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Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s23_qp_629709_s23_ms_629709/62/M/J/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
In a survey of 200 randomly chosen students from a certain college, 23% of the students said that they owned a car. Calculate an approximate 93% confidence interval for the proportion of students from the college who own a car. [3]
Mark scheme
1 0.23×(1− 0.23) M1 Expression of correct form. Any z, but z = 0.8328 scores 0.23 ± z × B0M0. 200 z = 1.811 or 1.812 B1 0.176 to 0.284 (3 sf) A1 Must be an interval. 3
Question
(a) The random variable W has a Poisson distribution. State the relationship between E W and Var W . [1] (b) The random variable X has the distribution B n, p. Jyothi wishes to use a Poisson distribution as an approximate distribution for X . Use the formulae for E X and Var X to explain why it is necessary for p to be close to 0 for this to be a reasonable approximation. [1] (c) Given that Y has the distribution B 20 000, 0.000 07, use a Poisson distribution to calculate an estimate of P Y > 2. [3]
Mark scheme
2(a) E(W) = Var(W). B1 Allow ‘they are the same’ OE. Must be = not ≈ (and not both = and ≈). Condone E(W) =λ and Var(W)= λ. 1 2(b) np ≈ np(1 – p), hence 1 – p must be close to 1 B1 OE. Must see formulae and q = 1 – p must be seen or implied and conclusion made. 1 2(c) λ = 1.4 B1 Seen. 2 1 – e–1.4(1 + 1.4 + 1.4 ) or 1 – e–1.4(1 + 1.4 +0.98) or 1- M1 Allow any λ; allow one end error. Expression must be 2 seen (accept correct sigma notation). (0.2466+0.3452+0.2417) = 0.167 (3 sf) or 0.166 A1 Use of Binomial scores SCB1 for 0.167 or 0.166 . No working: 0.167 [or 0.166] SC B1. Note: λ=1.4 and 0.167 with no working seen scores SC B1B1. Use of Normal scores B0M0. 3
Question
The masses, in kilograms, of newborn babies in country A are represented by the random variable X , with mean - and variance 3 2 . The masses of a random sample of 500 newborn babies in this country were found and the results are summarised below. n = 500 Σ x = 1625 Σ x2 = 5663.5 (a) Calculate unbiased estimates of - and 32 . [3] A researcher wishes to test whether the mean mass of newborn babies in a neighbouring country, B, is different from that in country A. He chooses a random sample of 60 newborn babies in country B and finds that their sample mean mass is 2.95 kg. Assume that your unbiased estimates in part (a) are the correct values for - and 3 2 . Assume also that the variance of the masses of newborn babies in country B is the same as in country A. (b) Carry out the test at the 1% significance level. [5]
Mark scheme
3(a) Est (μ) = 3.25 = 13/4 or 1625/500 B1 1 1625 2 M1 Expression of correct form. Est(σ ) = 500 ( 5663.5 − "3.25"2 ) or 2 5663.5 − 499 500 499 500 = 0.766 (3 sf) or 1529/1996 A1 Biased variance of 0.7645 scores M0A0. 3 3(b) H0: Pop mean (or μ) = ‘3.25’ B1FT Not just ‘mean’. H1: Pop mean (or μ) ≠ ‘3.25’ FT their 3.25 . 2.95−"3.25" M1 Standardising with their values. "0.766"÷60 Must have √60. = –2.655 A1 Or P(𝑋 < 2.95) = 0.0039 or 0.00396 or 0.00397 . SC FT their biased est(σ2), i.e. 0.7645 to give z = 2.658 A1. ‘2.655’ > 2.576 or ‘–2.655’ < –2.576 M1 For valid comparison, e.g. 0.0039 or 0.00396 or 0.00397 < 0.005, or 0.0078 < 0.01, or 0.00792 < 0.01 . [Reject H0] A1FT OE. Must be in context and not definite, e.g., not ‘Mean There is evidence that (mean) mass in (country B) is different (from country mass is not different’, No contradictions. A). Context needs either ‘mass’ or ‘countries’ OE. SC, Use of one-tail test. ‘2.655’ > 2.326 or 0.0039 < 0.01 M1A0 (Max B0M1A1M1A0 3/5). Accept critical value method. Either: Xcrit=2.959 M1A1 2.95<2.959 M1A1FT with correct conclusion, or Xcrit=3.241 M1A1 3.25>3,241 M1A1FT with correct conclusion. 5
Question
The number, X , of books received at a charity shop has a constant mean of 5.1 per day. (a) State, in context, one condition for X to be modelled by a Poisson distribution. [1] Assume now that X can be modelled by a Poisson distribution. (b) Find the probability that exactly 10 books are received in a 3-day period. [2] (c) Use a suitable approximating distribution to find the probability that more than 180 books are received in a 30-day period. [4] The number of DVDs received at the same shop is modelled by an independent Poisson distribution with mean 2.5 per day. (d) Find the probability that the total number of books and DVDs that are received at the shop in 1 day is more than 3. [3]
Mark scheme
4(a) Books received independently or singly or randomly. B1 OE. Must be in context. If more than one condition given, ignore extras. 1 4(b) 10 M1 Allow incorrect λ. e−15.3 × 15.3 10! = 0.0439 (3sf) A1 SC No working shown but correct answer seen scores B1. 2 4(c) N(153, 153) B1 Seen or implied. 180.5−153 M1 For standardising with their values (can be implied). [= 2.223] Allow with wrong or missing continuity correction. 153 1 ̶ ɸ(‘2.223’) M1 For correct probability area consistent with their values. = 0.0131 (3sf) A1 4 4(d) (λ =) 5.1 + 2.5 [= 7.6] B1 Give at early stage (seen or implied). −7.6 (1 + 7.6 + 7.62 + 7.63 ) M1 Allow incorrect λ. Allow one end error. 1– e = 1– e-7.6(1+7.6+28.88+73.16) Must see an expression (accept correct sigma notation). 2 3! = 1 – (0.0005005+0.003803+0.01445+0.03661) = 0.945 (3sf) A1 SC No working, 0.945 B1(could be implied) SC B1. 3
