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Mathematics 9709/52 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_529709_s23_ms_529709/52/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
9709/52/M/J/23
9709_s23_ms_52.pdf

9709/52 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 16 marks

Question

The random variable X takes the values −2, 2 and 3. It is given that P X = x = k x2 − 1, where k is a constant. (a) Draw up the probability distribution table for X , giving the probabilities as numerical fractions. [3] (b) Find E X and Var X . [3]

Mark scheme

1(a) 1 B1 [3k + 3k + 8k = 1, so] k = 14 B1 FT Table with correct values of x, and at least one correct x −2 2 3 probability linked with outcome. FT their k. Condone any additional X values if probability stated as 3 3 8 P(x) , 0.214 ,0.214 ,0.571 0. 14 14 14 B1 FT The outcomes in the table must be –2, 2 and 3. 2 further correct probabilities in table or 3 correct probabilities not in table linked to outcomes, or 3 correct FT probabilities in table using their k, or 3 incorrect probabilities summing to 1 in table if k not stated. If k not calculated, SC B1 for the below. x −2 2 3 P(x) 3k 3k 8k 3 1(b) 3 3 8 M1 Accept unsimplified expression. May be calculated in E ( X ) = −2 × 14 + 2 × 14 + 3 × 14 = variance. FT their table with 3 probabilities summing to 0.999 ⩽ 6 6 24 total ⩽ 1 (0 < p < 1) or in terms of k. − + + 14 14 14 3 3 8 2 M1 Appropriate variance formula using their (E(X))2 value. Var ( X ) = (−2) × 14 + 2 × 14 + 3 × 14 − ( their E ( X ) ) = 2 2 2 FT their table with 3 or more probabilities (0 < p < 1) 2 which need not sum to 1, or in terms of k with an 3 3 8 12 expression no more evaluated than shown. 4× + 4 × + 9 × − their 14 14 14 7 12 + 12 + 72 12 2 − their 14 7 12 5 A1 Answers for E(X) and Var(X) must be identified. E(X) E(X) = , 1.71, 1 may be identified by correct use in Variance (condone E, 7 7 V, μ, σ2, etc.). 192 45 Var(X) = , 3.92, 3 If A0 earned, SC B1 for identified correct final answers. 49 49 3

Question 26 marks

Question

A sports event is taking place for 4 days, beginning on Sunday. The probability that it will rain on Sunday is 0.4. On any subsequent day, the probability that it will rain is 0.7 if it rained on the previous day and 0.2 if it did not rain on the previous day. (a) Find the probability that it does not rain on any of the 4 days of the event. [1] (b) Find the probability that the first day on which it rains during the event is Tuesday. [2] (c) Find the probability that it rains on exactly one of the 4 days of the event. [3]

Mark scheme

2(a) 192 B1 Exact value required [P(no rain) = 0.6 × ( 0.8) =] 0.3072, 3 625 1 2(b) 0.6 × 0.8 × 0.2 M1 a × b × c where a, b = 0.6, 0.8, c = 0.2, 0.4, 0.7 . Condone including Wednesday with both 0.3 and 0.7 used. 12 A1 = 0.096[0], 125 2 2(c) 48 B1 Correct probability for one clearly identified outcome P ( RDDD ) = 0.4 × 0.3 × 0.8 × 0.8 = 0.0768, evaluated accept unsimplified. 625 18 A correct unsimplified expression is not sufficient. P ( DRDD ) = 0.6 × 0.2 × 0.3 × 0.8 = 0.0288, 625 M1 Add 4 probability values, 0 < p < 1, for appropriate 18 P ( DDRD ) = 0.6 × 0.8 × 0.2 × 0.3 = 0.0288, identified scenarios. Accept unsimplified. 625 48 Ways of identifying scenarios for this mark: P ( DDDR ) = 0.6 × 0.8 × 0.8 × 0.2 = 0.0768, 625 Stating the days. All the unsimplified probability calculations exactly as stated in the mark scheme. Identifying the correct branches on a tree diagram and linking with the values. No repeated scenarios. No incorrect scenarios. 132 A1 Accept 0.211 0.2112, 625 132 If 0/3 scored SC B1 for 0.2112, . 625 3

