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Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w20_qp_629709_w20_ms_629709/62/O/N/20
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
On average, 1 in 50 000 people have a certain gene. Use a suitable approximating distribution to find the probability that more than 2 people in a random sample of 150 000 have the gene. [3]
Mark scheme
1 Poisson, any λ M1 Used 32 M1 Allow one end error 1 – e–3 1 + 3 + 2 = 0.577 (3sf) A1 SC Use of Binomial (or unsupported correct answer) scores B1 only 3
Question
A six-sided die has faces marked 1, 2, 3, 4, 5, 6. When the die is thrown 300 times it shows a six on 56 throws. (a) Calculate an approximate 96% confidence interval for the probability that the die shows a six on one throw. [3] (b) Maroulla claims that the die is biased. Use your answer to part (a) to comment on this claim. [1]
Mark scheme
2(a) 56 244 M1 For expression of the correct form. Must be a z value × 56 300 300 ± z× 300 300 z = 2.054 or 2.055 B1 0.14(0) to 0.233 (3sf) or 0.141 to 0.233 (3sf) A1 Must be an interval 3 2(b) 1 B1 FT 1 56 (= 0.167) Note if confidence interval set up with , it should be the 6 6 300 This is within confidence interval, so no reason to believe die is value used here. biased. FT their confidence interval. Not definite, e.g. not ‘Die not biased’. 1
Question
f x c x −1 0 1 2 3 4 A random variable X takes values between 0 and 3 only and has probability density function as shown in the diagram, where c is a constant. (a) Show that c = 23 . [1] (b) Find P X > 2. [2] (c) Calculate E X . [4]
Mark scheme
3(a) 1 B1 Must see this line, oe, and result ×3×c=1 2 −c (Alternative method involving equation of line y = x + c 2 3 (c = AG) 3 must have all relevant working shown) 1 3(b) 1 2 M1 1 Allow M1 for seen as a linear scale factor 3 3 or Attempt to find equation of line (of form y = mx + c where c ≠ 0) and interval from 2 to 3 OE 2 or Attempt to find the point (2, ) using the equation of the line 9 (of form y = mx +c where c ≠ 0) and then use area of triangle 1 A1 = or 0.111(3sf) 9 2 3(c) 2 2 *M1 OE Must be of form y = mx + c (c ≠ 0). Equation of line is y = − ÷ 3 x 3 3 3 2 2 DM1 Attempt integration x × their f(x), ignore limits E(X) = x − x 2 dx 0 3 9 x 2 2 x3 3 A1 FT Correct integration and limits. FT their equation of line = − 3 27 0 =1 A1 4
Question
The areas, X cm2, of petals of a certain kind of flower have mean - cm2. In the past it has been found that - = 8.9. Following a change in the climate, a botanist claims that the mean is no longer 8.9. The areas of a random sample of 200 petals from this kind of flower are measured, and the results are summarized by Σ x = 1850, Σ x2 = 17 850. Test the botanist’s claim at the 2.5% significance level. [8]
Mark scheme
4 1850 B1 est(μ) = or 9.25 200 200 17850 1850 2 1 18502 M1 est(σ2) = − or 17850 − 199 200 200 199 200 1475 A1 = 3.71 or 3.7060 or 398 H0: μ = 8.9 B1 Accept Population mean (not just mean) H1: μ ≠ 8.9 1850 M1 Use of biased variance (3.6875) still scores M1 − 8.9 200 "3.706" 200 = 2.57(3sf) (or using areas 0.00507 – 0.0051) A1 Accept 2.58 (3sf) or using areas 0.0049–0.005 where biased variance used. 2.24 < 2.57 or 0.00507 < 0.0125 M1 For valid comparison with 2.240 or 2.241 or valid comparison with 0.0125 Accept 2.24 < 2.58 or 0.00496 < 0.0125 where biased variance used (Reject H0) There is evidence that μ is not 8.9 A1 FT Not definite, e.g. NOT ‘μ ≠ 8.9’ Must be in context. No contradictions. (Accept cv method) (Note: Use of 1 tail test scores Max B1M1A1B0M1A1M1A0, max 6 out of 8) 8
Question
Customers arrive at a shop at a constant average rate of 2.3 per minute. (a) State another condition for the number of customers arriving per minute to have a Poisson distribution. [1] It is now given that the number of customers arriving per minute has the distribution Po 2.3. (b) Find the probability that exactly 3 customers arrive during a 1-minute period. [2] (c) Find the probability that more than 3 customers arrive during a 2-minute period. [3] (d) Five 1-minute periods are chosen at random. Find the probability that no customers arrive during exactly 2 of these 5 periods. [3]
Mark scheme
