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Mathematics 9709/51 May/June 2024: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_s24_qp_519709_s24_ms_519709/51/M/J/24

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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/24
9709_s24_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 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) 439 + 470 M1 439 + 470 909 [For all 45 values Mean =] + 30 or seen. 45 45 45 = 50.2 A1 If M0 awarded, SC B1 50.2 WWW. Alternative Method for Question 1(a) 25 × 30 + 470 + 20 × 30 + 439 (M1) 1220 + 1039 2259 [For all 45 values Mean =] or seen. 45 45 45 = 50.2 (A1) If M0 awarded, SC B1 50.2 WWW. 2 1(b) 12405 +11346 909 2 M1 their (12405 + 11346 ) or 23751 909 2 For all 45 values Sd = 2 − − their 45 45 45 45 sd [= 119.76 ] = 10.9 A1 If M0 awarded, SC B1 10.9 WWW. 2

Question 27 marks

Question

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

Question 36 marks

Question

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 1 160 160 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) M1 f 16 cw 20 10 10 5 20 At least four frequency densities calculated e.g. . cw 20 fd 0.8 3.2 7.6 12.8 0.6 f Condone if unsimplified. cw ± 0.5 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 h < 170 [UQ:] 170 h < 175 M1 170 h < 175 − 160 h < 170 175 – 160 = 15 UQ and LQ classes seen. A1 175 – 160 = 15 If M0 scored, SC B1 for 175 – 160 = 15. 2

Question 46 marks

Question

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. [1] (c) Xeno and Yao each have three turns. Find the probability that Xeno scores 2 more points than Yao. [3]

Mark scheme

4(a) Method 1 3 37 3 M1 3 1 1− (s) , s = 3 [Probability of 4 in 3 throws is] 1 − = or . 4 64 4 4 A1 AG Method 2 1 1 3 1 3 37 2 (M1) 1 3 t + t (1 − t ) + t (1 − t ) , t = 2 [Probability of 4 in 3 throws is] + × + × = or . 4 4 4 4 4 64 4 4 (A1) AG Method 3 1 3 2 1 3 2 1 37 3 (M1) 3 C1 × × + 3C2 × × + 3C3 × = 4 4 4 4 4 64 (A1) AG 2 4(b) Method 1 B1FT (1 − their (a) ) , accept unsimplified. 4 4 37 1 − = 0.0317 64 1 Method 2 3 3 6 6 (B1FT) Accept unsimplified. [Probability no 4s is] × = 0.0317 4 4 1 4(c) 3 2 B1 Correct probability for 1 identified scenario. X3 37 37 27 Accept unsimplified. Y1 × × ×3 [= 0.059645] 64 64 64 M1 Add values of 2 correct scenarios. Identification may be implied by 2 3 X2 37 27 27 correct unsimplified expressions (condone omission of × 3). Values Y0 × × ×3 [= 0.03176] may not be probabilities. 64 64 64 Probability = 0.0914 A1 If A0 scored, SC B1 for 0.0914 WWW. 3

Question 510 marks

Question

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. [3] 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. [2] 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. [5]

Mark scheme

5(a) Method 1 [P(5, 6, 7) =] M1 One term 7Cx ( p ) x (1 − p )7 − x , with 0 < p < 1, x ≠ 0 or 7. 7 C5 0.75 0.32 + 7C6 0.76 0.31 + 0.77 A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. [ = 0.31765 + 0.24706 + 0.08235] = 0.647 B1 0.647 ⩽ p < 0.6475 Method 2 [P(5, 6, 7) = 1 – P(0, 1, 2, 3, 4) =] (M1) One term 7Cx ( p ) x (1 − p )7 − x , with 0 < p < 1, x ≠ 0 or 7. 1 – {0.37 +7C1 0.71 0.36 +7C2 0.72 0.35 +7C3 0.73 0.34 +7C4 0.74 0.33} (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 5(b) Method 1 [1 – P(0 white weeks) =] M1 1 – p3, 0 < p < 1, p = 1 – their (a), or correct. 1 – (1 – 0.647)3 0.956 A1 Method 2 [P(1, 2, 3 white weeks) = ] (M1) 3 × q × (1 − q ) + 3 × q 2 × (1 − q ) + q3 , q = their (a), or correct. 2 3 × 0.647 × 0.3532 + 3 × 0.6472 × 0.353 + 0.6473 0.956 (A1) 2 5(c) [Mean = 60 ×0.8 =] 48 B1 3 48 48 and 9.6, 9 , seen, allow unsimplified. [Variance = 6 0 × 0.8 × 0.2 =] 9.6 5 5 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. 46.5 − 48 M1 Substituting their µ and σ into ± standardising formula (any number P(X < 47) = P Z < 9.6 for 46.5), not their σ2 or their σ . M1 Use continuity correction 46.5 or 47.5 in their standardised formula. ±1.5 ±1.5 Note: or seen gains M2 BOD. 9.6 3.098 [P( Z < −0.4841) = 1 − Φ ( 0.4841)] M1 Appropriate area Φ, from final process, must be a probability. Expect final answer < 0.5. 1 – 0.6858 Note: appropriate final answer implies this M1. = 0.314 A1 0.314 ⩽ p < 0.3145 5

Question 67 marks

Question

Harry has three coins: • One coin is biased so that the probability of obtaining a head when it is thrown is 13 . • The second coin is biased so that the probability of obtaining a head when it is thrown is 14 . • The third coin is biased so that the probability of obtaining a head when it is thrown is 15 . 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. [4] 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. [3]

