Cambridge A Level Mathematics 9709 — 2023 May/June Paper 6 · Variant 3

2 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 OR Only a finite number of digits used B1 Already used these numbers, so therefore not random. Does not include numbers not in the list, therefore not random (not random or biased needs a reason). 1

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3 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 100 100 (1 ) 100 x x z    = 0.07896 M1 OE. Equation of correct form. Accept p = x/100. Any z. Allow missing factor of 2. [x(100– x) = 1003  0.078962  1.6452] x2 – 100x  2304 = 0 A1 Any correct (likely scalar multiple) three-term quadratic equation in x or p with simplified coefficients. 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 Answer Marks Guidance 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' '90.72' [= –1.722] no mixed methods M1 For standardising with their values. No mixed methods. 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. N(65.2, 2 3.6 7 ). 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' '226800' [= –1.722] no mixed methods M1 For standardising with their values. No mixed methods. ɸ(–‘1.722’) = 1 – ɸ(‘1.722’) M1 For correct probability area consistent with their values. = 0.0425 or 0.0426 A1 6

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5 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 Σt = 1700 Σt2 = 59 050 (a) Find unbiased estimates of the population mean and variance. 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(b) Test, at the 2% significance level, whether the mean delivery time has changed since last year. 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(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 Est(σ2) = 2 50 59050 34 49 50        or 2 1 1700 59050 49 50        M1 Est(σ2) = 2 59050 – 34 50 biased scores M0. = 25.5 (3 sf) or 1250 49 A1 = 25 scores A0. 3 5(b) H0: Population mean time = 32.4 H1: Population mean time ≠ 32.4 B1 Not just ‘mean’ but allow just ‘μ’. 34 – 32.4 '25.5' 50 M1 Must have 50 and not 50. FT their mean and var. Can be implied. = 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). Question Answer Marks Guidance 5(c) Distribution of times in the population is normal B1 Accept answers with no context here. Accept underlying distribution for population. 1

More questions on Hypothesis tests

7 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 1 (b) Show that angle AOB is 4π radians and hence find P X > π . 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Mark scheme: 7(a) 2 1 2 π 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 1 π π cos 4 2 π              B1 AG. Accept alternative approaches, e.g. using Pythagoras, tangent, or isosceles right-angle triangles. Answer should be convincingly obtained and all correct. Area of sector = 1 4 B1 Area of triangle AOB = 1 2OA OB  = 1 1 2 1 2 π π π    or Area of triangle AOB = 1 sin( ) 2OA OB AOB   = 1 1 2 π sin 2 π π 4   M1 Accept alternative approaches. Note: AB = 2 2 0.7979 0.5642  [= 0.5642] Allow values to 3sf. 1 2π or 0.1592 A1 ‘ 1 4 ’ – ‘ 1 2π ’ or ‘0.25’ – ‘0.1592’ M1 Attempt area of sector – area of triangle AOB. = 1 4 – 1 2π or 0.0908 (3sf) A1 Question Answer Marks Guidance 7(b) Alternative Method for Question Q7(b): Using integration Find equation of curve 2 2 2 π x y   M1 2 2 π y x   A1 Attempt to integrate (any limits) M1 Use of correct limits 1 π to 2 π B1 Correct integration with correct limits A1 = 1 4 – 1 2π or 0.0908 (3sf) A1 Correct final answer. 6

More questions on Probability

8 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] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) H0: Pop mean no. people = 3.03 or 1.01 (per 20 min) H1: 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’. Use of PO(3.03) M1 = 1 – e–3.03(1 + 3.03 + 2 3 4 5 3.03 3.03 3.03 3.03 + + + 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) M1 Allow incorrect λ. Allow one end error. Must see Poisson expression used. = 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 increased A1FT Conclusion stated must be in context, not definite and 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. Question Answer Marks Guidance 8(b) "0.0869727" – e–3.03 × 6 3.03 6! or 0.869727 – e–3.03(1.0748) or 0.869727 – 0.05193 or 1 – e–3.03(1 + 3.03 + 2 3 4 5 6 3.03 3.03 3.03 3.03 3.03 2 3! 4! 5! 6!     ) M1 OE. Must see Poisson expression (may be in part (a)). 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 evening) has increased when it has not B1 OE. Conclusion must be in context. 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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