Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 6 · Variant 3

2 The mean mass of packets of Trueleaf tea is supposed to be 500 grams. An inspector wishes to test whether this value is correct. He weighs 60 randomly chosen packets and notes the mass, x grams, of each packet. The results are summarised as follows. n = 60 / x = 29 970 / x2 = 14 970 300 Test, at the 5% significance level, whether the population mean mass is 500 grams. [8] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 est (μ) = 499.5 or 29970/60 B1 60 14970 300 M1 Biased var = 4.75 M0. est (σ2) = ( – ‘499.5’2) or 1/59(14970300 – (29970)2 /60 ) 59 60 = 4.83 or 285/59 A1 H0: Pop mean (or μ) = 500 B1 H1: Pop mean (or μ) ≠ 500 Both. Not just ‘mean’. '499.5' − 500 M1 For standardising with their values. '4.83' Must have ÷√60 . Ignore cc s. 60 = –1.762 A1 Allow −1.778 (from biased variance) accept 3 sf if nothing better. ‘1.762’ < 1.96 or −’1.762’ > −1.96 M1 For valid comparison. 0.0390 > 0.025 Allow 1.778 < 1.96 0.0377 > 0.025. There is insufficient evidence that [mean ] mass is not 500g A1FT In context. Not definite. No contradictions. FT from incorrect z. µ only accepted if defined. Accept CV method 499.5 > 499.44. Using biased var or using incorrect hypothesis can score a maximum of 6/8. 8

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3 The data produced by a certain data entry firm always include a small number of incorrect characters that occur at random. The proportion of incorrect characters is denoted by p, and experience has shown that p = 0.0001. A particular data set from the firm contains 14 500 characters, of which X characters are incorrect. (a) Use a suitable approximating distribution to find P( X 1 4) . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The firm’s management wishes to decrease the value of p by giving their employees some training. Their aim is that, for a data set containing 14 500 characters, the value of P( X = 0 ) for the new value of p should be double the value of P( X = 0 ) when p = 0.0001. (b) Use a suitable approximating distribution to find the new value of p. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) λ = 1.45 B1 Need indication of Poisson. 2 3 M1 Allow one term extra or omitted or incorrect. 1.45 1.45 e−1.45 (1 + 1.45 + + ) = e−1.45(1+1.45+1.05125+0.5081041) Allow incorrect lambda. 2! 3! = 0.94[0] or 0.941 A1 SC use of Binomial(14500, 0.0001) to find prob (< 4) = 0.940485 = 0.940 scores B2. Unjustified 0.94[0] or 0.941scores B1B1. 3 3(b) e−14500p seen or 2e−1.45 [= 0.46914 ] seen B1FT FT their λ from 3(a). e−14500p = 2e−1.45 [  −14500p = ln 2 + (−1.45)] M1 Forming an equation in p and attempting to solve. FT their e−14500p . Factor of 2 on the wrong side of the equation can still score M1. p = 0.0000522 (3 sf) A1 OE. Use of binomial reaching 0.0000522 scores SCB2. 3

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4 The masses of a certain species of animal are known to be normally distributed with standard deviation v kg. A researcher obtains the masses of a random sample of n animals of this species and uses these masses to find two confidence intervals (a% and 90%) for the population mean. The width of the a% confidence interval is .1414 # the width of the 90% confidence interval. (a) Find a. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that the 90% confidence interval contains the population mean given that the a% confidence interval contains the population mean. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a)   M1 Form an equation in z (accept z = 1.414 × 1.645 [2 ×] z × = 1.414 × [2 ×] 1.645 × [ z = 2.326 ] n n OE). Factor of 1.414 on the wrong side of the equation can still score M1. Condone 1.282 instead of 1.645 for M1. 2Φ(‘2.326’) –1 M1 OE. α = 98 (3 sf) A1 Allow 98%. 3 4(b) 90 45 B1FT FT their α. = or 0.918 (3 sf) '98' 49 no ft if alpha < 90 in 4(a). 1