Question 37 marks

Question

The following back-to-back stem-and-leaf diagram represents the monthly salaries, in dollars, of 27 employees at each of two companies, A and B. Company A Company B 5 4 1 1 0 25 4 4 5 6 6 7 9 9 8 7 2 1 0 26 0 1 3 5 5 7 9 9 8 6 4 2 1 0 27 1 3 4 6 6 8 8 6 5 4 2 0 28 0 1 2 2 2 9 8 5 29 1 30 9 . . Key: 1 27 6 means $2710 for company A and $2760 for company B (a) Find the median and the interquartile range of the monthly salaries of employees in company A. [3] The lower quartile, median and upper quartile for company B are $2600, $2690 and $2780 respectively. (b) Draw two box-and-whisker plots in a single diagram to represent the information for the salaries of employees at companies A and B. [3] (c) Comment on whether the mean would be a more appropriate measure than the median for comparing the given information for the two companies. [1]

Mark scheme

3(a) Median = 2710 B1 Must be identified, condone Q2. Ignore units throughout. 2840 – 2610 M1 2820 ⩽ UQ ⩽ 2850 – 2600 ⩽ LQ ⩽ 2620. 230 A1 www If M0 scored SC B1 for 230 www. If key ignored consistently: B0 Median = 271 SC M1 282 ⩽ UQ ⩽ 285 – 260 ⩽ LQ ⩽ 262 SC A1 23. 3 3(b) Box-and-whisker plot on provided grid. B1 All 5 key values for B plotted accurately in standard format using a linear scale with 3 identified values. Labelled B. Scale at least 1 cm = $100. B: 2540 2600 2690 2780 3090 B1FT All 5 key values for A, FT from (a), plotted accurately in A: 2500 2610 2710 2840 3010 standard format using a linear scale with 3 identified values. Labelled A. Scale at least 1cm = $100 B1 Whiskers not through box for both, not drawn at corners of boxes, single linear scale for the diagram and labelled ‘salaries’ (oe) and $. 3 3(c) Examples: B1 Must refer to company B, may be implied by appropriate Mean less appropriate than median because of extreme value for company B use of $3090. [at $3090]. Must include an indication that the mean is not No, extreme value in company B. appropriate. No, $3090 is an anomaly. No contradictory statements can be present, e.g. acceptable comment with ‘but mean could be used for company A’. Condone reference to $309. 1

Question 49 marks

Question

A fair 5-sided spinner has sides labelled 1, 2, 3, 4, 5. The spinner is spun repeatedly until a 2 is obtained on the side on which the spinner lands. The random variable X denotes the number of spins required. (a) Find P X = 4. [1] (b) Find P X < 6. [2] Two fair 5-sided spinners, each with sides labelled 1, 2, 3, 4, 5, are spun at the same time. If the numbers obtained are equal, the score is 0. Otherwise, the score is the higher number minus the lower number. (c) Find the probability that the score is greater than 0 given that the score is not equal to 2. [3] The two spinners are spun at the same time repeatedly . (d) For 9 randomly chosen spins of the two spinners, find the probability that the score is greater than 2 on at least 3 occasions. [3]