Question
(a) Two random variables X and Y have the independent distributions N 7, 3 and N 6, 2 respectively. A random value of each variable is taken. Find the probability that the two values differ by more than 2. [5] (b) Each candidate’s overall score in a science test is calculated as follows. The mark for theory is denoted by T , the mark for practical is denoted by P, and the overall score is given by T + 1.5P. The variables T and P are assumed to be independent with distributions N 62, 158 and N 42, 108 respectively. You should assume that no continuity corrections are needed when using these distributions. (i) A pass is awarded to candidates whose overall score is at least 90. Find the proportion of candidates who pass. [5] (ii) Comment on the assumption that the variables T and P are independent. [1]
Mark scheme
5(a) E(X – Y) = 1 Var(X – Y) = 5 B1 Seen or implied, OE e.g. X – Y – 2. 2 −1 [= 0.447] −2 −1 [= –1.342] M1 Standardising with their values must come from a 5 5 combination. 1 – Φ('0.447') Φ('–1.342') = 1 – Φ('1.342) M1 Correct probability area consistent with their values. = 0.327 or 0.328 = 0.0898 or 0.0899 A1 Seen or implied. Probability that difference is more than 2 = 0.417 (3 sf) or 0.418 A1 5 5(b)(i) E(X) = 62 + 1.5(42) [= 125] B1 OE. Var(X) = 158 + 1.52 × 108 [= 401] B1 Correct expression OE. 90−"125" M1 Correct standardisation using their E(X) and Var(X). [= ̶ 1.748] Must both be from a combination attempt. "401" Ignore any attempted continuity correction. Φ(‘1.748’) M1 Correct probability area consistent with their stated values. = 0.960 or 96.0% (3 sf) A1 Allow 0.96 or 96%. 5 5(b)(ii) Unlikely. A candidate who does well in Theory is likely to do well in B1 Need both. Practical. Accept ‘unlikely’, ‘not independent’, ‘dependent’, ‘not realistic’, or similar; and accept ‘both testing knowledge from the same syllabus’, ‘theory and practical share same content’ or similar statement. 1
Question
When a child completes an online exercise called a Mathlit, they might be awarded a medal. The publishers claim that the probability that a randomly chosen child who completes a Mathlit will be awarded a medal is 13 . Asha wishes to test this claim. She decides that if she is awarded no medals while completing 10 Mathlits, she will conclude that the true probability is less than 13 . (a) Use a binomial distribution to find the probability of a Type I error. [2] The true probability of being awarded a medal is denoted by p. (b) Given that the probability of a Type II error is 0.8926, find the value of p. [3]
Mark scheme
6(a) (1 – 1 )10 M1 3 = 0.0173 (3 sf) A1 No working scores SC B1. 2 6(b) 1 – (1 – p)10 = 0.8926 M1 Accept 1 – q10 = 0.8926 . Equation must be in p or in q but not both. 1 – p = 0.10740.1 [= 0.800] M1 For valid attempt to solve their (binomial) equation in p10 or q10. p = 0.200 (3 sf) or 0.2 A1 3
Question
(a) f x x 0 4 The diagram shows the graph of the probability density function, f, of a random variable X which takes values between 0 and 4 only. Between these two values the graph is a straight line. (i) Show that f x = kx for 0 ≤ x ≤ 4, where k is a constant to be determined. [2] (ii) Hence, or otherwise, find E X . [3] (b) g w w 0 a The diagram shows the graph of the probability density function, g, of a random variable W which takes values between 0 and a only, where a > 0. Between these two values the graph is a straight line. Given that the median of W is 1, find the value of a. [3] 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)(i) 1 ×4×a=1 M1 For use of area = 1 2 4 or let f(x) = kx and attempt kx dx = 1 . 0 [a = 1 ] f(x) = 1 x A1 2 4 2 8 k x = 1; 8k = 1; k = 1 . 2 0 8 f(x) = 1 x or k = 1 . 8 8 2 7(a)(ii) 4 M1 Attempt to integrate x × their f(x). Ignore limits accept x × x dx 0 1 8 in terms of k. x3 4 A1ft Their integral and correct limits accept in terms of k. 24 0 = 8 or 2.67 (3 sf) A1 Note: Final answer of 64k/3 scores 2/3. 3 3 7(b) a −1 1 M1 1 1 = Or attempt g( w)dw = 1 i.e. ( 2 − 22 w)dw = 1 , a 2 2 a a 2 0 0 or integral from 1 to a. g(w) must be linear of form g(w) = mw (+c). Or area attempt: attempt to calculate heights using their linear equation (h1=2/a and h2= –2/a2 + 2/a) and use in either area trapezium = 0.5, or area trapezium =area small triangle or area small triangle = 0.5 . Area trapezium =1/2 × 1 (2/a + –2/a2 + 2/a) Area triangle = 1/2(a – 1)(–2/a2 + 2/a)) Note: alternative expression for h1 = (a – 2)/(a – 1). a 2 − 2 =a A1 Or a2 – 4a + 2 = 0. Any correct equation in a, a not in denominator. a = 2 + 2 = 3.41 A1 3
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Mathematics 9709/62 for May/June 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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