Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 6 · Variant 5

1 A student notes the length, t minutes, of certain lectures. The results for a random sample of 80 lectures are summarised as follows. n = 80 / t = 6430 / t2 = 519 740 (a) Calculate a 96% confidence interval for the population mean length of the lectures. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The method used in part (a) is valid because the sample size was large. Explain why the method would not be valid if the sample size were small. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 6430 643 B1 Accept 80.4 if nothing better seen. Est( μ ) = or or 80.375 80 8 36.61 scores M0A0.  519740  6430  2  1  6430 2  M1 Biased  2 80  = = 519740 −     −   Est( )   79 80 79 80  80      11715 A1 = 37.073  37.1 3 sf  or 316 z = 2.054 or 2.055 B1 6430 '37.073' M1 Use of correct formula, any z .  z  80 80 = 79. 0 to 81.9 or 81.8 [3 sf] A1 Must be an interval. Use of biased [79.0 to 81.8] can score B1M0A0B1M1A1. 6 1(b) CLT would not apply B1 Or distribution of the sample means is not [necessarily] normal. 1

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2 A researcher is investigating whether the proportion of families who do not own a car in his town is different from the proportion of the population in the whole country, which is 10.1%. He takes a large random sample of families in his town and finds the proportion of families that do not own a car. (a) Explain why a two-tailed test is appropriate in this context. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State suitable null and alternative hypotheses for the test. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The researcher calculates the value of the test statistic z and finds that z = 1.82. He carries out the test at the 5% significance level. (c) State the conclusion of the test, explaining your answer. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) Researcher is looking for evidence of ‘different from’ or B1 OE. researcher is not looking for ‘less than’ or ‘more than’ 1 2(b) H0: p = 0.101 H1: p  0.101 B1 [where p is proportion [in his town] not owning a car] 1 2(c) 1.82  1.96 M1 Valid comparison, accept 0.034  0.025 OE. [Do not reject H0] A1 In context, not definite, no contradictions. Insufficient evidence that [town] proportion is different from 10.1% If p is used, it must be defined. SC: if in 2(b) H1: p  0.101 [or p  0.101 ], then 1.82  1.645 or 0.0344  0.05 scores M1A0. 2

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3 A certain website receives an average of n hits per hour. In the past the value of n was 14.4. After making some improvements, the owner of the website wishes to test whether the value of n has increased. He chooses a 10-minute period at random and finds that there were 6 hits during this period. You may assume that the number of hits the website receives in any given time period follows a Poisson distribution. (a) Carry out the test at the 2.5% significance level. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Explain whether it is possible that a Type I error or a Type II error or both may have been made in carrying out the test. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) H0: = 2.4 or = 14.4 B1 OE. H1:  2.4 or  14.4 = 2.4 soi B1 −2.4  2.4 2 2.43 2.4 4 2.45  M1 Any . 1 − e  1 + 2.4 + + + +  = Allow one end error.  2! 3! 4! 5!  This expression must be seen, accept fully correct sigma 1 − e−2.4 (1 + 2.4 + 2.88 + 2.304 + 1.3824 + 0.6636 ) = notation. 1 − ( 0.09072 + 0.2177 + 0.2613 + 0.2090 + 0.1254 + 0.060196 ) = 0.0357 (3 sf) A1 SC unjustified answer of 0.0357 scores B1M0B1. '0.0357'  0.025 M1 Valid comparison. [Do not reject H0] A1FT In context, not definite, no contradictions. Insufficient evidence that  has increased Condone sufficient evidence that  has not increased OE. 6 3(b) H0 not rejected, so Type II error may have been made. B1FT OE. H0 not rejected, so Type I error cannot have been made. B1FT OE. ‘H0 not rejected’ is necessary, but only needs to be stated once 2

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4 A sports fan produces a magazine each month. (a) On average 1 in 540 characters in the magazine is incorrect. (i) Use an appropriate approximating distribution to find the probability that, in a magazine containing 2430 characters, there are at least 4 incorrect characters. [3] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Justify your approximating distribution. [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (b) On average the number of copies, X, of the magazine sold per month is 123.4. (i) State one condition for X to have a Poisson distribution. [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... You are now given that X has a Poisson distribution. (ii) Use an appropriate approximating distribution to find the probability that in a randomly chosen month, more than 130 copies of the magazine are sold. [5] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

