Cambridge A Level Mathematics - Further 9231 — 2021 Oct/Nov Paper 4 · Variant 2

1 The number, x, of pine trees was counted in each of 40 randomly chosen regions of equal size in country A. The number, y, of pine trees was counted in each of 60 randomly chosen regions of the same equal size in country B. The results are summarised as follows. / x = 752 / x 2 = 14320 / y = 1548 / y 2 = 40200 Find a 95% confidence interval for the difference between the mean number of pine trees in regions of this size in countries A and B. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 2 2 1 752 14320 4.677 39 40   = − =     xs = 304 65 M1 One correct unsimplified. 2 2 1 1548 40200 4.434 59 60   = − =     ys = 1308 295 A1 Both correct to at least 3sf. 2 4.6769 4.4339 40 60 = + s M1 0.1908 3659 19175   =     A1 CI = 752 1548 1.96 40 60   − ±     s M1A1 Correct form with a z-value with 1.96 s can be unsimplified. 7.0 0.856 − ± = [−7.86, −6.14] A1 Accept in either form, ISW. Accept [6.14, 7.86]. Accept inequality form. 7

More questions on Inference using normal and t-distributions

2 It is claimed that the heights of a particular age group of boys follow a normal distribution with mean 125 cm and standard deviation 12 cm. Observations for a randomly chosen group of 60 boys in this age group are summarised in the following table. The table also gives the expected frequencies, correct to 2 decimal places, based on the normal distribution with mean 125 cm and standard deviation 12 cm. Height, x 1 100 100 G x 1 110 110 G x 1 120 120 G x 1 130 130 G x 1 140 x H 140 x cm Observed 0 3 15 23 11 8 frequency Expected 1.12 5.22 13.97 19.38 13.97 6.34 frequency (a) Show how the expected frequency for 130 G x 1 140 is obtained. 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(b) Carry out a goodness of fit test, at the 5% significance level, to determine whether the claim is supported by the data. 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Mark scheme: 2(a) P(130 140 < x  ) = P 130 125 140 125 12 12 Z − −   <     ≤ P( 0.4167 1.25) < < Z Must see 0.4167 and 1.25, may be implied by 0.8944 and 0.6616 Accept 0.417. 0.8944 0.6616 0.2328 − = Multiply by 60 leads to 13.97 A1 AG Accept 0.6615. Need to see 0.2328 or 0.2329 or 13.968 or ( ) 0.8944 0.6616 60 − × . 2 2(b) 0 3 15 23 11 8 1.12 5.22 13.97 19.38 13.97 6.34 M1 Combine first two values. Test stat = 1.7588 + 0.0761 + 0.6743 + 0.6309 + 0.4352 M1 3.58 A1 Accept 3.57 – 3.58 . SC: 3.88, if values not combined, scores M1A0. 0 : distribution fits data H N(125, 122) is a good model for the data oe B1 Must mention distribution and data. 4 degrees of freedom, so tabular value = 9.488 3.58 < 9.488 Accept 0 H . M1 Compare their value with 9.488 (9.49) and correct FT conclusion. Or, if values not combined, compare their value with 11.07 and correct FT conclusion for M1A0. There is sufficient evidence that the normal distribution fits the data There is sufficient evidence that the claim is supported by the data A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6

More questions on χ²-tests

3 The continuous random variable X has probability density function f given by a + 15 x 0 G x 1 1 , G x G 2 , f ( x) = 2a - 15 x 1 * 0 otherwise, where a is a constant. (a) Find the value of a. 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(b) Find E ( X 2 ) . 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(c) Find the cumulative distribution function of X. 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Mark scheme: 3(a) Total prob = 1, so 1 2 0 1 1 1 2 1 5 5     + + − =           a x dx a x dx 2 2 1 1 2 10 10     + + −         ax x ax x = 1 *M1 Correct expression integrated and equated to 1. Powers of x correct. 1 3 1 5 − = a DM1 Limits used and attempt to solve. Solve to give 2 5 = a A1 3 3(b) ( ) ( ) 1 2 2 2 0 1 1 1 2 4 5 5 + + −   x x dx x x dx 3 4 3 4 1 2 1 1 4 1 5 3 4 5 3 4     + + −         x x x x M1 Correct expressions integrated, FT their a. 13 10 A1 11 67 60 60 + 2 Question Answer Marks Guidance 3(c) 2 2 0 < 0 1 1 2 + 0 <1, 5 2 F( ) = 1 1 4 1 1 2, 5 2 1 > 2. x x x x x x x x x ìïïïï æ ö ï ÷ ç ï ÷ ç ï ÷ çè ø ïïíï æ ö ï ÷ ç ï ÷ ç ÷ ï çè ø ï - ï - ïï ïî    M1 Integration of their PDF. A1 Middle 2 parts correct. ( ) 0 0 and 1 ( 2) x x = < = > A1 First and last parts correct and all domains correct, with no gaps. 3

