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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_639709_s23_ms_639709/63/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
A random variable X has probability density function f, where T3 1 − x2 0 ≤ x ≤ 1, f x = 2 0 otherwise. Find E X . [3]
Mark scheme
1 3 1 M1 Attempt to integrate xf(x); ignore limits. 2 0 ( x − x3 )dx 1 A1 Correct integration and limits. 3 x2 x4 = − 2 2 4 0 3 A1 = 8 3
Question
A club has 264 members, numbered from 1 to 264. Donash wants to choose a random sample of members for a survey. In order to choose the members for the sample he uses his calculator to generate random digits. His first 20 random digits are as follows. 10612 11801 21473 22759 (a) The numbers of the first two members in the sample are 106 and 121. Write down the numbers of the next two members in the sample. [2] (b) To obtain the numbers for members after the 4th member, Donash starts with the second random digit, 0, and obtains the numbers 061 and 211. Explain why this method will not produce a random sample. [1]
Mark scheme
2(a) 180, 227 B1 One correct. Ignore incorrect numbers. B1 Both correct and no extra numbers seen. (Allow other correct use of list of digits). 2 2(b) These numbers are not independent of the previous numbers B1 Already used these numbers, so therefore not random. Does not include numbers not in the list, therefore not OR random (not random or biased needs a reason). Only a finite number of digits used 1
Question
In a random sample of 100 students at Luciana’s college, x students said that they liked exams. Luciana used this result to find an approximate 90% confidence interval for the proportion, p, of all students at her college who liked exams. Her confidence interval had width 0.157 92. (a) Find the two possible values of x. [4] Suzma independently took another random sample and found another approximate 90% confidence interval for p. (b) Find the probability that neither of the two confidence intervals contains the true value of p. [1]
Mark scheme
3(a) z = 1.645 B1 x × (1 − 100 x ) M1 OE. Equation of correct form. Accept p = x/100. Any z. z× 100 = 0.07896 Allow missing factor of 2. 100 [x(100– x) = 1003 × 0.078962 ÷ 1.6452] A1 Any correct (likely scalar multiple) three-term quadratic equation in x or p with simplified coefficients. x2 – 100x + 2304 = 0 Accept p2 – p + 0.2304 = 0 or p(1–p) = 0.2304 . x = 36 or 64 A1 4 3(b) 0.12 = 0.01 B1 Accept either. 1
Question
The mass, in tonnes, of steel produced per day at a factory is normally distributed with mean 65.2 and standard deviation 3.6. It can be assumed that the mass of steel produced each day is independent of other days. The factory makes $50 profit on each tonne of steel produced. Find the probability that the total profit made in a randomly chosen 7-day week is less than $22 000. [6]
Mark scheme
4 Method 1: Based on mass Mean = 7 × 65.2 = 456.4 B1 Var = 7 × 3.62 [= 90.72] M1 22 000/50 = 440 used in standardising equation M1 '440' – '456.4' M1 For standardising with their values. No mixed methods. [= –1.722] no mixed methods '90.72' ɸ(–‘1.722’) = 1 – ɸ(‘1.722’) M1 For correct probability area consistent with their values. = 0.0425 or 0.0426 A1 Note: accept alt method using per day. 2 N(65.2, 3.67 ). No mixed methods. Method 2: Based on profit Mean = 7 × 65.2 × 50 = 22 820 B1 Var = 7 × 3.62 M1 Var = 502 × ‘90.72’ [= 226 800] M1 22000 – '22820' M1 For standardising with their values. No mixed methods. [= –1.722] no mixed methods '226800' ɸ(–‘1.722’) = 1 – ɸ(‘1.722’) M1 For correct probability area consistent with their values. = 0.0425 or 0.0426 A1 6
Question
Last year the mean time for pizza deliveries from Pete’s Pizza Pit was 32.4 minutes. This year the time, t minutes, for pizza deliveries from Pete’s Pizza Pit was recorded for a random sample of 50 deliveries. The results were as follows. n = 50 Σ t = 1700 Σ t2 = 59 050 (a) Find unbiased estimates of the population mean and variance. [3] (b) Test, at the 2% significance level, whether the mean delivery time has changed since last year. [5] (c) Under what circumstances would it not be necessary to use the Central Limit Theorem in answering (b)? [1]
Mark scheme
5(a) x = 1700/50 = 34 B1 50 59050 1 17002 M1 59050 2 Est(σ2) = – 342 biased scores M0. Est(σ2) = − 34 or 59050 − 49 50 49 50 50 1250 A1 = 25 scores A0. = 25.5 (3 sf) or 49 3 5(b) H0: Population mean time = 32.4 B1 Not just ‘mean’ but allow just ‘μ’. H1: Population mean time ≠ 32.4 34 – 32.4 M1 Must have 50 and not 50. FT their mean and var. '25.5' Can be implied. 50 = 2.24 (3 sf) A1 or P( T > 34) = 0.0125. SC use of biased var (25) z = 2.26 or p = 0.0119, allow M1A1. ‘2.24’ < 2.326 M1 Or 0.0125 > 0.01 for a valid comparison. [Not reject H0] Insufficient evidence that (mean) time has changed A1FT In context, not definite, e.g. not ‘Time not changed’. No contradictions. Note: accept CV method xcri = 34.06 for M1A1. Compares 34 < 34.06 for M1, conclusion for A1. Condone x = 32.34 M1A1: compares 32.4 > 32.34 for M1, conclusion for A1. 5 SC for using a one-tail method. Award max 3/5 (B0 M1 A1 M1 A0). 5(c) Distribution of times in the population is normal B1 Accept answers with no context here. Accept underlying distribution for population. 1
