Cambridge A Level Mathematics 9709 — 2024 May/June Paper 6 · Variant 2

1 A random variable X has the distribution Po ( 145) . (a) Use a suitable approximating distribution to calculate P ( X G 150 ) . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Justify the use of your approximating distribution in this case. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) N(145, 145) B1 Stated or implied. ± 150.5 145 145  [= ±0.457] M1 Condone incorrect or omitted continuity correction. Φ(‘0.457’) M1 For area consistent with their working. = 0.676 (3sf) A1 SC: Unsupported answer of 0.676 scores B3. Unsupported answer of 0.646 or 0.661 scores B2. Unsupported answer of 0.6799 scores B1. 4 1(b) 145 > 15 B1 Explicit. λ > 15 B0 if λ = 145 not stated. Accept ⩾ Accept mean for λ. 1

More questions on The Poisson distribution

2 Henri wants to choose a random sample from the 804 students at his college. He numbers the students from 1 to 804 and then uses random numbers generated by his calculator. The first 20 random digits produced by his calculator are as follows. 5 6 7 1 0 9 8 4 3 1 0 9 6 6 5 0 2 1 7 6 Henri’s first two student numbers are 567 and 109. (a) Use Henri’s digits to find the numbers of the next two students in the sample. [2] ............................................................................................................................................................ ............................................................................................................................................................ There were 30 students in Henri’s sample. He asked each of them how much time, X hours, they spent on social media each week, on average. He summarised the results as follows. n = 30 Rx = 610 Rx 2 = 12405 (b) Use this information to calculate an unbiased estimate of the mean of X and show that an unbiased estimate of the variance of X is less than 0.1 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Henri’s friend claims that Henri has probably made a mistake in his calculation of Rx or Rx2 . Use your answer to part (b) to comment on this claim. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) [567, 109], 665, 21 B2 B1 for each. Allow 021. If more than 2 answers given, count first two and ISW. 2 Question Answer Marks Guidance 2(b) Est(µ) = 610 30 or 61 3 B1 OE or 20.3. Est(σ2) = 2 30 610 29 30 12405 30 ( ( )  ) or   2 610 1 29 30 12405  M1 Use of correct formula. = 0.0575 (3sf) A1 Accept 5 87 . 3 2(c) Variance is [unrealistically] small so Henri has [probably] made a mistake/claim is [probably] correct B1 FT Need both parts. Need ‘small’ OE, not just < 0.1. FT their < 0.1 variance value (not –ve), e.g. 0.0556 (if omit 30 29). Accept ‘s.d. = 0.24 is small, so Henri has probably made a mistake’. Note: ‘mean is large/small’ scores B0, but ‘mean large compared to variance so Henri prob made a mistake’ scores B1. 1

More questions on Discrete random variables

3 A student wishes to estimate the proportion, p, of students at her college who have exactly one brother. She surveys a random sample of 50 students at her college and finds that 18 of them have exactly one brother. She calculates an approximate a% confidence interval for p and finds that the lower limit of the confidence interval is 0.244 correct to 3 significant figures. Find a correct to the nearest integer. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 18 50 − z × 18 18 50 50 (1 ) 50  = 0.244 z = 1.709 or 1.708 A1 Accept 1.71 if nothing better seen. ɸ–1('1.709') = 0.956 ; 1 − 2(1 – ‘0.956’) [= 0.912] M1 Attempt area above or below their 1.709 and use correct method to find α. α = 91 A1 Allow α = 91% 0.91 or 91.2 score A0. 4

More questions on Discrete random variables

4 A random variable X has the distribution N(10, 12). Two independent values of X, denoted by X1 and X2, are chosen at random. (a) Write down the value of P ( X 1 2 X 2 ) . [1] ............................................................................................................................................................ ............................................................................................................................................................ (b) Find P ( X 1 2 2X 2 - 3) . [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 0.5 B1 1 4(b) E(X1 − 2X2 + 3) =10-20+3 [= −7] or E(2X2 − X1 − 3) = 20 − 10 – 3 [= 7] B1 Or equivalent using X1 − 2X2 =10 – 20 [= –10] or 2X2 − X1 = 20 – 10 [= +10]. Var(X1 − 2X2 + 3) = 12 + 22×12 + 0 [= 60] B1 0 (' 7') '60'  [= 0.904] M1 Or numerator 3–‘10’ or –3–(‘–10’), but not ‘–3 –10’ (i.e. numerator must be ‘7’ or ‘–7’). 1 − Φ(‘0.904’) M1 For area consistent with their working. = 0.183 A1 5