Mark scheme

6(a) B1 Table with correct X values and at least one probability. x 0 1 2 3 Values need not be in order, lines may not be drawn, may be vertical, X and P(X) may be omitted. P(X = x) 24 26 9 1 Condone any additional X values if probability stated as 0. 60 60 60 60 B1 P(X = 1) or P(X = 2) correct and identified, need not be in table, 2 13 3 1 accept unsimplified. 5 30 20 60 B1 Two more correct and identified probabilities, need not be in table, 0.4 0.433 0.15 0.0167 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 6(b) 2 3 4 B1 2 3 4 2 × × × (1 − p ) ; × × × (1 − p ) , not × (1 − p ) ; 2 2 2 [P(Y = 0) =] Either 3 4 5 3 4 5 5 1 1 1 1 1 1 1 [P(Y = 5) =] × × × p 2 or × × × p 2 , not × p2. 3 4 5 3 4 5 60 2 3 4 1 1 1 M1 Equating and forming a 3 term quadratic equation. × × × (1 − p ) = 6 × × × × p 2 2 3 4 5 3 4 5 Their P(Y = 0) = 6 × their P(Y = 5). 24 (1 − p ) = 6 × p 2 2 3 p2 − 8 p + 4 = 0 2 A1 Not dependent on B1. p= A0 if p = 2 seen and not clearly rejected. 3 2 SC B1 if p = obtained from a correct quadratic with more than 3 three terms. If p = 2 seen and not clearly rejected, SC B0. 3

Question 710 marks

Question

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. [4] 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. [5] 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! B1 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! M1 6! − 4! × r − s , r = 1, 2 and s a positive integer (including 0). 3! 3! B1 4! Seen either alone or in t – 4!, t an integer value > 24. M1 6! × r − 4! × u , r = 1, 2 and u = 1, 2. 3! = 96 A1 4 7(c) Method 1 M1 One correct calculation, unsimplified for an identified scenario 2s 4s 1,3,5 containing 1, 2 and/or 1, 4. 3 0 0 3 C3 1 A1 Two correct outcomes evaluated, accept unsimplified. 0 1 2 3 C1 × 3C2 9 M1 Four correct scenarios added. 2 1 0 2 3 C1 × C2 6 1 1 1 2 C1 × 3C1 × 3C1 18 [Total 34 ways] [Total number of selections = ] 8C3 [= 56] B1 Used as denominator of probability expression. 34 17 A1 [Probability =] 8 = , 0.607 C3 28 7(c) Method 2 Combinations of 3 numbers (M1) One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. 1,2,3 1 C1 × 2C1 × 1C1 2 (A1) Five correct outcomes evaluated, accept unsimplified. 1,2,4 1 C1 × 2C1 × 3C1 6 (M1) Ten correct scenarios added. 1,2,5 1 C1 × 2C1 × 1C1 2 1,3,4 1 C1 × 1C1 × 3C1 3 1,3,5 1 C1 × 1C1 × 1C1 1 1,4,5 1 C1 × 3C1 × 1C1 3 2,3,4 2 C1 × 1C1 × 3C1 6 2,3,5 2 C1 × 1C1 × 1C1 2 2,4,5 2 C1 × 3C1 × 1C1 6 3,4,5 1 C1 × 3C1 × 1C1 3 8 [Total 34 ways] (B1) C3 or 56 as denominator of probability expression. [Total number of selections = ] 8C3 [= 56] 34 17 (A1) [Probability =] 8 = , 0.607 C3 28 7(c) Method 3 (M1) One correct calculation, unsimplified for an identified scenario 1,2,3 1 2 1 12 containing 1, 2 and/or 1, 4. × × × 3! 8 7 6 336 (A1) Five correct outcomes evaluated, accept unsimplified. 1,2,4 1 2 3 36 × × × 3! 8 7 6 336 (M1) Ten correct scenarios added. 1,2,5 1 2 1 12 × × × 3! 8 7 6 336 1,3,4 1 1 3 18 × × × 3! 8 7 6 336 1,3,5 1 1 1 6 × × × 3! 8 7 6 336 1,4,5 1 3 1 18 × × × 3! 8 7 6 336 2,3,4 2 1 3 36 × × × 3! 8 7 6 336 2,3,5 2 1 1 12 × × × 3! 8 7 6 336 2,4,5 2 3 1 36 × × × 3! 8 7 6 336 3,4,5 2 3 1 18 × × × 3! 8 7 6 336 7(c) (B1) 336 or 8×7×6 seen as a denominator. 17 (A1) [Probability ] = , 0.607 28 Method 4 (M1) 1-1 correct calculation, unsimplified for an identified scenario not 444 3 2 1 6 containing three 4s. × × 8 7 6 336 (A1) Five correct probabilities evaluated, accept unsimplified. 445 3 2 1 18 × × ×3 7(c) 8 7 6 336 (M1) Nine correct scenarios subtracted. 443 2 3 1 18 × × ×3 8 7 6 336 442 2 3 1 36 × × ×3 8 7 6 336 441 2 3 1 18 × × ×3 8 7 6 336 225 2 3 1 6 × × ×3 8 7 6 336 224 2 3 1 18 × × ×3 8 7 6 336 223 2 3 1 6 × × ×3 8 7 6 336 221 2 3 1 6 × × ×3 8 7 6 336 (B1) 336 or 8 × 7 × 6 seen as a denominator. 132 204 (A1) [Probability] = 1 − , , 0.607 336 336 5

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Mathematics 9709/51 for May/June 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.

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