Question
It is known that 1 in 5000 people in Atalia have a certain condition. A random sample of 12 500 people from Atalia is chosen for a medical trial. The number having the condition is denoted by X . (a) Use an appropriate approximating distribution to find P X ≤ 3. [3] (b) Find the values of E X and Var X , and explain how your answers suggest that the approximating distribution used in (a) is likely to be appropriate. [2]
Mark scheme
6(a) X ~ Po(2.5) B1 SOI. 2 3 e–2.5(1 + 2.5 + 2.52 + 2.5 ) M1 Any λ. Allow one end error. 3! = 0.758 (3 sf) A1 SC use of binomial B1 for 0.758. SC when no working is shown, X ~ Po(2.5) seen scores B1, 0.758 seen also scores B1. 3 6(b) E(X) = 5 or 2.5 , Var(X) = 4999 or 2.4995 *B1 Just an answer of 2.5 for the variance is not sufficient. 2 2000 However, 2.4995 is sufficient. These are almost equal DB1 Condone ‘equal’. 2
Question
f x B A x _ _ O 1 2 π π A random variable X has probability _ density function f, where the graph of y = f x is a semicircle 2 with centre 0, 0 and radius , entirely above the x-axis. Elsewhere f x = 0 (see diagram). π (a) Verify that f can be a probability density function. [2] _ 1 A and B are the points where the line x = meets the x-axis and the semicircle respectively. π P _ Q 1 (b) Show that angle AOB is 41 π radians and hence find P X> . [6] π
Mark scheme
7(a) 2 M1 1 π 2 2 π = 1, which is the area under a PDF [and f(x) ≥ 0] A1 Result and statement are both needed. 2 7(b) 1 B1 AG. π Accept alternative approaches, e.g. using Pythagoras, cos −1 π = tangent, or isosceles right-angle triangles. 2 4 Answer should be convincingly obtained and all correct. π Area of sector = 1 B1 4 1 1 1 2 1 M1 Accept alternative approaches. Area of triangle AOB = OA × OB = × × − 2 2 π π π Note: AB = 0.79792 − 0.56422 [= 0.5642] or Allow values to 3sf. 1 1 1 2 π Area of triangle AOB = OA × OB × sin( AOB) = × × sin 2 2 π π 4 1 or 0.1592 A1 2π ‘ 1 ’ – ‘ 1 ’ or ‘0.25’ – ‘0.1592’ M1 Attempt area of sector – area of triangle AOB. 4 2π = 1 – 1 or 0.0908 (3sf) A1 4 2π 7(b) Alternative Method for Question Q7(b): Using integration 2 M1 Find equation of curve x 2 + y 2 = π 2 A1 y= − x2 π Attempt to integrate (any limits) M1 1 2 B1 Use of correct limits to π π Correct integration with correct limits A1 = 1 – 1 or 0.0908 (3sf) A1 Correct final answer. 4 2π 6
Question
A new light was installed on a certain footpath. A town councillor decided to use a hypothesis test to investigate whether the number of people using the path in the evening had increased. Before the light was installed, the mean number of people using the path during any 20-minute period during the evening was 1.01. After the light was installed, the total number, n, of people using the path during 3 randomly chosen 20-minute periods during the evening was noted. (a) Given that the value of n was 6, use a Poisson distribution to carry out the test at the 5% significance level. [6] (b) Later a similar test, at the 5% significance level, was carried out using another 3 randomly chosen 20-minute periods during the evening. Find the probability of a Type I error. [2] (c) State what is meant by a Type I error in this context. [1] (d) State, in context, what further information would be needed in order to find the probability of a Type II error. Do not carry out any further calculation. [2] 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
8(a) H0: Pop mean no. people = 3.03 or 1.01 (per 20 min) B1 These must not just be ‘mean’, but allow just ‘λ’ or ‘μ’. H1: Pop mean no. people > 3.03 or 1.01 (per 20 min) Use of PO(3.03) M1 2 3 4 5 M1 Allow incorrect λ. Allow one end error. Must see = 1 – e–3.03(1 + 3.03 + 3.03 + 3.03 + 3.03 + 3.03 ) Poisson expression used. 2 3! 4! 5! = 1– e–3.03(1 + 3.03 + 4.5905 + 4.6364 + 3.5120 + 2.128) = 1– (0.04832 + 0.1464 + 0.2218 + 0.2240 + 0.1697 + 0.1028) = 0.0870 (3sf) [0.0869727] A1 Allow 0.087 . 0.0870 > 0.05 M1 For a valid comparison. (Do not reject H0) Insufficient evidence to believe (mean) number of people has A1FT Conclusion stated must be in context, not definite and increased include no contradictions (e.g. not ‘mean number people has not increased’). 6 If only P(x = 6) award max 2/6 (single term not valid). SC No working B1 B2 M1 A1. Award maximum 5/6. 8(b) M1 OE. Must see Poisson expression (may be in part (a)). "0.0869727" – e–3.03 × 3.03 6 6! or 0.869727 – e–3.03(1.0748) or 0.869727 – 0.05193 2 3 4 5 6 or 1 – e–3.03(1 + 3.03 + 3.03 2 + 3.03 3! + 3.03 4! + 3.03 5! + 3.03 6! ) 0.0350 or 0.0351 A1 Accept 0.035. SC no working seen, award B1 for 0.0350, 0.0351 or 0.035. 2 8(c) Concluding that the (mean) number of people (using the path per 20 mins in the B1 OE. Conclusion must be in context. evening) has increased when it has not 1 8(d) A value for the true mean B1 Allow without context for this mark. Number of people using the path per 20 mins in the evening. B1 Condone equivalent comment on three randomly chosen 20-minute periods. 2
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Mathematics 9709/63 for May/June 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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