Mark scheme

4(a) 64 B1 Condone 0.102 . [P(X = 4) = ( 0.8 ) ( 0.2 ) =] 0.1024, 3 625 1 4(b) [P(X < 6 ) =] 1 − 0.85 M1 1 − 0.8d , d = 5, 6. 2101 A1 0.67232 to at least 3SF. = 0.672, 3125 2101 If M0 awarded, SC B1 for or 0.67232 only. 3125 Alternative Method for Question 4(b) 2 3 4 1 4 1 4 1 4 1 4 1 M1 If answer correct, condone omission of 2 from 3 middle [P(X < 6 ) = ] + + + + terms. 5 5 5 5 5 5 5 5 5 Allow M1 for 2 3 4 5 1 4 1 4 1 4 1 4 1 4 1 + + + + + 5 5 5 5 5 5 5 5 5 5 5 2101 A1 0.67232 to at least 3SF. = 0.672, 3125 2101 If M0 awarded, SC B1 for or 0.67232 only. 3125 2 4(c) P ( X > 0 X ≠ 2) M1 14 P ( X > 0|X ≠ 2 ) = = [P( X > 0 ∩ X ≠ 2) = ] , 0.56[0] seen as numerator or P ( X ≠ 2) 25 denominator of conditional probability fraction. 14 = 25 M1 [P(X ≠ 2) = ] 19 , 0.76[0] seen as denominator of 19 25 25 conditional probability fraction. 14 = , 0.737 19 A1 14 Final answer = , 0.7368421… to at least 3SF. 19 If A0, SC B1 for correct final answer www. Alternative Method for Question 4(c) 1 2 3 4 5 M1 Number of outcome ( X > 0 X ≠ 2 ) = 14 seen as 1 0 1 2 3 4 numerator or denominator of conditional probability 2 1 0 1 2 3 fraction. 3 2 1 0 1 2 4 3 2 1 0 1 M1 Number of outcome ( X ≠ 2 ) =19 seen as denominator 5 4 3 2 1 0 of conditional probability fraction. Number of outcome ( X > 0 X ≠ 2 ) P ( X > 0|X ≠ 2 ) = = A1 14 Number of outcomes X ≠ 2 Final answer = , 0.7368421… to at least 3SF. 19 14 , 0.737 19 3 4(d) 6 M1 One term 9Cx ( p ) x (1 − p )9− x , 0 < p < 1, 0 < x < 9. [P(X >2) = 1 – P(0, 1, 2) with p = ] 25 9 1 8 2 7 19 6 19 6 19 A1 1 – (9C0 (1− p )9 + 9C1 ( p )1 (1− p )8 1 – (9C0 + 9C1 + 9C2 ) 25 25 25 25 25 + 9C2 ( p ) (1 − p ) ) , 0 < p < 1 . 2 7 [1 – (0.08459 + 0.2404 + 0.3037)] Correct expression from their p, accept unsimplified, no terms omitted leading to final answer. Condone omission of last bracket only. 0.371 B1 0.371 ⩽ p < 0.3715 . Alternative Method for Question 4(d) 6 M1 One term 9Cx ( p ) x (1 − p )9− x , 0 < p < 1, 0 < x < 9. [P(X>2) = P(3,4,5,6,7,8,9) with p = ] 25 3 6 4 5 8 1 9 9 6 19 6 19 6 19 6 A1 9C3 ( p )3 (1− p )6 + 9C4 ( p )4 (1− p )5 + … + 9C8 C3 + 9C4 + … + 9C8 + 9C9 25 25 25 25 25 25 25 ( p ) (1− p ) + 9C9 ( p ) , 0 < p < 1 . 8 1 9 [0.2238 + 0.1060 + … + 7.529 × 10 + 2.642 × 10 ] –5 –6 Correct expression from their p, accept unsimplified, no terms omitted leading to final answer. 0.371 B1 0.371 ⩽ p < 0.3715 . 3

Question 512 marks

Question

The lengths of Western bluebirds are normally distributed with mean 16.5 cm and standard deviation 0.6 cm. A random sample of 150 of these birds is selected. (a) How many of these 150 birds would you expect to have length between 15.4 cm and 16.8 cm? [4] The lengths of Eastern bluebirds are normally distributed with mean 18.4 cm and standard deviation 3 cm. It is known that 72% of Eastern bluebirds have length greater than 17.1 cm. (b) Find the value of 3. [3] A random sample of 120 Eastern bluebirds is chosen. (c) Use an approximation to find the probability that fewer than 80 of these 120 bluebirds have length greater than 17.1 cm. [5]

Mark scheme

5(a) 15.4 − 16.5 16.8 − 16.5 M1 Use of ± standardisation formula once with 16.5, 0.6 and [P(15.4 < X < 16.8 ) =] P( <Z < ) either 15.4 or 16.8 substituted. 0.6 0.6 [= P( −1.833 < Z < 0.5) ] [= Φ ( 0.5 ) + Φ (1.833) − 1 = ] M1 Calculating the appropriate probability area (leading to 0.6915 + 0.9666 – 1 their final answer, expect > 0.5). 0.6915 – (1 – 0.9666) or (0.6915 – 0.5) + (0.9666 – 0.5) OE are alternatives. = 0.658 A1 0.658 ≤ p < 0.6585 . If A0 scored, SC B1 for 0.658 ⩽ p < 0.6585 . [Expected number =] 0.6581 × 150 B1 FT FT their 4SF (or better) probability from a normal = 98, 99 calculation. Must be a positive single integer answer. No approximation notation. 4 5(b) 17.1 − 18.4 B1 0.5825 < z ⩽ 0.583 or – 0.583 ⩽ z < – 0.5825 seen. P Z > σ = 0.72 M1 Use of the ± standardisation formula with 17.1, 18.4, σ 17.1 − 18.4 and a z-value (not 0.28, 0.72, 0.4175, 0.2358, 0.7642, = −0.583 σ 0.6103, 0.3897, …). Condone continuity correct ± 0.05, not σ 2 , σ . σ = 2.23 A1 AWRT 3 5(c) [Mean = 120 × 0.72 =] 86.4 B1 2 24 86.4, 84 and 24 , 24.192 to at least 3SF seen, allow [Var = 120 × 0.72 × 0.28 =] 24.192 5 125 unsimplified. May be seen in standardisation formula. (4.918 ⩽ σ ⩽ 4.919 implies correct variance) Incorrect notation is penalised. 79.5 − 86.4 M1 Substituting their mean (not 18.4) and their positive P(X < 80) = P( Z < ) 4.9185 into ± standardisation formula (any number for 24.192 79.5), condone their 4.9182 and √their 4.918 . M1 Using continuity correction 79.5 or 80.5 in their standardisation formula. [P( Z < −1. 4029) = 1− Φ (1.403) ] M1 Appropriate area Φ, from final process, must be a probability. Expect final answer < 0.5 . 1 − 0.9196 Note: correct final answer implies this M1. 0.0804 A1 0.0803 ⩽ p ⩽ 0.0804 5

