 120 = ]78 B1 Seen or implied by use on graph. 28.5 B1 28 < ans < 29 2 1 1 2 0.28 × ( 0.25) = 0.0175 2 1 2 1 0.28 × ( 0.25) = 0.0175 2 1 1 1 0.28 × ( 0.25) = 0.0175 2 1 1 1 0.28 × ( 0.25) = 0.0175 2 1 1 2 0.28 × ( 0.25) = 0.0175 2 1 1 3 0.28 × ( 0.25) = 0.0175 2 1 1 4 0.28 × ( 0.25) = 0.0175 2 1 2 1 0.28 × ( 0.25) = 0.0175 2 1 2 2 0.28 × ( 0.25) = 0.0175 2 1 3 1 0.28 × ( 0.25) = 0.0175 2 1 4 1 0.28 × ( 0.25) = 0.0175 2

Question 27 marks

Question

Hazeem repeatedly throws two ordinary fair 6-sided dice at the same time. On each occasion, the score is the sum of the two numbers that she obtains. (a) Find the probability that it takes exactly 5 throws of the two dice for Hazeem to obtain a score of 8 or more. [2] (b) Find the probability that it takes no more than 4 throws of the two dice for Hazeem to obtain a score of 8 or more. [2] (c) For 8 randomly chosen throws of the two dice, find the probability that Hazeem obtains a score of 8 or more on fewer than 3 occasions. [3]

Mark scheme

2(a) M1 (1 − p ) × p , 0 < p < 1 4 4 21 15 36 36 12005 A1 0.0482454… to at least 3SF. = , 0.0482 248832 2 2(b) Method 1 21 4 M1 1 − b r , b = their (1 − p) in 2(a) or correct; r = 4, 5. [P( X 4) =] 1 − 36 18335 A1 0.884211… to at least 3SF. = , 0.884 20736 2 Method 2 2 15 15 21 15 21 15 21 3 M1 p + p(1 – p) + p(1 – p)2 + p(1 – p)3 [P(X ⩽ 4) =] + × + × + × 36 36 36 36 36 36 36 ) + p (1 − p )4 ] FT from 2(a) or correct. 18335 A1 0.884211… to at least 3SF. = , 0.884 20736 2 2(c) Method 1 x( ) ( ) 0 8 1 7 2 6 M1 One term 8C q x 1 − q 8− x , 0 < q < 1, x ≠ 0,8. 5 7 8 5 7 8 5 7 [P(0,1,2) = ] 8C0 + C1 + C2 12 12 12 12 12 12 A1 FT Correct expression, accept unsimplified, no terms omitted leading to 0.01341 + 0.07661 + 0.1915 final answer. FT only with unsimplified expression. = 0.282 B1 0.2815 ⩽ q ⩽ 0.282 Method 2 x( ) ( ) 3 5 4 4 M1 One term 8C q x 1 − q 8− x , 0 < q < 1, x ≠ 0,8. 5 7 5 7 [1 – P(3,4,5,6,7,8) = ] 1 – ( C3 + 8C4 + … + 8C7 8 12 12 12 12 7 1 8 0 A1 FT Correct expression, accept unsimplified, no terms omitted leading to 5 7 8 5 7 final answer. + C8 ) 12 12 12 12 FT only with unsimplified expression. = 1 – (0.2736 + 0.2443 + … + 0.01017 + 9.084×10 ) -4 = 0.282 B1 0.2815 ⩽ q ⩽ 0.282 3 2 1 1 0.12 × ( 0.25) = 0.0075 2 0.06 A1 If A0 scored, SC B1 for 0.06 www. 3

Question 311 marks

Question

A farmer sells eggs. The weights, in grams, of the eggs can be modelled by a normal distribution with mean 80.5 and standard deviation 6.6. Eggs are classified as small, medium or large according to their weight. A small egg weighs less than 76 grams and 40% of the eggs are classified as medium. (a) Find the percentage of eggs that are classified as small. [3] (b) Find the least possible weight of an egg classified as large. [3] 150 of the eggs for sale last week were weighed. (c) Use an approximation to find the probability that more than 68 of these eggs were classified as medium. [5]

