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Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w23_qp_639709_w23_ms_639709/63/O/N/23
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
A random variable X has the distribution N 410, 400. Find the probability that the mean of a random sample of 36 values of X is less than 405. [3]
Mark scheme
1 405 − 410 [= –1.5] M1 For standardising, must have 36. 20 6 14580 − 14760 Allow totals method . 14400 No mixed methods. ɸ('–1.5') = 1 – ɸ('1.5') M1 For area consistent with their working. = 0.0668 A1 3
Question
In a survey of 300 randomly chosen adults in Rickton, 134 said that they exercised regularly. This information was used to calculate an !% confidence interval for the proportion of adults in Rickton who exercise regularly. The upper bound of the confidence interval was found to be 0.487, correct to 3 significant figures. Find the value of ! correct to the nearest integer. [4]
Mark scheme
2 134 × 166 M1 For expression of the correct form. 134 + z 300 300 = 0.487 300 300 z = 1.405 A1 Accept 1.404, or anything that rounds to 1.39 to 1.41. ɸ–1('1.405') = 0.9199 or 0.92; 1 − 2(1 – 0.92) M1 Attempt area above or below their 1.405 and convert to a confidence level. α = 84 A1 Allow α = 84%. cwo Note: final answer 0.84 scores A0. 4
Question
A website owner finds that, on average, his website receives 0.3 hits per minute. He believes that the number of hits per minute follows a Poisson distribution. (a) Assume that the owner is correct. (i) Find the probability that there will be at least 4 hits during a 10-minute period. [3] (ii) Use a suitable approximating distribution to find the probability that there will be fewer than 40 hits during a 3-hour period. [4] A friend agrees that the website receives, on average, 0.3 hits per minute. However, she notices that the number of hits during the day-time (9.00 am to 9.00 pm) is usually about twice the number of hits during the night-time (9.00 pm to 9.00 am). (b) (i) Explain why this fact contradicts the owner’s belief that the number of hits per minute follows a Poisson distribution. [1] (ii) Specify separate Poisson distributions that might be suitable models for the number of hits during the day-time and during the night-time. [2]
Mark scheme
3(a)(i) λ=3 B1 For mean = 3. 2 3 1 – e–3(1 + 3 + 32 + 33! ) or 1 – e–3(1 + 3 + 4.5 + 4.5) M1 Any λ. Allow one end error. or 1 – (0.04979 + 0.14936 + 0.22404 + 0.22404) = 0.353 (3 sf) A1 No working scores B1. 3 3(a)(ii) N(54, 54) M1 soi 39.5 − 54 (= –1.973) M1 Allow with wrong or no continuity correction. 54 For standardising with their mean and variance. 1 – ɸ ('1.973') M1 For area consistent with their working. = 0.0242 (3 sf) A1 Special case: if no working seen, 0.0242 scores SC B3, 0.0284 scores SC B2. 4 3(b)(i) ‘Mean not constant’ or’ ‘number of hits per minute not constant’ or ‘not a B1 constant rate’ 1 3(b)(ii) 2p + p = 2 × 0.3 [ p = 0.2] M1 May be implied by answer. [where p is the rate per minute for night time] [During day-time]: Po(0.4). [During night-time]: Po(0.2) A1 Accept Po(24) [per daytime hour], Po(12) [per night time hour]. Accept Po(288) [per day time shift], Po(144)[ per night time shift]. Note: Po(432), Po(216) scores M0A0. 2
Question
The masses, in kilograms, of chemicals A and B produced per day by a factory are modelled by the independent random variables X and Y respectively, where X ∼ N 10.3, 5.76 and Y ∼ N 11.4, 9.61. The income generated by the chemicals is $2.50 per kilogram for A and $3.25 per kilogram for B. (a) Find the mean and variance of the daily income generated by chemical A. [2] (b) Find the probability that, on a randomly chosen day, the income generated by chemical A is greater than the income generated by chemical B. [6]
Mark scheme
4(a) E(A income) = [10.3 × 2.50 ] = 25.75 [$] B1 Accept 3sf. Var(A income) = [5.76 × 2.502 ] = 36 [$2] B1 2 4(b) B income ~ N(37.05, 101.506) B1 Or N(37.1, 102) soi. or E(B income) = 37.05 and Var(B income) = 101.51 A income – B income ~ N(‘25.75’ – ‘37.05’, ‘36’ + ‘101.506’) M1 Ft their values for A and B. = N(–11.3, 137.506) A1 Accept 3sf. 0 − ( − '11.3') M1 Standardising with their values from attempt at A [= 0.964] income – B income. '137.506' 1 – ɸ('0.964') M1 For area consistent with their values. = 0.168 or 0.167 (3 sf) A1 cwo 6
Question
In the past the number of enquiries per minute at a customer service desk has been modelled by a random variable with distribution Po 0.31. Following a change in the position of the desk, it is expected that the mean number of enquiries per minute will increase. In order to test whether this is the case, the total number of enquiries during a randomly chosen 5-minute period is noted. You should assume that a Poisson model is still appropriate. Given that the total number of enquiries is 5, carry out the test at the 2.5% significance level. [5]
Mark scheme
5 H0: Population mean no. enquiries = 1.55 Or “population mean no. enquiries = 0.31 (per minute)” H1: Population mean no. enquiries > 1.55 B1 oe. Allow 'λ = 1.55’ or µ = ‘1.55’. 2 3 4 M1 Allow one end error, e.g. extra term: e-1.55 × 1.555 . P(X ⩾ 5) = 1 – e-1.55(1 + 1.55 + 1.55 + 1.55 + 1.55 ) 5! 2! 3! 4! or 1 – e-1.55(1 + 1.55 + 1.20125 + 0.62065 + 0.24050) or 1 – (0.21225 + 0.32898 + 0.25496 + 0.13173 + 0.05105) = 0.0210 (3 sf) A1 Allow 0.021. SC B1 no working scores B1 instead of M1A1. 0.0210 < 0.025 M1 For valid comparison. [Reject H0] There is sufficient evidence [at 2.5% level] to suggest that mean no. A1 FT In context, not definite, of enquiries has increased. e.g., not "Mean no. of enquiries has increased". No contradictions. 5 5 Note: e-1.55× 1.55 = 0.0158 < 0.025: scores max B1 5!