Question 610 marks

Question

In a group of 25 people there are 6 swimmers, 8 cyclists and 11 runners. Each person competes in only one of these sports. A team of 7 people is selected from these 25 people to take part in a competition. (a) Find the number of different ways in which the team of 7 can be selected if it consists of exactly 1 swimmer, at least 4 cyclists and at most 2 runners. [4] For another competition, a team of 9 people consists of 2 swimmers, 3 cyclists and 4 runners. The team members stand in a line for a photograph. (b) How many different arrangements are there of the 9 people if the swimmers stand together, the cyclists stand together and the runners stand together? [2] (c) How many different arrangements are there of the 9 people if none of the cyclists stand next to each other? [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) S + 4C + 2R 6C1 × 8C4 × 11C2 [= 6 × 70 × 55] = 23 100 M1 6Ce × 8Cf × 11Cg, with e + f + g = 7 seen. S + 5C + 1R 6C1 × 8C5 × 11C1 [= 6 × 56 × 11] = 3696 S + 6C [+ 0R] 6C1 × 8C6 [× 11C0][= 6 × 28] = 168 B1 Correct outcome/value for 1 identified scenario, accept unsimplified, www. M1 Add values of 3 correct scenarios. No incorrect scenarios, no repeated scenarios. Condone 6Ce × 8Cf × 11Cg, with e + f + g = 7 to identify S, C, R. [Total =] 26964 A1 cao 4 6(b) 2! × 3! × 4! × 6 M1 2! × 3! × 4! × k, k an integer > 0. 1 can be implied. =1728 A1 If A0 scored SC B1 for 1728 www. 2 6(c) Method 1 6! × 7 × 6 × 5 M1 6! × k, k an integer > 0. 1 can be implied. M1 m! × 7 × n × r; 6 ⩽ m ⩽ 9; a = 1, 2; b = 1, 4; a! × b! 1 ⩽ n, r ⩽ 6, n ≠ r. M1 m! × 7 × 6 × 5; 6 ⩽ m ⩽ 9; a =1, 2; b = 1, 4. a! × b! 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. Method 2 6! × 7P3 M1 6! × k, k an integer > 0. 1 can be implied. M1 m! m! × 7Pq, or × 7 Cq × q ! ; 6 ≤ m ≤ 9; a =1, 2; a! × b! a! × b! b = 1, 4; 1 ≤ q ≤ 6. M1 m! m! × 7P3, or × 7 C3 × 3! ; 6 ≤ m ≤ 9; a =1, 2; a! × b! a! × b! b = 1, 4. 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. 6(c) Method 3 6! × 35 × 3! M1 6! × k, k an integer > 0. 1 can be implied. M1 m! × 35 × q!; 6 ⩽ m ⩽ 9; a =1, 2; b = 1, 4; a! × b! 1 ⩽ q ⩽ 3. M1 m! × 35 × 6; 6 ⩽ m ⩽ 9; a =1, 2; b = 1, 4. a! × b! 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. Method 4 9! – 7!3! – 3P2 × 6! × 7 × 6 M1 9! – 7!r! – q, r an integer > 1, q an integer ⩽ 0. 0 and 1 Or may be implied. 9! – 7!3! – 3! × 7! × 6 M1 s! – 7!3! – q; s = 8, 9; a =1, 2; b = 1, 3; c = 1, 4; [= 362 880 – 30 240 – 181 440] a ! × b! × c! q an integer ⩾ 0. 0 and 1 may be implied. M1 s! – 7!3! – 3P2 × 6! × 6 × 7, 6 ⩽ s ⩽ 9, a ! × b! × c! s! or – 7!3! – 3! × 7! × 6, 6 ⩽ s ⩽ 9. a ! × b! × c! a =1, 2 b = 1, 3 c = 1, 4. 1 may be implied. 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. 4

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Mathematics 9709/52 for May/June 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.

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