Mark scheme

3(a) 76 − 80.5 M1 Use of ± standardisation formula with 76, 80.5 and 6.6, condone [P(X < 76) =] P( Z < ) 6.6 6.62 or 6.6, no continuity correction. [= Φ(−0.6818) = 1 −Φ(0.6818) = ] M1 Calculating the appropriate probability area (leading to their final answer). 1 – 0.7524 = 0.2476 24.8% A1 24.75% < ans ⩽ 24.8% (percentage value required). If A0 scored, SC B1 for 24.75% < ans ⩽ 24.8% www. 3 3(b) [% of large eggs = 100 – 40 – 24.76 = 35.24] B1 0.378 ⩽ z < 0.3791 or −0.3791 < z ⩽ −0.378 seen. x − 80.5 [P(Z > ) = 0.40 + 0.2476 = 0.6476] M1 Use of ± standardisation formula with x, 80.5, 6.6 and a z-value (not 6.6 x − 80.5 0.6476, 0.3524, 0.4, 0.2476) (treat ±0.38 as a z-value), not 6.62 , not = 0.378 6.6 6.6 , no continuity correction. x = 83[.0] A1 awrt 83.0 3 3(c) Mean = 150 × 0.4 = 60 B1 60 and 36 seen, allow unsimplified. Var = 150 × 0.4 × 0.6 = 36 68.5 − 60 M1 Substituting their 60 and their 6 into ± standardisation formula (any P(X > 68) = P ( Z > ) number for 68.5), condone their σ2 and their √σ. 36 M1 Using continuity correction 67.5 or 68.5 in their standardisation formula. P (Z > 1.417) = 1 − Φ (1.417 ) M1 Appropriate area Φ, from final process, must be a probability. [= 1 – 0.9217] 0.0783 A1 0.07825 < p ⩽ 0.0783 If A0 scored, SC B1 for 0.07825 < p ⩽ 0.0783. 5 3 1 1 0.24 × ( 0.25) = 0.015 2 [P(product of 3 scores ⩽ 4 ∩ X is odd) = ] 0.28× ( 0.25) × 8 + 0.24 × ( 0.25 ) M1 0.28× ( 0.25) × x + 0.24× ( 0.25) , 2 2 2 2 or 0.0175 × x + 0.015 where x = 4, 5, 6, 7, or 8. Seen alone or as numerator/denominator of a conditional probability fraction.

Question 49 marks

Question

The times, to the nearest minute, of 150 athletes taking part in a charity run are recorded. The results are summarised in the table. Time in minutes 101 − 120 121 − 130 131 − 135 136 − 145 146 − 160 Frequency 18 48 34 32 18 (a) Draw a histogram to represent this information. [4] (b) Calculate estimates for the mean and standard deviation of the times taken by the athletes. [5]

