Cambridge A Level Mathematics 9709 — 2025 Feb/March Paper 6 · Variant 2

1 The random variables X and Y have the independent distributions N(44, 16) and N(30, 9) respectively. Find P ( X - Y 1 15) . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 X − Y ~ N(44-30, …..) N(14, ….) B1 or X–Y–15~ N(–1, ….) soi give at early stage. Var(X − Y) = 16 + 9 = 25 or sd = 5 B1 SOI give at early stage. 15−'14' M1 Standardising with their values. [= 0.2] '25' Φ(‘0.2’) M1 For area consistent with their values. = 0.579 (3 sf) A1 5

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2 A researcher records the time, T seconds, taken by adults to complete a questionnaire. The results for a random sample of 60 adults who completed the questionnaire this year are summarised as follows. n = 60 / t = 3678 / t 2 = 226 313 .36 (a) Find an unbiased estimate of E(T ), and show that an unbiased estimate of Var(T ) is 14.44. 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In the past, the population mean time was 62.4 seconds. (b) Test at the 2% significance level whether the population mean time for this year is less than 62.4 seconds. 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(c) State, with a reason, whether it was necessary to use the Central Limit Theorem in your answer to part (b). 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Mark scheme: 2(a) ˆ = 61.3 = 3678/60 = 613/10 B1 2 60 226313.36 2 M1  = 59 ( 60 − '61.3' ) = 1/59 (226313.36 -36782 /60) [=361/25] = 14.44 AG A1 from correct expression must see 14.44. 3 2(b) H0: Population mean = 62.4 B1 Allow ‘μ’ but not just ‘mean’. H1: Population mean < 62.4 '61.3' − 62.4 M1 Standardise with their mean. 14.44 Ignore cc for M1. Must have √60. 60 = −2.242 (accept ±) A1 Accept 3sf if nothing better seen. 2.242 > 2.054 or −2.242 < −2.054 (accept 2.055) M1 or compare 1 – ɸ("2.242") with 0.02. i.e. 1 − 0.9876 = 0.0124 or 0.0125 < 0.02. [Reject H0 ] A1ft OE. Not definite, e.g. not ‘The mean time has There is sufficient evidence to suggest that (at 2% level) that the (mean) time has decreased’ No contradictions. In context. decreased. Accept cv method 61.392 M1 A1 61.392>61.3 M1 A1 OR 62.308 M1 A1 62.4>62.308 M1 A1 (3sf accuracy) SC For 2-tail test B0 M1 A1 M1(with 2.326 OE) A0 5 2(c) Yes, because population distribution of times is unknown B1 OE. Allow ‘ … is not normal’ (Accept ‘parent’ dist Accept underlying distribution). 1

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3 The random variable X has the distribution Po(1.5). (a) Find P ( X H 3) . 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(b) Find the probability that the sum of three independent values of X is between 3 and 5 inclusive. 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(c) The sum of a large number, n, of values of X is denoted by T. Using a suitable approximation, it was found that P ( T 2 330) = 0.0391, correct to 3 significant figures. Find the value of n. 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Mark scheme: 3(a) 1 − e-1.5(1 + 1.5 + 1.52 ) = 1- e-1.5(1 + 1.5 +1.125) = 1-(0.22313 +0.334695 M1 Allow one end error. Accept fully correct sigma 2 notation. +0.25102) = 0.191 A1 SC unsupported correct answer scores B1. 2 3(b) λ = 4.5 B1 4.53 4.54 4.55 M1 Any λ. Allow one end error. Accept fully correct sigma e-4.5( + + )= e-4.5 (15.1875 +17.0859 + 15.3773) 3! 4! 5! notation. = 0.1687 + 0.1898+0.1708 = 0.529 A1 SC. Unsupported correct answer scores B1 B1. 3 3(c) T ~ N(1.5n, 1.5n) B1 May be implied. Φ−1(1 − 0.0391) [= 1.761] M1 Attempted. 330.5 −1.5 n M1* Attempt to standardise and = Φ−1(1 − 0.0391). = 1.761 Allow no or incorrect cc instead of 330.5. 1.5 n Note: (330.5–1.5n)/√1.5 scores either B1M0* or B0M1* [1.5n +1.761 1.5n − 330.5 = 0 or 1.5n + 2.1568 n − 330.5 = 0 or 2.25n2 - M1dep Correctly forming and valid attempt to solve a quadratic 996.1516n +109230.25 = 0] equation in n or n or √(1.5n) OE. SOI by correct answer. [ n = 14.14 [or −15.58] ]  n = 200 (only) A1 Dep 330.5 used. 5