More questions on Probability

5 The number of goals scored by a sports team in the first half of any match has the distribution X + Po ( 3. 1) . The number of goals scored by the same team in the second half of any match has the distribution Y + Po ( 2.4) . You may assume that the distributions of X and Y are independent. (a) Find P ( X 1 4) . 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(b) Find the probability that, in a randomly chosen match, the team scores at least 5 goals. 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(c) Given that the team scores a total of 5 goals in a randomly chosen match, find the probability that they score exactly 3 goals in the first half. 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Mark scheme: 5(a) e−3.1(1 + 3.1 + 2 3 3.1 3.1 2! 3!  ) or e−3.1(1 + 3.1 + 4.805 + 4.965) or 0.0450 + 0.1397 + 0.2165 + 0.22368 M1 Condone one end error. Any λ. Accept fully correct Σ notation. Expression must be seen. = 0.625 (3sf) A1 Correct answer with no working scores SC B1. 2 Question Answer Marks Guidance 5(b) [λ]= 5.5 B1 SOI 1 − e−5.5(1 + 5.5 + 2 3 4 5.5 5.5 5.5 2! 3! 4!   ) or 1 − e−5.5(1 + 5.5 + 15.125 + 27.7292 + 38.1276) or 1 − e−5.5(0.004087 + 0.0224772 + 0.061812 + 0.113323 + 0.155819) M1 Condone one end error. Any λ. Accept fully correct Σ notation. Expression must be seen. = 0.642 or 0.643 (3sf) A1 Correct answer with no working scores SC B1 B1. 3 5(c) [P(X = 3) × P(Y = 2) = ] = e−3.1× 3 3.1 3! × e−2.4 × 2 2.4 2! or 0.223676 × 0.261267 [= 0.05844] M1 Find P(3 in first half AND 2 in second half). Must see expression. [P(total 5) = ] 5 5.5 5.5 5! e  or 0.17140 M1 Use of 5.5 to find P(5). P(P(exactly 3 in 1st half given total 5) = st exactly 3 in 1 half and total 5) total 5) P( P( M1 Attempt at conditional probability; numerator = their 0.05844 and denominator = P(total 5) Note: ( 3 3.1 3! × 2 2.4 2! )÷( 5 5.5 5! ) scores M1 M1 M1. [= '0.05844' '0.17140' ] = 0.341 (3sf) A1 4

More questions on The Poisson distribution

6 The masses of cereal boxes filled by a certain machine have mean 510 grams. An adjustment is made to the machine and an inspector wishes to test whether the mean mass of cereal boxes filled by the machine has decreased. After the adjustment is made, he chooses a random sample of 120 cereal boxes. The mean mass of these boxes is found to be 508 grams. Assume that the standard deviation of the masses is 10 grams. (a) Test at the 2.5% significance level whether the mean mass of cereal boxes filled by the machine has decreased. 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Later the inspector carries out a similar test at the 2.5% significance level, using the same hypotheses and another 120 randomly chosen cereal boxes. (b) Given that the mean mass is now actually 506 grams, find the probability of a Type II error. 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Mark scheme: 6(a) H0: Population mean mass = 510 g H1: Population mean mass < 510 g B1 Allow ‘μ’ but not just ‘mean’. ± 508 510 10 120   M1 Standardising must have 120. = ± −2.191 or −2.190 A1 −2.191 < −1.96 or 2.191 > 1.96 Area comparison: 0.0143 or 0.0142 < 0.025 M1 OE For valid comparison. Inequality sign the wrong way round scores M1 A0. [Reject H0] There is sufficient evidence to suggest that the [mean] mass has decreased A1FT OE In context (must be ‘decreased’ OE, not ‘changed’); not definite. No contradictions. Condone ‘there is sufficient evidence to support the inspector’s claim’. NB: Accept alternative method using critical value (= 508.21) and comparison with 508. Condone 509.79 compared with 510. Two tail test scores maximum B0 M1 A1 M1 A0; must have comparison with 0.0125 or 2.24/2.241. 5 Question Answer Marks Guidance 6(b) cv 510 10 120   = −1.96 M1 Standardising to find critical value (must use 510 and 10÷ 120) . Accept ± 1.96. cv = 508.21 A1 Accept 3 sf if nothing better seen. Note: cv could be found in (a). z = ± 508.21 506 10 120   [= 2.421] M1 Standardising with their 508.21 and 506 (must use 10÷ 120) . P (X > 508.21 | µ = 506) = 1 − Φ(‘2.421’) M1 For area consistent with their working. = 0.0077 to 0.0080 (2sf) A1 Note: 510 506 10 120   scores max M0 A0 M1 M1 A0. 5