Mark scheme

4(a) M1 f 18 Class 20 10 5 10 15 At least 4 frequency densities calculated by e.g. (condone width cw 20 f if unsimplified). Frequency 0.9 4.8 6.8 3.2 1.2 cw ± 0.5 density Accept unsimplified, may be read from graph using their scale, no lower than 1cm = 1 fd. A1 All bar heights correct on graph (no FT), using their suitable linear scale with at least 3 values indicated, no lower than 1cm = 1 fd. B1 Bar ends at 120.5, 130.5, 135.5, 145.5, 160.5. 5 bars drawn with a horizontal linear scale, no lower than 1 cm = 10 min, with at least 3 values indicated. 100 ⩽ horizontal scale ⩽ 160. B1 Axes labelled frequency density (fd), time (t) and minutes (min, m) oe, or an appropriate title. (Axes may be reversed). 4 4(b) [Midpoints ] 110.5 125.5 133 140.5 153 B1 At least 4 correct mid-points seen, may be by data table or used in formula. 18 ×110.5 + 48 ×125.5 + 34 ×133 + 32 ×140.5 + 18 ×153 M1 Correct formula for mean using midpoints ±0.5, condone 1 midpoint Mean = error within class. 150 1989 + 6024 + 4522 + 4496 + 2754 = 150 = 131.9 A1 9 1319 Accept 132, 131 , or . Must be identified. 10 10 Variance = M1 Appropriate variance formula with their 5 midpoints within class 18 ×110.5 + 48 × 125.5 + 34 × 133 + 32 × 140.5 + 18 × 153 2 2 2 2 2 (not upper bound, lower bound, class width, frequency density, − ( their 131.9 ) 2 frequency or cumulative frequency). Condone 1 error. 150 If correct midpoints seen, accept 3200 + 41400 + 194400 + 157300 + 153600 2630272.5 or 150 150 −{131.9 or 17397.61}. 2 [ = 137.54] A1 11.7277448… to at least 3SF. [Standard deviation =] 11.7 Accept 11.6 ⩽ σ < 11.95 www. If M0 awarded, SC B1 11.6 ⩽ σ < 11.95 www. 5

Question 59 marks

Question

A red spinner has four sides labelled 1, 2, 3, 4. When the spinner is spun, the score is the number on the side on which it lands. The random variable X denotes this score. The probability distribution table for X is given below. x 1 2 3 4 P X = x 0.28 p 2p 3p (a) Show that p = 0.12. [1] A fair blue spinner and a fair green spinner each have four sides labelled 1, 2, 3, 4. All three spinners (red, blue and green) are spun at the same time. (b) Find the probability that the sum of the three scores is 4 or less. [3] (c) Find the probability that the product of the three scores is 4 or less given that X is odd. [5]

Mark scheme

5(a) 0.28 + 6 p = 1, p = 0.12 B1 Using sum of probabilities = 1 to form an equation. Accept 0.28 + p + 2p + 3p = 1, p = 0.12. Substitution of 0.12 into the expression scores B0. 1 5(b) [For fair spinners (blue and green), probability of any score is 0.25 B1 Correct probability for 1 identified scenario, accept unsimplified, Scenarios to give total 4 or less:] www. R B G M1 Add values of 4 correct scenarios, may be implied by correct unsimplified expressions. No incorrect/repeated scenarios. 5(c) [P(X is odd) = 0.28 + 2×0.12 or 0.24 ]= 0.52[0] B1 Seen alone or as the denominator of a conditional probability fraction. Accept unsimplified. M1 Values of at least 5 identified correct scenarios added, accept R B G unsimplified, condone incorrect scenarios in calculation. 5(c) P ( product of 3 scores 4 X is odd ) M1 0.28× ( 0.25) × x + 0.24 × ( 0.25) 2 2 P ( product of 3 scores 4 | X is odd ) = = x = 4, 5, 6, 7, 8 P ( X is odd ) 0.28 + 0.24 or 0.155 0.52 their identified P ( product of 3 scores is 4 or less and X is odd ) . their identified P ( odd ) 155 31 A1 0.2980769… to at least 3SF. = 0.298, , 520 104 5 5 5

Question 610 marks

Question

Table X Table Y In a restaurant, the tables are rectangular. Each table seats four people: two along each of the longer sides of the table (see diagram). Eight friends have booked two tables, X and Y . Rajid, Sue and Tan are three of these friends. (a) The eight friends will be divided into two groups of 4, one group for table X and one group for table Y . Find the number of ways in which this can be done if Rajid and Sue must sit at the same table as each other and Tan must sit at the other table. [3] When the friends arrive at the restaurant, Rajid and Sue now decide to sit at table X on the same side as each other. Tan decides that he does not mind at which table he sits. (b) Find the number of different seating arrangements for the 8 friends. [3] As they leave the restaurant, the 8 friends stand in a line for a photograph. (c) Find the number of different arrangements if Rajid and Sue stand next to each other, but neither is at an end of the line. [4] 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

