Cambridge A Level Mathematics 9709 — 2025 May/June Paper 6 · Variant 2

1 (a) One of a group of three students is to be chosen at random. Explain how a single throw of a fair six-sided dice could be used to make the choice. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The times, in minutes, taken by students to complete a test are normally distributed with mean 125 and variance 50. Two students are chosen at random. (b) Find the probability that the difference between the times taken by these two students to complete the test is more than 12 minutes. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) E.g. 1–2: choose student 1; 3–4: choose student 2; 5–6: choose student 3. B1 Other correct methods may be seen. Must be un-ambiguous i.e. if no example, need to say 2 different numbers per Note: must be a single throw. person AND different to the other two people so that all 6 numbers on dice used. 1 1(b) E(D) = 0, Var(D) = 100 B1 Or E(D)= ±12. 12 − 0 M1 For standardising with their E(D) and Var(D) Must [= 1.2] '100' be from a combination attempt, ignore cc attempts. 1 −Φ(‘1.2’) M1 For finding area consistent with their values. = 0.115 (3sf) A1 SOI by correct final answer. P(Difference > 12 minutes) = 0.23[0] A1FT FT their 0.115 (as long as ‘0.115’ < 0.5 i.e. final prob not bigger than 1). 5

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2 The height of a certain species of plant is denoted by H cm. The heights of a random sample of 100 plants were measured, and the following results were found. • The mean, h , for the sample was 80.2. • An unbiased estimate of the population variance of H was 15.6. Calculate the value of Rh2. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 100  Σh 2 2  M1 For attempt biased or unbiased = 15.6. 15.6 = − 80.2 OR 15.6 = 1/99 (Σh2 – 80202/100) 99  100    A1 For correct expression =15.6. Σh2 = 644748.4 or 645000 (3 sf) or 3223742/5 A1 3

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3 The random variable X has the distribution Po ( 15) . (a) Write down an expression in terms of e for P ( X = 12) . [1] ............................................................................................................................................................ ............................................................................................................................................................ It is given that P ( X = n) = P ( X = n + 1 ) . (b) Write down an equation in n, and hence find the value of n. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 12 B1 Seen. 15 e−15 × 12! 1 3(b) n n +1 B1 If brackets (n+1)! missing allow benefit of doubt. 15 15 e-15 × = e-15 × n ! ( n +1)! 15 M1 OE. 1 = n+1 Attempt to legitimately remove powers and factorials i.e. powers reduced to 15/λ seen, and factorials reduced to n + 1 seen. n = 14 A1 Note: Trial and error solutions: B1B2 for 14. 3

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4 A biased spinner has four sides. Each side is of a different colour: yellow, red, green or black. The probability, p, that the spinner will land on red is unknown. The spinner was spun 200 times, and the proportion, a, of times that it landed on red was noted. This proportion was used to calculate an approximate 90% confidence interval for p. The width of this confidence interval was 0.1066 correct to 4 significant figures. Find the two possible values of a. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 M1 Allow any z and/or omit ‘2 ×’. a (1 − a ) 2 × 1.645 × = 0.1066 a 200 Condone confusion between a and . 200 a  a   1 −  200  200  E.g. 2  1.645  = 0.1066 200 z = 1.645 B1 a2 – a + 0.20997 = 0 M1 Or for valid attempt to reach quadratic in a by  a  confusing between a and   .  200  E.g. a2 –200a + 8398.7 = 0 a = 0.3(00) 3sf or 0.7(00) 3sf A1 Note: Do not ISW a = 0.3  200 or 0.7  200 score A0. 4

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5 The amount of time, in minutes, spent by a customer on one visit to a certain shop is modelled by the random variable X + N ( n, v 2 ) . In the past, the values of n and v were 10.5 and 3.8 respectively. The shop has recently moved to a new location, and the manager hopes that the new value of n will be greater than 10.5. He takes a random sample of 10 customers and notes the time they each spend in the shop. He then calculates the sample mean x for these 10 times. Using a hypothesis test at the 5% significance level, the manager finds that there is sufficient evidence to conclude that the new value of n is greater than 10.5. Stating a necessary assumption, find the smallest possible value of x. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 sd (σ) remains at 3.8, or is unchanged B1 Or Var unchanged. = 1.645 M1 Any z (M0 if not a z value) Must have 10 . x −10.5 B1 z = = ±1.645 3.8 10 [Smallest value of] x = 12.5 (3 sf) A1 ISW after 12.476… or 12.5 seen. Accept x >12.476 or x >12.5 as final answer. 4

