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Mathematics 9709/52 October/November 2024: question paper and mark scheme

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

9709_w24_qp_529709_w24_ms_529709/52/O/N/24

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/52/O/N/24
9709_w24_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

At a college, the students choose exactly one of tennis, hockey or netball to play. The table shows the numbers of students in Year 1 and Year 2 at the college playing each of these sports. Tennis Hockey Netball Year 1 16 22 12 Year 2 24 18 28 One student is chosen at random from the 120 students. Events X and N are defined as follows: X: the student is in Year 1 N: the student plays netball. (a) Find P ( X N ) . [1] (b) Find P (N X ) . [1] (c) Determine whether or not X and N are independent events. [1] One of the students who plays netball takes 8 shots at goal. On each shot, the probability that she will succeed is 0.15, independently of all other shots. (d) Find the probability that she succeeds on fewer than 3 of these shots. [3]

Mark scheme

1(a) 12 B1 3 [P(X|N) =] 0.3, , 30% OE. 40 10 1 1(b) 12 B1 6 [P(N|X) =] 0.24, OE. 50 25 1 1(c) 12 40 50 B1 P ( N ) , P ( X ) and P ( N X ) or P ( N and X ) notation seen P(N X ) = , P( N ) = , P( X ) = 120 120 120 and equated to the values for P ( N ) , P ( X ) and P ( N X ) or P ( N and X ) . Calculation stated and evaluated. 40 50 5 12 × = , 0.138[8] ≠ , 0.1 Not independent Not independent clearly stated. 120 120 36 120 5 12 ≠ does not need to be stated. 36 120 All values OE. Condone consistent use of A, B etc. If values for P(N), P(X) stated, accept 5 P(N)×P(X) = . 36 1 1(d) Method 1 [P(0, 1, 2) =] (0.85)8 + 8C1(0.85)7(0.15) + 8C2(0.85)6(0.15)2 M1 One term of form 8Cx ( p ) x (1 − p )8− x . [= 0.27249 + 0.38469 + 0.23760] With 0 < p < 1, x ≠ 0 or 8. A1 Correct unsimplified expression, no terms omitted leading to final answer. = 0.895 B1 0.8945 ⩽ p ⩽ 0.895. Method 2 [P(0, 1, 2) =] 1 – {8C3(0.85)5(0.15)3 + 8C4(0.85)4(0.15)4 + 8C5(0.85)3(0.15)5 M1 One term of form 8Cx ( p ) x (1 − p )8− x + 8C6(0.85)2(0.15)6 + 8C7(0.85)(0.15)7 + (0.15)8} With 0 < p < 1, x ≠ 0 or 8. A1 Correct unsimplified expression. Condone omission of final bracket ‘}’. If other brackets omitted, allow recovery if 1 – 0.1052[…] seen. = 0.895 B1 0.8945 ⩽ p ⩽ 0.895. 3

Question 24 marks

Question

(a) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC. [1] (b) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC in which there are no more than two letters between the two As. [3]

Mark scheme

2(a) 9! B1 Exact value must be seen. 2! = 181440 CAO. 1 2(b) Method 1 B1 Correct outcome/value for 1 identified scenario, accept Scenario number of unsimplified. letters between As M1 Add values of 3 correct scenarios, no incorrect/repeated A A [^ ^ ^ ^ ^ ^ ^] 7! × 8 or 8! [ = 40320] scenarios. A ^ A [^ ^ ^ ^ ^ ^] 7! × 7 [ = 35280] A ^ ^ A [^ ^ ^ ^ ^] 7! × 6 [ = 30240] Total: 7! × (8 + 7 + 6) = 105 840 A1 If M1 not awarded, SC B1 for 105840 WWW. Method 2 B1 Correct outcome/value for 1 identified scenario, accept Scenario number of unsimplified. letters between As M1 Add values of 3 correct scenarios, no incorrect/repeated A A [^ ^ ^ ^ ^ ^ ^] 8! [ = 40320] scenarios. 7 A ^ A [^ ^ ^ ^ ^ ^] P1 × 7! or [ = 35280] 7 C1 × 7! 7 A ^ ^ A [^ ^ ^ ^ ^] P2 × 6! or [ = 30240] 7 C2 × 2 × 6! Total: 8! + 7P1 × 7! +7P2 × 6! or 8! + 7C1 × 7! +7C2 × 2 × 6! =105 840 A1 If M1 not awarded, SC B1 for 105840 WWW. 2(b) Method 3 B1 Correct outcome/value for 1 identified scenario, accept Scenario number of unsimplified. letters between As M1 their 2(a), or correct, subtract values of 5 correct scenarios, A ^ ^ ^ A [^ ^ ^ ^] 7! × 5 [ = 25200] no incorrect/repeated scenarios. A ^ ^ ^ ^ A [^ ^ ^] 7! × 4 [ = 20160] A ^ ^ ^ ^ ^ A [^ ^] 7! × 3 [ = 15120] A ^ ^ ^ ^ ^ ^ A [^] 7! × 2 [ = 10080] A^^^^^^^A 7! [× 1] [ = 5040] 9! Total = − 7!× ( 5 + 4 + 3 + 2 + 1) 2! =105 840 A1 If M1 not awarded, SC B1 for 105840 WWW. 3