More questions on Hypothesis tests

7 The probability density function, f, of a random variable X is given by k ( 1 + cos x) 0 G x G r, f ( x) = )0 otherwise, where k is a constant. 1 (a) Show that k = . 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(b) Verify that the median of X lies between 0.83 and 0.84 . 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(c) Find the exact value of E(X ). 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Mark scheme: 7(a) k 0 (1 cos )d x x    = 1 M1 Attempt integrate f(x) with correct limits and equate to 1. k 0 sin x x   = 1 A1 Correct integration. [e.g. k(π + sin π − (0 + 0)) = 1], kπ = 1, k = 1  A1 AG Some evidence of substitution of limits, i.e. at least one interim step (e.g. kπ = 1) as minimum requirement. Convincingly obtained; no errors seen. 3 Question Answer Marks Guidance 7(b) 1   0.83 0 sin x x  or 1 (0.83 + sin 0.83) 1   0.84 0 sin x x  or 1 (0.84 + sin 0.84) M1 Substitute correct limits into their integral. OR1: integrate 0 to 0.83 and 0.84 to π. OR2: use g(m) = m + sin m – (π/2) and find g(0.83) and g(0.84). OR3: use h(m) = m + sin m and find h(0.83) and h(0.84). Both attempted. = 0.499 (3 sf) = 0.504 (3 sf) A1 OR1: 0.499 and 0.496. OR2: g(0.83) = – 0.00286/7 and g(0.84) = 0.0138/9. OR3: h(0.83) = 1.57 and h(0.84) = 1.58 or 1.59. Both correct. ‘0.499’ < 0.5 < ‘0.504’ hence 0.83 < median < 0.84 Equivalent to –0.000912 < 0 < 0.00441 hence 0.83 < median < 0.84 A1FT FT their areas; dep 0.5 is between their areas OE. OR1: 0.499 < 0.5 and 0.496 < 0.5, so 0.83 < m < 0.84. OR2: g(0.83) > 0 g(0.84) < 0 OE, so 0.83 < m < 0.84. OR3: h(0.83) < 2  h(0.84) > 2 , so 0.83 < m < 0.84. Both statements needed. Note: A score of M1 A0 A1FT is possible. If 0 scored, SC: 1 (m + sin m) = 0.5 B1 and m = 0.831 to 0.832, so 0.83< m < 0.84 B1. 3 Question Answer Marks Guidance 7(c) 0 1 ( cos )d x x x x     M1* Attempt integrate xf(x). Ignore limits. =     2 0 0 0 1 sin sin d 2 x x x x x                      DM1 OE Attempt to integrate (using ‘parts’) with correct limits, reaching an expression of the form ax2 + uv – ∫v du. OR using parts to integrate x(1+cosx) reaching an expression of the form uv – ∫v du i.e. 1 (x2 + x sin x – ∫(x + sin x) dx)) = 2 1 sin cos 2 x x x x          or e.g. 1 0 2 (0 [ cos ] ) x      or e.g. 2  + 1 ((−1 −1)) A1 Integration fully correct. 2 2     A1 OE ISW after correct exact value seen. SC1: Unsupported answer of 2 2    scores B3. SC2: Unsupported answer of 0.934 scores B2. 4

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