5(a) Customers arrive independently or singly or at random B1 Any one of these in context 1 5(b) 2.33 M1 Attempt correct expression seen e −2.3 × 3! = 0.203 (3sf) A1 2 5(c) Po(4.6) B1 SOI 4.62 4.63 M1 Correct expression, with any λ, allow one end error 1 − e−4.6 1 + 4.6 + + 2! 3! = 0.674 (3sf) A1 As final answer. 3 5(d) P(none arrive) = e–2.3 (= 0.10026) M1 Must be clearly their P(none arrive) 5 C2(e–2.3)2(1 – e–2.3)3 M1 FT their e–2.3 = 0.0732 or 0.0733 (3sf) A1 3
Question
A biscuit manufacturer claims that, on average, 1 in 3 packets of biscuits contain a prize offer. Gerry suspects that the proportion of packets containing the prize offer is less than 1 in 3. In order to test the manufacturer’s claim, he buys 20 randomly selected packets. He finds that exactly 2 of these packets contain the prize offer. (a) Carry out the test at the 10% significance level. [5] (b) Maria also suspects that the proportion of packets containing the prize offer is less than 1 in 3. She also carries out a significance test at the 10% level using 20 randomly selected packets. She will reject the manufacturer’s claim if she finds that there are 3 or fewer packets containing the prize offer. Find the probability of a Type II error in Maria’s test if the proportion of packets containing the prize offer is actually 1 in 7. [3] (c) Explain what is meant by a Type II error in this context. [1]
Mark scheme
6(a) 1 B1 Allow p for P(contains offer) but not just proportion H0: P(contains offer) = 3 1 H1: P(contains offer) < 3 P(0,1 or 2 offers in 20 | H0) M1 20 19 18 2 2 2 1 2 1 = + 20 + 20C2 3 3 3 3 3 = 0.0176 (3sf) A1 ‘0.0176’ < 0.1 M1 For valid comparison. SC comparison of 0.982(4) > 0.9 scores M1 and recovers the previous M1 A1 (Reject H0 ) No evidence (at 10% level) to support A1 FT In context. Not definite. No contradictions. manufacturers claim (Note 2 tail test scores max B0M1A1M1A0, max 3 out of 5) Accept critical region method: M1 A1 for correctly finding critical region of < 4 ; 2 in critical region M1; A1 conclusion 20 40 SC Use of Normal approximation N , scores B1 M1 A0 3 9 20 2.5 − 3 M1 A1 max; the first M1 for requires use of correct 40 9 continuity correction and the comparison 0.024 < 0.1 OE must be a valid comparison 5 6(b) 1 – P(X ⩽ 3) M1 M1 for 1 – (one term omitted or extra or incorrect) or omit '1 − ' 6 20 19 6 1 18 6 1 2 6 1 17 3 A1 for all correct expression = 1 − + 20 + 20 C2 + 20 C3 7 7 7 7 7 7 7 = 0.318 (3sf) A1 As final answer. 3 6(c) Concluding that prop is 1 in 3 when it is actually less(1 in 7) B1 OE, in context. 1
Question
Before a certain type of book is published it is checked for errors, which are then corrected. For costing purposes each error is classified as either minor or major. The numbers of minor and major errors in a book are modelled by the independent distributions N 380, 140 and N 210, 80 respectively. You should assume that no continuity corrections are needed when using these models. A book of this type is chosen at random. (a) Find the probability that the number of minor errors is at least 200 more than the number of major errors. [5] The costs of correcting a minor error and a major error are 20 cents and 50 cents respectively. (b) Find the probability that the total cost of correcting the errors in the book is less than $190. [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) P(S > L + 200) = P(S – L > 200) B1 These may be implied by next line E(S – L) = 380–210 (=170) or E(S–L–200)= 380–210–200 (=–30) Var(S – L) = 140+80 (=220) or Var(S–L–200)= 140+80 (=220) B1 200 − "170" 0 − "− 30" M1 Standardising with their values (must be from a combination or (= 2.023) attempt) Allow with attempted continuity correction. "220" "220" 1 – ɸ("2.023") M1 Area consistent with their values = 0.0216 (3sf) A1 (0.0234 with continuity correction) 5 7(b) E(total cost) = 380 × 20 + 210 × 50 (= 18 100) B1 or $181 Var(total cost) = 140 × 202 + 80 × 502 (= 256 000) B1 or 25.6 (dollar2) These may be implied by next line 19000 − "18100" 190 − "181" M1 Standardising with their values (must be from a combination or (= 1.778) attempt). No mixed methods. "256000" "25.6" Allow with attempted continuity correction. ɸ("1.778") M1 Area consistent with their values = 0.962 or 0.963 (3 sf) A1 (0.953 or 0.954 with continuity correction) 5
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Mathematics 9709/62 for October/November 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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