Question 36 marks

Question

A fair coin and an ordinary fair six-sided dice are thrown at the same time. The random variable X is defined as follows. • If the coin shows a tail, X is twice the score on the dice. • If the coin shows a head, X is the score on the dice if the score is even and X is 0 otherwise. (a) Draw up the probability distribution table for X. [3] (b) Find Var (X ) . [3]

Mark scheme

3(a) B1 Table with correct x values and at least 2 correct x 0 2 4 6 8 10 12 probabilities. Condone any additional x values if probability stated as 0. P(X = x) 3 2 2 2 1 1 1 , , 12 12 12 12 12 12 12 1 1 0.0833 4 6 0.25 0.167 B1 Four more probabilities correctly linked to the correct x value, need not be in table, accept unsimplified. B1 7 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3 SF. SC B1 7 or more probabilities summing to 1 placed in a probability distribution table. 3 3(b) 2 2 2 1 1 1 M1 Accept unsimplified expression. May be calculated in [E(X) =] [0] + 2 × + 4 × + 6 × + 8 × + 10 × + 12 × variance. 12 12 12 12 12 12 1 2 2 5 4 8 12 8 10 12 54 Accept + + 1 + + + 1 OE for the M mark FT their [ 0] + 12 + 12 + 12 + 12 + 12 + 12 = 12 = 4.5 3 3 3 6 table with 7 or 8 probabilities summing to 0.999 ⩽ total ⩽ 1 (0 < p < 1). FT acceptable at the bold partially evaluated stage. [Var(X) =] M1 Appropriate variance formula using their (E(X))2 value. 2 2 2 1 1 1 FT their table with 6 or more probabilities. [ 0] + 2 2 × + 42 × + 62 × + 82 × + 102 × + 122 × − (their 4.5)2 (0 < p < 1) which need not sum to 1 or with an expression 12 12 12 12 12 12 no more evaluated than shown. [0] + 2 × 4 + 2 ×16 + 2 × 36 + [1×] 64 + [1×]100 + [1×]144 81 − FT acceptable at the bold partially evaluated stage with their 12 4 probabilities. = 35 − 4.52 3 A1 CAO = 14.75, 14 59 4 Accept . 4 If either or both M marks not awarded, SC B1 for correct answer WWW 3

Question 47 marks

Question

The heights, in metres, of white pine trees are normally distributed with mean 19.8 and standard deviation 2.4 . In a certain forest there are 450 white pine trees. (a) How many of these trees would you expect to have height less than 18.2 metres? [4] The heights, in metres, of red pine trees are normally distributed with mean 23.4 and standard deviation v. It is known that 26% of red pine trees have height greater than 25.5 metres. (b) Find the value of v. [3]

Mark scheme

4(a) 18.2 − 19.8 M1 Use of ± standardisation formula with 18.2, 19.8 and 2.4 [P(X < 18.2) = P( Z <] ) substituted appropriately, no continuity correction. Condone 2.4 σ 2 ( 2.42 ) or σ ( 2.4 ) . = Φ ( −0.6667 ) = 1 − Φ ( 0.6667 ) M1 Calculating the appropriate probability areas (leading to their final answer, expect < 0.5). = 1 – 0.7477 Note: 0.432 is z-value of 0.667 so is not an appropriate probability area (M0). = 0.252(3) A1 AWRT 0.252 SOI, accept 0.2525. If one or both M marks not awarded, SC B1 for AWRT 0.252 SOI, accept 0.2525. [Expected number = 0.2523 ×450 = 113.5, ] B1FT Strict FT their at least 4 figure probability × 450. = 113 or 114 (Check with calculator). One integer answer only. No indication of ‘approximation’, e.g. ≈, , about . 4 4(b) 25.5 − 23.4 B1 0.643 ⩽ z ⩽ 0.6435 or −0.6435 ⩽ z ⩽ −0.643 seen. [P(X >25.5) = 0.26 so P(Z > ) = 0.26 ] σ M1 ± standardisation formula with 25.5, 23.4, σ equating to a z- 25.5 − 23.4 = 0.643 value, (not 1 – their z-value…). σ Condone continuity correction ±0.05, not σ 2 , not σ . σ = 3.27 A1 3.26 ⩽ σ ⩽ 3.27. Do not award for improper fractions. 3