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4 (a) f(x) b a x 0 2 The diagram shows the graph of the probability density function, f, of a random variable X. The graph is a straight line from (0, a) to (2, b), where a and b are positive constants. Elsewhere, f ( )x = 0 . (i) Show that b = 1 - a . [2] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Given that E ( X ) = 1.2 , find the value of a. 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(b) A random variable T has probability density function given by 1 cos t - 1 r G t G 1 r, g ( t) = * 2 2 2 0 otherwise. Find the value of c such that P ( - c 1 t 1 c) = 1 . 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Mark scheme: 4(a)(i) a + b  2 = 1 or ½(b - a) x 2 + 2a = 1 or 2b - 1/2 x 2(b - a) = 1 M1 2 or 2 ( a + b −2 a x )dx = 1 and attempt to integrate  0 eg a + b = 1 or 2a + b − a = 1 or 2b – b + a = 1  b = 1 − a A1 Must see correct intermediate step and answer correctly obtained no errors seen. 2 4(a)(ii) 1−a2 b − a B1 Could be seen in 4(a)(i). y = a + 2 x or y = a + 2 x 2 2 M1 Attempt to integrate xf(x) ignore limits, ft their line ( ax + b −2 a x 2 )dx equation (of form y = mx + c and in terms of a or a and  ( ax + 1−22 a x 2 )dx or  0 0 b). 2 A1 Correct integration, ignore limits. 2  ax 2 ( b − a ) x 3   ax 2 (1− 2 a ) x 3  + + 6 [=1.2] 2 6 or 2   0   0 [2a + 43 (1 −a2 ) =1.2] M1 Substitute correct limits into their integral in terms of a, not a and b (their integral must come from xf(x)) and equated to 1.2. 4 −a2 A1 [ 3 = 1.2] a = 0.2 5 4(b) c c M1 OE. 1 cos t dt = 1 2 4 or 12 cos t dt = 12 Attempt integrate cos t with correct limits and RHS.   0 − c A1 OE. c c 1 1 1 1 sin t  = 2 Correct integral with correct limits and RHS. 2  sin t  = 4 or 2  0 − c sin c = 1 A1 2 π A1 Alone. Accept c = 0.5236. c = 6 4

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5 Amir believes that 20% of the students at his college are left-handed. His friend believes that the true proportion, p, is less than 20%. Amir plans to use the binomial distribution to test the null hypothesis, H0 : p = 0 .2 , against the alternative hypothesis, H1 : p 1 0 .2 . He decides to choose 35 students at random. If 3 or fewer of these students are left-handed, Amir will reject his belief. (a) Find the significance level of the test. 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(b) State the probability of a Type I error. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ It is now given that the true value of p is 0.05. (c) Find the probability of a Type II error. 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Mark scheme: 5(a) B(35, 0.2) used B1 May be implied. 0.835 + 35×0.834×0.2 + 35C2×0.833×0.22 + 35C3×0.832×0.23 OR M1 No end errors. Accept fully correct sigma notation. 0.0004056 + 0.0035494 + 0.015085 + 0.0414838 = 0.0605 [so significance level is] 6.05% or 6.1% or 6% (or accept anything in A1 As final answer. Must see 0.0605. range 6.05% to 14.3%) SC Unsupported correct answer 6.05% scores B1 B1. 3 5(b) 0.0605 B1 Correct or FT their <=3 (from Bin) in 5(a) must be 3sf. 1 5(c) B(35, 0.05) used B1 May be implied. 1 − (0.9535 + 35×0.9534×0.05 + 35C2× 0.9533×0.052 + 35C3× 0.9532×0.053) M1 No end errors. Accept fully correct sigma notation. 1 - (0.1661+0.3059 +0.2737 + 0.1585) = 0.0958 (3 sf) accept 0.0957 A1 SC Unsupported correct answer scores B1 B1. 3

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6 Nikki is investigating the views of students at her school about the school sports facilities. She plans to give a survey to a sample of students. Nikki’s friend says, “This survey is about sports facilities, so you should choose a sample of students from the school sports teams.” (a) State, with a reason, whether you agree with Nikki’s friend. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Nikki chooses an appropriate random sample of 60 students. She finds that 45 of these students think that the sports facilities are good. (b) Calculate an approximate 95% confidence interval for the proportion of students who think that the sports facilities are good. 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For a different investigation, Nikki uses another large random sample to calculate a 99% confidence interval and an x% confidence interval. The width of the 99% confidence interval is double the width of the x% confidence interval. (c) Calculate the value of x. 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Mark scheme: 6(a) e.g. No. The views of students in sports may be different from other students. B1 No and any sensible reason for disagree. Allow just No, because biased (or not random) or not representative. 1 6(b) 45 15 M1 Any z.  60 60 45 z 60 60 z = 1.96 B1 0.640 to 0.860 (3 sf) or 0.64 to 0.86 A1 Must be an interval. Mark at the most accurate. 3 6(c) Φ−1(0.995) [= 2.574 to 2.579] M1 Allow Φ−1(0.99). Φ(‘2.576’ ÷ 2) [= Φ(‘1.288’) = 0.901 to 0.9015] M1 FT their 2.576. ‘0.9012’ − (1 − ‘0.9012’) M1 OE. [= 0.802 to 0.803]  x = 80.2 to 80.3 or x = 80 A1 Allow x = 80%. 4

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