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Mathematics 9709/51 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_519709_s23_ms_519709/51/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
9709/51/M/J/23
9709_s23_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

A summary of 50 values of x gives Σ x − q = 700, Σ x − q2 = 14 235, where q is a constant. (a) Find the standard deviation of these values of x. [2] (b) Given that Σx = 2865, find the value of q. [2]

Mark scheme

1(a) Σ ( x − q )2 Σ ( x − q ) 2 14235 700 2 M1 14235 700 2 Var = − = − − ; where a = 49, 50, 51. a a 50 50 50 50 [ = 284.7 – 196 = 88.7] [sd = 88.7 =] 9.42 A1 9.4180677 rounded to at least 3SF. 2 1(b) Σ x − 50q = 700 M1 Forming equation with Σx, 50q and 700. [2865 – 50q = 700] 3 A1 If M0 scored, SC B1 for 43.3 WWW. q = 43.3, 43 10 2

Question 25 marks

Question

(a) Find the number of ways in which a committee of 6 people can be chosen from 6 men and 8 women if it must include 3 men and 3 women. [2] A different committee of 6 people is to be chosen from 6 men and 8 women. Three of the 6 men are brothers. (b) Find the number of ways in which this committee can be chosen if there are no restrictions on the numbers of men and women, but it must include no more than two of the brothers. [3]

Mark scheme

2(a) 6 C3 × 8C3 M1 6 C3 × b or c × 8C3 seen. b, c integers ⩾ 1 (1 may be implied). 1120 A1 2 2(b) Method 1 0 brothers [3C0] × 11C6 462 B1 3 Cx × 11C6 – x, with x = 1 or 2 seen. 1 brother 3 C1 × 11C5 1386 2 brothers 3 C2 × 11C4 990 M1 Add values of 3 correct scenarios, (may be identified by the appropriate calculations) no incorrect/repeated scenarios, condone use of permutations. 2838 A1 Only dependent on the M mark. SC B1 for the correct calculation or 2838 seen WWW. Method 2 14 C6 – 11C3 B1 14 C6 – d, where d a positive integer. 3003 – 165 M1 e – 11C3, where e is a positive integer >165. = 2838 A1 3

Question 36 marks

Question

(a) Find the number of different arrangements of the 8 letters in the word COCOONED. [1] (b) Find the number of different arrangements of the 8 letters in the word COCOONED in which the first letter is O and the last letter is N. [2] (c) Find the probability that a randomly chosen arrangement of the 8 letters in the word COCOONED has all three Os together given that the two Cs are next to each other. [3]

Mark scheme

3(a) 8! B1 2!3! = 3360 1 3(b) 6! M1 6! ; f = 1, 2, 3. 2!2! 2! f ! 180 A1 2 3(c) P ( OOO CC ) M1 5! g a positive integer, g ≠ 3360, 1. P ( OOO|CC ) = = P ( CC ) g 5! 5! Condone numerator of . 7! 3360g 3! M1 h h or , where h is a positive integer. 7! 8! 3! 3! Condone division by 3360 in denominator. 120 1 A1 0.1428571… to at least 3SF. = , , 0.143 1 840 7 If M0 scored SC B1 for WWW. 7 3

Question 49 marks

Question

A mathematical puzzle is given to a large number of students. The times taken to complete the puzzle are normally distributed with mean 14.6 minutes and standard deviation 5.2 minutes. (a) In a random sample of 250 of the students, how many would you expect to have taken more than 20 minutes to complete the puzzle? [4] All the students are given a second puzzle to complete. Their times, in minutes, are normally distributed with mean - and standard deviation 3 . It is found that 20% of the students have times less than 14.5 minutes and 67% of the students have times greater than 18.5 minutes. (b) Find the value of - and the value of 3 . [5]