Question 56 marks

Question

In a class of 21 students, there are 10 violinists, 6 guitarists and 5 pianists. A group of 7 is to be chosen from these 21 students. The group will consist of 4 violinists, 2 guitarists and 1 pianist. (a) In how many ways can the group of 7 be chosen? [2] On another occasion a group of 5 will be chosen from the 21 students. The group must contain at least 2 violinists, at least 1 guitarist and at most 1 pianist. (b) In how many ways can the group of 5 be chosen? [4]

Mark scheme

5(a) 10 C4 × 6C2 × 5C1 M1 Ca × 6Cb × 5Cc , a + b + c = 7, a,b,c integers. 10 No other terms present but condone × 6 or × 3!. [ = 210 ×15 × 5] = 15750 A1 2 5(b) M1 One product using 2 or 3 combinations with upper numbers Scenario VGP correct and lower numbers summing to 5 and linked to a correct identified scenario. VVVVG 4 1 0 10 C4 × 6C1 [× 5C0] [1260 Condone the consistent use of permutations. ] B1 2 identified outcomes evaluated accurately, accept VVVGG 3 2 0 10 C3 × 6C2 [× 5C0] [1800 unsimplified. ] M1 Add values of 5 correct scenarios, no incorrect/repeated VVGGG 2 3 0 10 C2 × 6C3 [× 5C0] [900] scenarios. VVVGP 311 10 C3 × 6C1 × 5C1 [3600 ] VVGGP 221 10 C2 × 6C2 × 5C1 [3375 ] Total = 10935 A1 If either or both Ms not awarded, SC B1 for 10935 WWW 4

Question 610 marks

Question

Teams of 15 runners took part in a charity run last Saturday. The times taken, in minutes, to complete the course by the runners from the Falcons and the runners from the Kites are shown in the table. Falcons 38 39 42 44 46 48 50 51 52 56 58 59 64 69 76 Kites 32 40 40 45 47 48 52 54 58 59 59 60 61 63 65 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with the Falcons on the left-hand side. [4] (b) Find the median and the interquartile range of the times for the Falcons. [3] Let x and y denote the times, in minutes, of a runner from the Falcons and a runner from the Kites respectively. It is given that / x = 792 , / x 2 = 43 504 , / y = 783 , / y 2 = 42 223 . (c) Find the mean and the standard deviation of the times taken by all 30 runners from the two teams. [3]

Mark scheme

6(a) B1 Correct stem, ignore extra values (not in reverse, not split). Falcons Kites If a split stem-and-leaf plot is used (i.e. stem values are 9 8 3 2 repeated) the remaining B marks are available. 8 6 4 2 4 0 0 5 7 8 B1 Correct Falcons labelled on left, leaves in order from right to left and lined up vertically, no commas or other 9 8 6 2 1 0 5 2 4 8 9 9 punctuation. 9 4 6 0 1 3 5 B1 Correct Kites labelled on same diagram, leaves in order and lined up vertically, no commas or other punctuation. 6 7 Penalise each error only once in question. E.g. commas in both sets of data. Key: 1|5|4 means 51 minutes for Falcons and 54 minutes for Kites B1 Correct key, for their diagram, need both teams names and ‘mins’ at least once here, or in leaf headings or title. If 2 separate diagrams drawn max marks B1 if both stems correct. B1 if Falcons correct to the left of the stem. B1 if both keys correct including ‘mins’ and team name. 4 6(b) Median = 51 [minutes] B1 Accept Q2, must be identified. [IQR =] 59 – 44 M1 58 ⩽ UQ ⩽ 64 – 42 ⩽ LQ ⩽ 46. Implied if both quartile values are stated and an appropriate IQR calculated accurately. = 15 [minutes] A1 WWW 3 6(c) 792 + 783 1575 1 105 B1 1575 mean = = = 52.5, 52 , is not acceptable for this mark. 30 30 2 2 30 Σ ' x 2 ' = 85727 M1 Accept unsimplified variance formula. FT their mean. ( 43504 + 42223) 792 + 783 2 sd2 = [ Variance =] − Ignore any square root leading to sd for this mark. 30 30 2 85727 1575 − [= 101.3167] 30 30 σ= ( 101.3167 =) 10.1 A1 AWRT. Must be identified, e.g. sd, s, std d, σ . Condone ‘short’ square root signs. If M1 not awarded, SC B1 for, 6079 σ = 101.3167 or oe =10.1 . 60 3

Question 711 marks

Question

In a game, players attempt to score a goal by kicking a ball into a net. The probability that Leno scores a goal is 0.4 on any attempt, independently of all other attempts. The random variable X denotes the number of attempts that it takes Leno to score a goal. (a) Find P (X = 5) . [1] (b) Find P (3 G X G 7) . [2] (c) Find the probability that Leno scores his second goal on or before his 5th attempt. [3] Leno has 75 attempts to score a goal. (d) Use a suitable approximation to find the probability that Leno scores more than 28 goals but fewer than 35 goals. [5] Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown.