Mark scheme: 4(a)(i) B1 SOI. = 2430 or = 4.5 540 −4.5  4.52 4.53  M1 Any . 1 − e  1 + 4.5 + +  = Allow one end error.  2! 3!  This expression must be seen, accept fully correct sigma 1 − e−4.5 (1 + 4.5 + 10.125 + 15.1875) = notation. 1 − ( 0.01111 + 0.04999 + 0.11248 + 0.16872 ) = 0.658 (3sf) A1 SC unjustified answer of 0.658 scores B1M0B1.  1  SC use of B  2430,  and probability = 0.658 scores  540  B2. Note: use of normal could score B1 only. 3 4(a)(ii)  n = 2430  50 B1 Explicit, both needed. Note: n large and p small is insufficient. and  np = 4.5  5 or  p = 0.00185  0.01 1 4b(i) Mean of X is constant B1 Any of these stated in context OE. Sales occur singly or randomly or independently of each other 1 4(b)(ii) N(123.4,123.4 ) B1 SOI. B1 for N(123.4,) B1 B1 for Var( X ) = 123.4 130.5 − '123.4' M1 For standardising with their values  = 0.639  Condone with no cc or incorrect cc. '123.4' 1 − Φ ( '0.639' ) M1 Correct area consistent with their working. = 0.261 (3 sf) A1 5

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5 The masses, in kg, of large and small bags of tomatoes have the distributions L ~ N(2.10, 0.12) and S ~ N(1.51, 0.09) respectively. (a) Find P( L 2 S + 0. 5) . [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The price of tomatoes is $4.30 per kg. (b) A large bag of tomatoes and a small bag of tomatoes are chosen at random. Find the probability that the total price of the tomatoes in the two bags is less than $16. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2.10 − 1.51,) OE.5(a) D ~ N ( 2.10 − 1.51, 0.12 + 0.09 ) = N ( 0.59, 0.21)  B1 B1 for N ( B1 B1 for Var( D) = 0.12 + 0.09 SOI. 0.5 − '0.59' M1 For standardising with their values.  = − 0.196  '0.21' ' ' M1 Correct area consistent with their working. 1 − Φ ' − 0.196 = Φ '0.196 ( ) ( ) = 0.578 (3 sf) A1 5 5(b) T ~ N B1 for N ( 4.3  ( 2.10 − 1.51) , .) ( 4.3  ( 2.10 + 1.51) , 4.32  ( 0.12 + 0.09 ) ) B1 =  N ( 15.532, 3.8829 )  M1 M1 for Var( T ) = 4.32  ( 0.12 + 0.09 ) SOI. 16 − '15.523' M1 For standardising with their values.  = 0.242  '3.8829' ' M1 Correct area consistent with their working. Φ '0.242 ( ) = 0.596 (3 sf) A1 16 − '3.61' 4.30 Accept alternative method: . '0.21' 5

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6 f ( )x a x O b 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 (b, 0) where a and b are constants. Elsewhere f ( )x = 0 . (a) Find an expression for b in terms of a. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that E ( X ) = 4 find the value of a. 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(c) Using the value of a found in part (b) find the value of k such that P( X 1 k) = 3 . [4] 4 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) 1 M1 Attempt area of triangle = 1 . ab = 1 2 b a  0 − b x + adx = 1 and attempt to integrate with correct limits to reach an equation in a and b . 2 A1 b = a 2 6(b) 2 B1FT FT their b . f ( x ) = a −a x or f ( x ) = a −a x SOI. 2 b 2 M1 Attempt to integrate xf ( x ) , ignore limits. a 2 2 b a x 2 2 2 ax − dx or x − x dx FT their linear f ( x ) .  2  b b 2 0 0 2 A1FT FT their b .  ax 2 a 2 x 3  a  2   x 2 2 x 3  b  1   −   =  or  − 2   = b  Correct integration and limits.  2 6  0  3 a   b 3b  0  3  Could be in terms of a or a and b . 2 4 1 4 2 M1 4 = or b = and b = ' ' Equate their integral to to obtain an equation in a . 3a 9 3 9 a 9 3 A1 a = 2 5 6(c) k 3 9 k 3 9 M1 Attempt to integrate f ( x ) , with limits 0 to k . ' ' − 0.5 ' − xdx  'xdx =  2 4 2 8 FT their linear f ( x ) and their a . 0 0  3 9 2  k  3  x − x =      2 16  0  4  3 9 2 3 2 M1 Attempting to find a quadratic equation in k in any form. k − k =  3k − 8k + 4 = 0    2 16 4 2 A1 ( 3k − 2 )( k − 2 ) = 0 k = 2 , k = 3 2 A1 4 k = only with explanation k  b = so reject k = 2 . 3 3 4

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