Mark scheme

4(a) 20 − 14.6 M1 Use of ± standardisation formula with 20, 14.6 and 5.2 not P( Z > ) = P( Z > 1.03846) 5.2 σ 2 , not σ , no continuity correction. 1 – 0.8504 M1 Calculating the appropriate probability area (leading to their final answer). 0.150 A1 0.1496, 0.149 < p ⩽ 0.15[0] . Only dependent on the 2nd M mark so M0M1A1 possible. SC B1 for 0.149 < p ⩽ 0.15[0] if M0M0A0 awarded. [250 × their 0.1496 =] 37, 38 B1 FT Strict FT their at least 4-figure probability seen anywhere (give BOD if they go on to use 0.150). Final answer must be positive integer, no approximation or rounding stated. 4 4(b) 14.5 − μ B1 −0.843 < z1 < −0.841 or 0.841 < z1 < 0.843 . z1 = = −0.842 σ B1 −0.441 < z2 < −0.439 or 0.439 < z2 < 0.441 . 18.5 − μ z2 = = −0.44 σ M1 Use of the ±standardisation formula once with μ, σ and a z-value (not 0.20, 0.80, 0.67, 0.23, 0.5793, 0.7881, 0.7486, 0.591 or 1 - z i.e. 0.158 etc.). Condone continuity correction ±0.05, not σ 2 , σ . Solve, obtaining values for μ and σ . M1 Solve using the elimination method, substitution method or other appropriate approach to obtain values for both μ μ = 22.9, σ = 9.95 and σ. A1 AWRT 22.9, 9.95 . 5

Question 57 marks

Question

The populations of 150 villages in the UK, to the nearest hundred, are summarised in the table. Population 100 − 800 900 − 1200 1300 − 2000 2100 − 3200 3300 − 4800 Number of villages 8 12 50 48 32 (a) Draw a histogram to represent this information. [4] (b) Write down the class interval which contains the median for this information. [1] (c) Find the greatest possible value of the interquartile range for the populations of the 150 villages. [2]

Mark scheme

5(a) cw 800 400 800 1200 1600 M1 At least 4 frequency densities calculated (F/cw, e.g. 8 8 fd 0.01 0.03 0.0625 0.04 0.02 condone , 799 ⩽ n ⩽ 801 ) Accept unsimplified, 800 n may be read from graph using their scale. A1 All heights correct on graph. B1 Bar ends at 50, 850, 1250, 2050, 3250, 4850 read at the axis with a horizontal linear scale with at least 3 values indicated. 50 ⩽ horizontal scale ⩽ 4850. B1 Axes labelled frequency density (fd) and population (pop) OE, or in a title. Linear vertical scale, with at least 3 values indicated. Vertical axis must cover at least the range 0 ⩽ vertical axis ⩽ 0.0625 . Axes may be reversed. 4 5(b) 2100 – 3200 B1 Accept 2050 – 3250 OE. Condone ‘4th interval’. 1 5(c) 3249 – 1250 M1 2050 ⩽ UQ ⩽ 3250 − 1250 ⩽ LQ ⩽ 2050. 1999 A1 Condone 3250 – 1250 = 2000. 2

Question 611 marks

Question

Eli has four fair 4-sided dice with sides labelled 1, 2, 3, 4. He throws all four dice at the same time. The random variable X denotes the number of 2s obtained. 3. (a) Show that P X = 3 = 64 [2] (b) Complete the following probability distribution table for X . [2] x 0 1 2 3 4 P X = x 81 3 1 256 64 256 (c) Find E X . [2] Eli throws the four dice at the same time on 96 occasions. (d) Use an approximation to find the probability that he obtains at least two 2s on fewer than 20 of these occasions. [5]