Mark scheme

7(a) 162 B1 [( 0.6) × 0.4 =] 0. 0518[4], 4 3125 1 7(b) Method 1 [P( X ≤ 7) − P( X ≤ 2 ) =] (1 – 0.67) – (1 – 0.62) M1 (1 − p ) − (1 − p ) or p − p seen, 0 < p <1. 7 2 2 7 [= 0.36 – 0.02799 ] A1 25938 25938 If M0 awarded SC B1 0.3320064 or CAO. = 0.332[0… ], 78125 78125 Method 2 [P(X = 3,4,5,6,7) = ] M1 (1 − p) × p2 + (1 − p) × p3 + (1 − p) × p4 + (1 − p) × p5 + (1 − p) × p6 0.4 × 0.62 + 0.4 × 0.63 + 0.4 × 0.64 + 0.4 × 0.65 + 0.4 × 0.66 seen, 0 < p <1. [= 0.144 + 0.0864 + 0.05184 + 0.031104 + 0.0186624] A1 25938 25938 If M0 awarded SC B1 0.3320064 or CAO. = 0.332[0… ], 78125 78125 Method 3 – geometric series [P(X = 3,4,5,6,7) = ] ( 0.144 1 − 0.65 ) M1 ( 0.144 1 − p 5 ) seen 0 < p < 1. 1 − 0.6 or 0.4 1− p 25938 A1 25938 =0.332[0…], If M0 awarded SC B1 0.3320064 or CAO. 78125 78125 2 7(c) Method 1 M1 2 correct unsimplified outcomes. 2nd goal scored on: Condone not identified but not incorrectly identified. 2nd attempt (0.4)2 [ = 0.16] M1 Add values for 4 identified correct scenarios. Condone adding values of 2nd, 3rd and 4th attempts only. 3rd attempt (0.4)2 (0.6) × (2 or 2C1) [ = 0.192] No incorrect scenarios. 4th attempt (0.4)2 (0.6)2 × (3 or 3C1) [ = 0.1728] 5th attempt (0.4)2 (0.6)3 × (4 or 4C1) [ =0.13824] 2072 A1 If either or both M marks not awarded, SC B1 for 0.663, = 0.663, 2072 3125 WWW condone 1 index error. 3125 Method 2 5 C2(0.4)2(0.6)3 +5C3(0.4)3(0.6)2 +5C4(0.4)4(0.6)1 +5C5(0.4)5 M1 At least 2 correct unsimplified terms. [0.3456 + 0.2304 + 0.0768 + 0.01024] or M1 Add values for 4 terms of the form 5Ca(0.4)a(0.6)5-a 1 – (5C0 (0.6)5 + 5C1(0.4)1(0.6)4) or 1 – sum of 2 terms of the form 5Ca(0.4)a(0.6)5-a. 2072 A1 If either or both M marks not awarded, SC B1 for 0.663 =0.663, www condone 1 index error. 3125 3 7(d) [Mean = 75 ×0.4 =] 30 B1 30 and 18 seen, allow unsimplified. May be seen in [Variance = 75 ×0.4 × 0.6 =] 18 standardisation formula. ( σ = 18, 3 2, 4.2426 σ 4.243 implies correct variance) Withold mark if variance clearly identified as standard deviation. 28.5 − 30 34.5 − 30 M1 Substituting their µ and positive σ into one ±standardising P(28 < X < 35) = P( <Z < ) formula (any number for 28.5 or 34.5), not σ2, not √σ. 18 18 M1 Using continuity corrections 27.5 or 28.5 and 34.5 or 35.5 in their 2 separate standardisation formula. Φ (1.0607 ) + Φ ( 0.3536) − 1 = M1 Appropriate area Φ, from final process. Must be a probability. = 0.8556 + 0.6383 – 1 Or 0.8556 – (1 – 0.6383) Or 0.8556 – 0.3617 Or (0.8556 – 0.5) + (0.6383 – 0.5) Or 0.3556 + 0.1383 = 0.494 A1 AWRT. 5

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Where can I find the 9709/52 October/November 2024 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, October/November 2024 (9709_w24_ms_52), and the question paper (9709_w24_qp_52), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/52 for October/November 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.

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