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5 It is known that 20% of households in a certain country contain more than 4 people. Laxmi believes that, in her town, the percentage is lower than 20%. She chooses a random sample of 40 households in her town and notes the number which contain more than 4 people. She then carries out a test at the 2.5% significance level using a binomial distribution. (a) Find the probability of a Type I error. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State the rejection region for the test. [1] ............................................................................................................................................................ ............................................................................................................................................................ Laxmi finds that exactly 2 households in her sample contain more than 4 people. (c) Explain why it is impossible for Laxmi to make a Type II error. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) P(X < 2) = 0.840 + 40×0.839×0.2 + 40C2×0.838×0.22 M1 Attempt P(X ⩽ 2) or P(X ⩽ 3) using B(40, 0.2). = 0.0001329 + 0.0013292 + 0.0064799 Need to see expressions. = 0.00794 ( < 0.025 ) A1 SC unjustified X ⩽ 2 = 0.00794 scores M0B1. P(X < 3) = [0.00794 + 40C3×0.837×0.23] B1FT Correct term P(X = 3) added to P(X ⩽ 2) and one = 0.0079421 + 0.0205199 = 0.0285 > 0.025 relevant comparison seen. Need to see expressions. P(Type I) = 0.00794 (3 sf) with both relevant comparisons seen B1 Unjustified final answer 0.00794 scores SCB1B1. 4 5(b) Rejection region is X ⩽ 2 B1 OE. 1 5(c) H0 will be rejected B1 Or result lies in the rejection region 1

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6 The masses, in kilograms, of large and small bags of potatoes have the independent distributions N(2.5, 0.05) and N(0.8, 0.02) respectively. (a) Find the probability that the total mass of a randomly chosen large bag of potatoes and a randomly chosen small bag of potatoes is more than 3.55 kg. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that the mass of a randomly chosen large bag of potatoes is less than 3 times the mass of a randomly chosen small bag of potatoes. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) N(2.5 + 0.8 , 0.05+ 0.02) = N(3.3, 0.07) B1B1 SOI B1 for N(3.3, …) B1 for 0.07 give at early stage OE. 3.55 − '3.3' M1 For standardising with their values. [= 0.945] '0.07' 1 – ɸ('0.945') M1 For area consistent with their values. = 0.172 (3 sf) A1 5 6(b) E(D) = 2.5 – 3 x 0.8 = 0.1 B1 Give at early stage. OE. Var(D) = 0.05 + 32×0.02 = 0.23 M1 Give at early stage. 0 − ('0.1') M1 For standardising with their values. [= −0.2085 or 0.209 or 0.208] Must have used the vars for denom. '0.23' Φ(‘−0.2085’) = 1 – Φ(‘0.2085’) M1 For area consistent with their values. = 0.417 (3 sf) or 0.418 A1 5

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7 The time, in minutes, taken by students to complete a test is modelled by the random variable X with probability density function - 3 ( x - 3 )( x - 5 ) 3 G x G 5 , f ( x) = * 4 0 otherwise. (a) Find the probability that a randomly chosen student takes longer than 4.5 minutes to complete the test. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Write down the median of X. [1] ............................................................................................................................................................ ............................................................................................................................................................ (c) Without performing an integration, use your answer to part (a) to find P( 3.5 1 X 1 4 .5 ). [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) 3 [f(x) = − (x2 − 8x + 15)] 4 5 M1 Attempt to integrate their f(x). 3 2 −  ( x − 8 x + 15)d x 4 4.5 Condone missing -3/4. 3  x3 2  5 A1 Correct integration and correct limits 4.5 and 5. = − − 4 x + 15 x OE e.g. 1– integration from 3 to 4.5  4   3  4.5 condone missing –3/4. 3  125 4.53 2  M1 Attempt substitute correct limits in correct integral − − 100 + 75 − ( −4 4.5 + 15  4.5) must have the –3/4. 4  3 3    OE. May be implied by correct answer. 5 A1 = or 0.15625 or 0.156 (3 sf) 32 4 7(b) Median = 4 B1 1 7(c) 5 5 M1 5 1 − 2 × ‘ ’ or 2(0.5 − ‘ ’) Must see working. FT their . 32 32 32 11 A1FT 5 or 0.6875 or 0.687 or 0.688 (3 sf) FT their . 16 32 SC not using 7(a) and integrating from 3.5 to 4.5 and getting 11/16 OE scores B1 only. 2

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