Mark scheme

6(a) 3 1 3 M1 3 1 3 [P(X = 3) =] × × 4 × × q ; q a positive integer (1 may be implied). 4 4 4 4 3 A1 AG. = 64 2 6(b) B1 27 x 0 1 2 3 4 Either P(1) = , 0.421875 64 81 27 27 3 1 27 P(X = x) or P(2) = , 0.2109375 correct to at least 3SF. 256 64 128 64 256 128 Condone not in table. B1 FT Both values in table. 81 FT P(1) + P(2) = , 0.6328125 . 128 2 6(c) 81 27 27 12 1 M1 Correct method from their probability distribution table [E(X) =] [ 0 × ] + 1× their + 2 × their + 3× + 4× with at least 4 terms, 0 < their P(x) < 1, accept partially 256 64 128 256 256 evaluated. 27 54 36 4 = 0+ + + + 64 128 256 256 =1 A1 2 6(d) 67 B1 1 Mean = 96 × = 25.125 25.125, 25 and 18.5493… to at least 3SF seen, allow 256 8 67 189 unsimplified (4.3068 ⩽ σ ⩽ 4.307 implies correct Var = 96 × × = 18.549 variance). 256 256 19.5 − 25.125 M1 Substituting their µ and σ into ± standardisation formula P(X < 20) = P( Z < ) 18.549 (any number for 19.5). Condone σ 2 and σ . M1 Using continuity correction 19.5 or 20.5 in their standardisation formula. ±5.625 Note: seen gains M2 BOD. 18.549 [= P(Z < −1.306) = 1 – Φ (1.306)=] 1 – 0.9042 = M1 Appropriate area Φ, from final process. Must be a probability. 0.0958 A1 0.0957 ⩽ p ⩽ 0.0958 . SC B1 for 0.0957 ⩽ p ⩽ 0.0958 if B1M0M0M1 scored. 5

Question 78 marks

Question

A children’s wildlife magazine is published every Monday. For the next 12 weeks it will include a model animal as a free gift. There are five different models: tiger, leopard, rhinoceros, elephant and buffalo, each with the same probability of being included in the magazine. Sahim buys one copy of the magazine every Monday. (a) Find the probability that the first time that the free gift is an elephant is before the 6th Monday. [2] (b) Find the probability that Sahim will get more than two leopards in the 12 magazines. [3] (c) Find the probability that after 5 weeks Sahim has exactly one of each animal. [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 [P(X < 6) = P(X ⩽ 5) =] 1 − 0.85 M1 1 – 0.8r, r = 5, 6. = 0.672 A1 Method 2 [P(X < 6) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) =] M1 4 1 Condone an extra term ( )5 × . 1 4 1 4 1 4 1 4 1 5 5 + × + 2 × + 3 × + 4 × 5 5 5 5 5 5 5 5 5 First, last and one of the 3 middle terms implies M1. = 0.672 A1 2 7(b) Method 1 [1 − P(0, 1, 2)] M1 One term 12Cx ( p ) x (1 − p )12 − x , 0 < p < 1, x ≠ 0, 1, 2. = 1 – (12C0 (0.8)12 + 12C1 (0.2)(0.8)11 + 12C2 (0.2)2 (0.8)10) [= 1 – (0.06872 + 0.20615 + 0.28347)] A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. Correct unsimplified expression or better. = 0.442 B1 0.411 < p ⩽ 0.442 WWW. Method 2 [P(3,4,5,6,7,8,9,10,11,12) = ] M1 One term 12Cx ( p ) x (1 − p )12 − x , 0 < p < 1, x ≠ 0, 1, 2. 12 C3 (0.2)3 (0.8)9 + 12C4 (0.2)4 (0.8)8 + … + 12C11 (0.2)11 (0.8)1 + 12C12 (0.2)12 [= 0.23622 + 0.13288 + … + 1.966 × 10–7 + 4.096 × 10–9] A1 Correct expression, accept unsimplified, leading to final answer. Accept first, last and 8 of the middle terms. =0.442 B1 0.411 < p ⩽ 0.442 . 3 7(c) ( 0.2 )5 × 5! M1 ( 0.2 )5 × s, s a positive integer. 1 may be implied. M1 t × 5! where 0 < t < 1. 24 A1 = 0.0384, 625 Alternative Method for Question 7(c) 5 C1 × 4C1 × 3C1 × 2C1 × [ 1C1 ] M1 ( 5C1 )5 or 55 as denominator. ( 5C1 )5 M1 5 C1 × 4C1 × 3C1 × 2C1 × [ 1C1 ] or 5! as numerator. 24 A1 = 0.0384, 625 3

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Where can I find the 9709/51 May/June 2023 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/51, May/June 2023 (9709_s23_ms_51), and the question paper (9709_s23_qp_51), to read or download, with each question's marking points written out underneath.

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

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