Cambridge A Level Mathematics 9709 — 2024 Feb/March Paper 6 · Variant 2

1 The lengths, X cm, of a sample of 100 insects of a certain type were summarised as follows. n = 100 / x = 36.8 / x 2 = 17.34 (a) Calculate unbiased estimates for the population mean and variance of X. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State a necessary condition for the estimates found in part (a) to be reliable. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 46 B1 Oe. est(μ) = 0.368 = 125 100  17.34 2  1  36.8 2  M1 For use of a correct formula (ft their µ). est(σ2) =  − their '0.368'  or  17.34 –   100  99  100  99   = 0.0384 (3 sf) A1 3 1(b) Must be a random sample B1 E.g. • Values must have been randomly selected. • Sample should be representative of the population. • All values should have equal chance of being selected. • It should be an unbiased sample. • Independent sample/insect lengths are independent of one another. ISW 1

More questions on Representation of data

2 A random sample of 250 people living in Barapet was chosen. It was found that 78 of these people owned a BETEC phone. (a) Calculate an approximate 98% confidence interval for the proportion of people living in Barapet who own a BETEC phone. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Manjit claims that more than 40% of the people living in Barapet own a BETEC phone. Use your answer to part (a) to comment on this claim. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 78 78 M1 Use of a correct formula (any z). −(1 ) 78 250 250 ± z × 250 250 z = 2.326 B1 = 0.244 to 0.38[0] (3 sf) A1 Must be an interval. 3 2(b) Unlikely to be true because confidence interval does not contain 0.4. B1 ft FT their confidence interval. Must include this reason and ‘unlikely’, oe. Allow “not true because 0.4 is not in the confidence interval.” But “Confidence interval only goes up to 0.38 so not true” and “‘it’ lies outside the confidence interval” both score B0. 1

More questions on Sampling and estimation

3 In a certain lottery, on average 1 in every 10 000 tickets is a prize-winning ticket. An agent sells 6000 tickets. (a) Use a suitable approximating distribution to find the probability that at least 3 of the tickets sold by the agent are prize-winning tickets. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Justify the use of your approximating distribution in this context. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) [λ =] 0.6 B1 Mean = 0.6 seen.  0.6 2  M1 Any λ Allow one end error. 1 – e─0.6  1 + 0.6 +  Must see expression.  2    Accept correct Σ notation. or 1 – e-0.6 (1 + 0.6 + 0.18) or 1 – (0.5488 + 0.3293 + 0.09879) = 0.0231 A1 SC 0.0231 and no working scores B1 (could be implied). SC use of binomial scores M1A1 for 0.0231. 3 3(b) 1 B1 Must state values of n and either np or p. 6000 > 50 and either np = 0.6 < 5 or < 0.1 10000 Note: ‘n large, p small’ is insufficient. 1

More questions on Probability

4 Each year a transport firm uses X litres of gasoline and Y litres of diesel fuel, where X and Y have the independent distributions X + N ( 10 700, 950 2 ) and Y + N ( 13400, 1210 2 ) . (a) Find the probability that in a randomly chosen year the firm uses more gasoline than diesel fuel. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The costs per litre of gasoline and diesel fuel are $0.80 and $0.85 respectively. (b) Find the probability that the total cost of gasoline and diesel fuel in a randomly chosen year is between $20 000 and $22 000. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) E(X − Y) = 10700 – 13400 [=−2700] B1 Oe, e.g. (Y − X). Var(X − Y) = 9502 + 12102 [= 2366600] M1 0 − (their '− 2700') M1 For standardising with their E and Var. [= 1.755] their '2366600' 1 − Φ(their ‘1.755’) M1 For area consistent with their values. = 0.0396 or 0.0397 (3 sf) A1 5 4(b) E(Total) = 10700 × 0.8 + 13400 × 0.85 [=19950] B1 Var(Total) = 9502 × 0.82 + 12102 × 0.852 [= 1635412.25] M1 22000 −their '19950' 20000 −their '19950' M1 For one standardisation with their E and Var. [= 1.603] or [= 0.0391] their '1635412.25' their '1635412.25' Φ(their ‘1.603’) − Φ(their ‘0.0391’) = 0.9455 – 0.5156 M1 For area consistent with their values. = 0.43[0] (3 sf) A1 5