Question
A continuous random variable X takes values from 0 to 6 only and has a probability distribution that is symmetrical. Two values, a and b, of X are such that P a < X < b = p and P b < X < 3 = 13 10 p, where p is a positive constant. 5. (a) Show that p ≤ 23 [1] (b) Find P b < X < 6 − a in terms of p. [2] It is now given that the probability density function of X is f, where T 1 6x − x2 0 ≤ x ≤ 6, f x = 36 0 otherwise. 5 , find the value of a. (c) Given that b = 2 and p = 27 [5]
Mark scheme
6(a) p + 13 p ⩽ 1 p ⩽ 5 AG B1 Allow ‘=’ in working but need an inequality in the 10 2 23 answer. Allow 0 < p ⩽ 5 . 23 1 6(b) e.g. 0.5 − 2.3p, p + 1.3p, 2 × 1.3p, 2.3p + 1.3p, M1 Any correct expression for the probability of a relevant 0 to a a to 3 2 × (b to 3) a to 3 + b to 3 region. 2p + 2.6p, 0.5 − 1.3p, 0.5 + 1.3p, 2 × (a to 3) 0 to b b to 6 18 p or 3.6p A1 5 2 6(c) 2 M1 Attempt to integrate with correct limits and equate to 1 (6 x − x 2 ) dx = 5 36 27 5 , oe. a 27 2 Integrate from 2 to 6 – a and equate 18 p = . 5 3 23 Integrate from a to 3 and equate to . 54 2 Integrate from 0 to a and equate to . 27 1 2 x3 2 5 1 8 a3 5 M1 For integrating and substitution of limits to form cubic 36 3x − 3 a = 27 36 12 − 3 − 3a + 3 = 27 2 in a. a3 – 9a2 + 8 = 0 A1 Any correct three term cubic equation in a. (a – 1)(a2 – 8a – 8) = 0 M1 Attempt to factorise their cubic equation. a = 8 ± 96 = 4 ± 24 or –0.899 or 8.90, [not between 0 and 6] 2 a = 1 only [other two values rejected] A1 SC B1 for a = 1 only, if no method seen for solving the cubic. 5
Question
A biologist wishes to test whether the mean concentration -, in suitable units, of a certain pollutant in a river is below the permitted level of 0.5. She measures the concentration, x, of the pollutant at 50 randomly chosen locations in the river. The results are summarised below. n = 50 Σ x = 23.0 Σ x2 = 13.02 (a) Carry out a test at the 5% significance level of the null hypothesis - = 0.5 against the alternative hypothesis - < 0.5. [7] Later, a similar test is carried out at the 5% significance level using another sample of size 50 and the same hypotheses as before. You should assume that the standard deviation is unchanged. (b) Given that, in fact, the value of - is 0.4, find the probability of a Type II error. [5] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) Est (μ) = 23/50 = 0.46 B1 ( ) 50 13.02 M1 For an expression of the correct form for unbiased Est (σ) = × − 0.462 or Est (σ2) = 50 × 13.02 − 0.462 oe standard deviation or variance. 49 50 49 50 1 (23.0) 2 Or estimated unbiased variance = 13.02 − 49 50 61 A1 Est (σ) = 0.22315 or Est (σ2) = 0.0497959 = or 0.0498 1225 0.46 − 0.5 M1 Standardising with their values. '0.22315' 50 = –1.268 or -1.267 or = –1.27 (3sf) A1 −1.268 > –1.645 or 0.102 to 0.103 > 0.05 M1 For a valid comparison. [Do not reject Ho] There is insufficient evidence [at 5% level] that the mean A1 FT In context, not definite. E.g., not ‘Mean concentration is concentration is less than 0.5. not less than 0.5’. No contradictions. 7 7(b) cv − 0.5 M1 '0.22315' = −1.645 50 cv = 0.448(1) or 0.448 (3 sf) A1 '0.448'− 0.4 M1 '0.22315' [=1.521 to 1.524] 50 1 – ɸ('1.524') M1 For area consistent with their working. = 0.0638 to 0.0642 A1 5
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Mathematics 9709/63 for October/November 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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