6(a) C2 × 2 M1 C2 × r, r = positive integer, 1 implied, no addition. M1 s × 2, s = 5C2 or 5P2 or if 5C2 or 5P2 not present, s = a single integer > 1 or t! × 2, 2 ⩽ t ⩽ 8, no other terms. 6(b) Method 1 6 6 C2 × 2 × 2× 2 × 4! M1 C2 × 2 × 2× 2 × t, t = positive integer ⩾ 1. 6 P2 × 2 × 2 × t, t = positive integer ⩾ 1. M1 u × 4!, u = positive integer > 1. 2880 A1 If A0 scored, SC B1 for 2880 nfww. Method 2 6! × 2 × 2 M1 6! × v, v = positive integer ⩾ 1. M1 w × 2 × 2, w = positive integer > 1. condone w × 4, w = positive integer > 1. 2880 A1 If A0 scored, SC B1 for 2880 nfww. 3 6(c) Method 1: Number of arrangements with Rajid and Sue together – Number of arrangements with Rajid and Sue together and at end of line 7! × 2 − 6! × 4 M1 7! × 2 – a, a = positive integer > 1. M1 b – 6! × 4, b = positive integer > 2880. M1 7! × c – 6! × d, c = 1,2 and d = 1, 4. = 7200 A1 If A0 scored, SC B1 for 7200 nfww. Method 2: Arrangements of 6 people and then place Rajid and Sue 6!× 2 × 5 M1 6! × e × f, e, f = positive integers ⩾ 1. M1 6! × 2 × f, f = positive integer ⩾ 1. If 5! Used, SC B1 5! × 2 × f, f = positive integer > 1. M1 6! × e × 5, e = positive integer ⩾ 1. 7200 A1 If A0, scored SC B1 for 7200 nfww. Method 3: Friends at ends picked first F ^ RS ^ ^ ^ F 6 6 P2 × 5! × 2 M1 P2 × e× f, e, f = positive integers ⩾ 1. 6 M1 P2 × 5! × f, f = positive integer ⩾ 1. Condone 6C2 × 5! × f, f = positive integer ⩾ 1. 6 M1 P2 × e × 2, e = positive integer ⩾ 1. Condone 6C2 × e × 2, e = positive integer ⩾ 1. 7200 A1 If A0 scored, SC B1 for 7200 nfww. 6(c) Method 4: RS placed in different possible positions Pn × a × ( 6 − n )! , a = positive integer, 1 ⩽ n ⩽ 5 seen once. 6 ^ RS ^ ^ ^ ^ ^ P1 × 2 × 5! = 1440 M1 6 6 ^ ^ RS ^ ^ ^ ^ P2 × 2 × 4! = 1440 6 ^ ^ ^ RS ^ ^ ^ P3 × 2 × 3! = 1440 ^ ^ ^ ^ RS ^ ^ 6 P4 × 2 × 2! = 1440 M1 6 Pn × 2 × ( 6 − n )! , a = positive integer, 1 ⩽ n ⩽ 5 seen at least 3 6 ^ ^ ^ ^ ^ RS ^ P5 × 2 × 1! = 1440 times in identified scenarios. M1 Add 5 values of appropriate scenarios only. No additional, incorrect or repeated scenarios. Accept unsimplified. 7200 A1 If A0 scored, SC B1 for 7200 nfww. 4

Question 20

Question

In the question paper above.

Mark scheme

20 A1 3

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On this page, free: the mark scheme for AS & A Level Mathematics 9709/51, October/November 2023 (9709_w23_ms_51), and the question paper (9709_w23_qp_51), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/51 for October/November 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.

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