More questions on Discrete random variables

5 A teacher models the numbers of girls and boys who arrive late for her class on any day by the independent random variables G + Po(0.10) and B + Po(0.15) respectively. (a) Find the probability that during a randomly chosen 2-day period no girls arrive late. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that during a randomly chosen 5-day period the total number of students who arrive late is less than 3. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) It is given that the values of P(G = r) and P(B = r) for r H 3 are very small and can be ignored. Find the probability that on a randomly chosen day more girls arrive late than boys. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Following a timetable change the teacher claims that on average more students arrive late than before the change. During a randomly chosen 5-day period a total of 4 students are late. (d) Test the teacher’s claim at the 5% significance level. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) [e−0.2] = 0.819 (3 sf) B1 Accept e–0.2 as final answer. 1 5(b) λ = 1.25 B1  1.25 2  M1 Any λ Allow one end error. e−1.25  1 + 1.25 +  Must see expression (in any form).  2    Accept correct Σ notation. or e−1.25(1 + 1.25 + 0.78125) or 0.2865 + 0.3581 + 0.2238 = 0.868 (3 sf) A1 SC Answer with no working seen scores B1 (could be implied). 3 5(c) e−0.15 × e−0.1(0.1) = 0.077879 M1 P(B = 0) × P(G = 1) 0.8607  0.09048  0 . 12  P(B = 0) × P(G = 2) 0.8607  0.004524 = 0.003894 e−0.15 × e−0.1   P(B = 1) × P(G = 2) 0.1291  0.004524   2   0 . 12 Note: P(B = 0)  P(G = 2) and P(B = 1)  P(G = 2) e−0.15 × 0.15 × e−0.1 × = 0.0005841 2 may be seen within P(G = 2)  P(B < 2). For one expression seen.  0 . 12  0 . 12 M1 P(B = 0) × P(G = 1) + P(B = 0) × P(G = 2) e−0.15 × e−0.1 (0.1) + e−0.15 × e−0.1   + e−0.15 × 0.15 × e−0.1 × + P(B = 1) × P(G = 2).  2  2   For the three Poisson terms added (must be from a = 0.077879 + 0.00389036 + 0.0005841 complete attempt at all 3 terms). = 0.0824 (3 sf) A1 Alternative method for Question 5(c) P(B = 0)  P(G > 0) M1 For one expression seen. e−0.15 × (1 − e−0.1) P(B = 1)  P(G > 1) e−0.15 × 0.15 × (1 − e−0.1(1 + 0.1)) e−0.15 × (1 − e−0.1) + e−0.15 × 0.15 × (1 − e−0.1(1 + 0.1)) M1 For adding their expressions. = 0.0824 (3 sf) A1 3 5(d) H0: λ = 1.25 or 0.25[per day] B1 Or µ or ‘population mean’. H1: λ > 1.25 or 0.25[per day]  1.252 1.253  M1 Any λ. No end errors. Expression must be seen (in P(> 4 late) = 1 − e−1.25  1 + 1.25 + +  any form). Accept correct Σ notation.  2 3!  or 1 – e−1.25(1 + 1.25 + 0.7813 + 0.3255) or 1 – (0.2865 + 0.3581 + 0.2238 + 0.09326) = 0.0383 A1 SC 0.0383 with no working scores B1. 0.0383 < 0.05 M1 For a valid comparison. [Reject H0] A1 FT No contradictions. In context and not definite, ‘Hence there is sufficient evidence to suggest that the teacher’s claim is true’ e.g. not ‘More students are late’ or ‘Claim is or ‘There is sufficient evidence to suggest that more students are late on correct’. average’. Ft their 0.0383. 5

More questions on The Poisson distribution

6 The graph of the probability density function f of a random variable X is symmetrical about the line x = 2 . It is given that P ( 2 1 X 1 5 ) = 117256 . (a) Using only this information show that P ( X 2- 1) = 245256 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ It is now given that, for x in a suitable domain, f ( x) = k ( 12 + 4x - x 2 ) , where k is a constant. (b) Find the value of k. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ + x - x The domain of(c) A different random variable X has probability density function g ( x) = 29 2 2 ` j. X is all values of x for which g ( x) H 0 . Find Var(X ). 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Mark scheme: 6(a) 1 117  11  M1 For use of symmetry about x = 2, oe. − = 2 256  256  E.g. (1 – 2 × 117 ) ÷ 2 or 1 + 117 256 2 256. 11 117 11 A1 Any correct numerical expression seen leading to 1 − or 2 × + AG. 256 256 256 245 = AG 256 2 6(b) 5 M1 Attempt to integrate f(x) with any limits. k (12 + 4 x − x 2 )d x  2  3 5    2 x  = k 12 x + 2 x −      3  2  117 M1 Use of limits 2 and 5 and equating their integration 39k = 256 117 attempt to . 256 Or limits –2 and 6 equated to 1. 245 Or limits –1 and 6 equated to . 256 234 Or limits –1 to 5 equated to . 256 Oe. No mixed methods. 3 A1 k = or 0.0117 256 3 6(c) [2 + x − x2 = 0] B1 x = −1 and x = 2 seen or implied [Domain is −1 ⩽ x ⩽ 2] Mean = 0.5 B1 2 *M1 Attempt to integrate x2 g(x) with any limits. 2 2 3 4 (2 x + x − x )d x 9  − 1  4 5 2  2  2 3 x x   =  x + −     9  3 4 5  −1  [= 0.7] their ‘0.7’ – their ‘0.5’2 DM1 Subtract their mean2 from their ∫ x2 g(x)dx (both must be numerical). 9 A1 = 0.45 or 20 5

More questions on Probability

7 The heights, in centimetres, of adult females in Litania have mean n and standard deviation v. It is known that in 2004 the values of n and v were 163.21 and 6.95 respectively. The government claims that the value of n this year is greater than it was in 2004. In order to test this claim a researcher plans to carry out a hypothesis test at the 1% significance level. He records the heights of a random sample of 300 adult females in Litania this year and finds the value of the sample mean. (a) State the probability of a Type I error. [1] ............................................................................................................................................................ ............................................................................................................................................................ You should assume that the value of v after 2004 remains at 6.95 . (b) Given that the value of n this year is actually 164.91, find the probability of a Type II error. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) 0.01 or 1% B1 Note: x ⩽ 0.01 scores B0. 1 7(b) h − 163.21 M1 Accept any z (±). 2.326 = 6.95  300 h = 164.14 A1 Accept 3 sf accuracy here. [Rejection region is h > 164.14] [P(Type II) = P( h < 164.14 | µ = 164.91)] their \'164.14'− 164.91 M1 For standardising 164.91 with their 164.14 (could [= −1.919] be 163.21). 6.95  300 Φ(their ‘−1.919’) = 1 − Φ(their ‘1.919’) M1 For area attempt consistent with their values. = 0.0275 or 0.0276 or 0.028[0] (3.s.f) A1 Accept anything in range 0.0275 to 0.028[0]. 5

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