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6 Use suitable approximating distributions to answer the following. (a) The random variable W has the distribution B ( 700 , 0 .005 ) . (i) Find P ( W H 4) . 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Two values of W are chosen at random. (ii) Find the probability that the sum of these two values is less than 3. 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(b) The random variable X has the distribution Po ( 200) . Use a suitable approximating distribution to find P ( X 2 205) . 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Mark scheme: 6(a)(i) (λ) = 3.5 B1 2 3 M1 Any λ. Expression or terms must be seen. 3.5 3.5 1 − e−3.5(1 + 3.5 + + ) = 1- e-3.5 (1 + 3.5 + 6.125 + 7.1458) 2! 3! Allow one end error. Accept fully correct Σ notation. = 1 – ( 0.030197 + 0.105691+ 0.1849589 + 0.215785) = 0.463 (3 sf) A1 SC Unsupported correct answer scores B1B1 SC Use of Binomial scores B1 for 0.464. Note: Use of Normal can score B1 for mean =3.5. 3 6(a)(ii) (λ =) 7 B1 Seen. 2 M1 Any λ. Expression or terms must be seen. 7 e−7(1 + 7 + ) = e-7 ( 1 + 7 + 24.5) = 0.0009119 + 0.006383 + 0.0223411 2! Allow one end error Accept fully correct Σ notation. Accept combination method for Poisson (6 combinations) allow 6 correct combinations identified B1, attempt to calculate and combine at least four correct combinations M1. = 0.0296 (3 sf) A1 SC Unsupported correct answer scores B1B1. SC Use of Bin B1 for 0.0294. Note: Use of Normal can score B1 for mean =7. 3 6(b) N(200, 200) M1 SOI. 205.5 − 200 M1 Allow with omitted or incorrect cc. [= 0.38891] 200 1 − Φ(“0.38891”) M1 For finding area consistent with their values. = 0.349 (3 sf) A1 4

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7 The random variable X has probability density function given by kx 2 0 G x G a , f ( x) = * a 2 0 otherwise, where k and a are positive constants. 3 (a) Show that k = . 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It is given that E ( X) = 1. (b) Find the value of a. 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(c) Find the median of X. 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Mark scheme: 7(a) a M1 Attempt to integrate f(x) and =1 ignore limits. k  x 2d x = 1 2 a 0  3  a A1 Correct integration and correct limits. kx   = 1  3a 2  0 ka  = 1  k = 3 A1 AG. 3 a Must see an intermediate step and answer. Convincingly obtained no errors seen. 3 7(b) a M1 Attempt to integrate xf(x) with limits 0 and a. 3 3  x dx Condone k instead of 3/a or missing ‘k’. 3 a 0  4  a A1 Correct expression after correct integration and = 1. 3 x 3 = 1 [ a = 1 ]   3 3 4 4 Condone k instead of . a   0 a 4 A1 a = or 1.33 (3 sf) 3 3 7(c) m M1 Attempt to integrate f(x), limits 0 and m or limits m 81 2  x dx = 0.5 and a. and =0.5 (accept in terms of a and/or k. 64 0 Condone missing k. FT their a. A1FT Correct expression after correct integration with a  3  m 27 3 81 x   = 0.5  m = 0.5 and k correctly substituted (at some point) not 64 3 0 64   necessarily simplified. FT their a. A1 OE. m = 3 32 or 1.06 (3 sf) 27 CWO. 3

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8 Birgitte has a six-sided dice. She suspects that the dice is biased so that the probability, p, that it will show a six on one throw is less than 1. She throws the dice 30 times and finds that it shows a six on 6 exactly 2 throws. (a) Use a binomial distribution with a 5% significance level to test Birgitte’s suspicion. 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Later, Birgitte carries out a similar test at the 5% significance level, using another 30 throws of the dice. (b) Calculate the probability of a Type I error. 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(c) Given that the value of p is actually 0.02, calculate the probability of a Type II error. 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Mark scheme: 8(a) 1 1 B1 H0: p = H1: p < 6 6 5 30 5 29 1 5 28 1 2 M1 Expression or terms must be seen. No end errors. ( ) + 30( ) ( ) + 30C2 ( ) ( ) = 0.00421272 + 0.0252763 + 0.07330 6 6 6 6 6 = 0.103 A1 SC Unsupported correct answer scores B1. ‘0.103’ > 0.05 M1 Valid comparison (must be a tail comparison and from a Bin but not necessarily correct Bin). 1 A1FT FT their 0.103. [Accept H0] Insufficient evidence to suggest that the probability is less than 6 No contradictions, in context and non-definite. Or Insufficient evidence to support Birgitte’s suspicion Note: Condone ‘Insufficient evidence to suggest that the dice is biased’ scores A1. 5 8(b) 5 30 5 29 1 M1 1 P(X ⩽ 1) = ( ) + 30( ) ( ) (= 0.029489...) P(X ⩽ 1) attempted using B(30, ). No end errors. 6 6 6 6 P(Type I error) = 0.0295 (3 sf) A1 P(0) and P(1) expression or terms may be seen in (a). If not in (a) unsupported answer of 0.0295 scores B1. 2 8(c) P(Type II error) = 1 − P(X ⩽ 1 | p = 0.02) M1 Use of B(30,0.02) to find 1– P(X ⩽ 1). May be implied. = 1− (0.9830 + 30 × 0.9829 × 0.02) = 1 – (0.54548 + 0.33397) M1 No end errors. Expression or terms must be seen. = 0.121 (3 sf) A1 SC Unsupported correct answer scores M1B1. 3

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