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4 Applicants for a particular college take a written test when they attend for interview. There are two different written tests, A and B, and each applicant takes one or the other. The interviewer wants to determine whether the medians of the distribution of marks obtained in the two tests are equal. The marks obtained by a random sample of 8 applicants who took test A and a random sample of 8 applicants who took test B are as follows. Test A 46 32 29 12 33 18 25 40 Test B 36 28 49 37 48 35 41 31 (a) Carry out a Wilcoxon rank-sum test at the 5% significance level to determine whether there is a difference in the population median marks obtained in the two tests. 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The interviewer considers using the given information to carry out a paired sample t-test to determine whether there is a difference in the population means for the two tests. (b) Give two reasons why it is not appropriate to use this test. 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Mark scheme: 4(a) 12 1 28 4 18 2 31 6 25 3 35 9 29 5 36 10 32 7 37 11 33 8 41 13 40 12 48 15 46 14 49 16 M1 Attempt at ranking. Test statistic: 52 A1 0 1 and : : x y x y H m m H m m = ≠ B1 Allow in words but ‘population’ must be included. Critical value for (8, 8) is 49. *B1 Allow 51 if clearly one-tail test in hypotheses. 52 > 49 Accept 0 H DM1 Compare their calculated value with 49 and correct FT conclusion. Insufficient evidence of difference in medians. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6 4(b) Not a paired sample. B1 Underlying distribution/population not (known to be) normal Underlying distribution/population unknown B1 B0 for ‘data is not normally distributed’ B0 for ‘marks are not normally distributed’ 2

More questions on Non-parametric tests

5 The random variable X is such that P ( X = r) = kr 2 for r = 1, 2, 3, 4, where k is a constant. (a) Find the value of k. 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(b) Find the probability generating function G X ( t ) of X. 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The random variable Y has probability generating function GY ( t ) 1 1 1 2 The random variable Z is the sum of X and Y. (c) Assuming that X and Y are independent, find the probability generating function GZ ( t ) of Z as a polynomial in t. 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(d) Given that E ( Z ) = 133 , use GZ ( t ) to find Var (Z). 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Mark scheme: 5(a) ( ) 1 1 4 9 16 1 so 30 k k + + + = = 1 5(b) ( ) 2 3 4 1 4 9 16 G 30 30 30 30 = + + + X t t t t t M1A1 Using their k in a polynomial, at least two terms correct for their k. 2 5(c) 2 3 4 2 1 4 9 16 1 1 1 30 30 30 30 4 2 4    + + + + +       t t t t t t M1 Method and attempt to multiply. ( ) 2 3 4 5 6 1 6 18 38 41 16 120 + + + + + t t t t t t M1A1 Multiplication to obtain single polynomial of order 6. 3 5(d) Given: G'(1) = 13/3 G''(t) = 2 3 4 1 (12 108 456 820 480 120 + + + + t t t t ) M1 Differentiate twice. Var(X) = G''(1) + 2 13 13 3 3   −    M1 Use correct formula. 1876 13 169 107 120 3 9 90 + − = or 1.19 A1 CAO 3

More questions on Probability generating functions

6 A scientist is investigating the masses of a particular type of fish found in lakes A and B. He chooses a random sample of 10 fish of this type from lake A and records their masses, x kg, as follows. 2.1 1.8 0.9 3.0 2.4 2.6 1.8 2.2 1.9 2.5 The scientist also chooses a random sample of 12 fish of this type from lake B, but he only has a summary of their masses, y kg, as follows. / y = 24.48 / y 2 = 53 .75 Test at the 10% significance level whether the mean mass of fish of this type in lake A is greater than the mean mass of fish of this type in lake B. You should state any assumptions that you need to make for the test to be valid. 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Mark scheme: 6 0 1 : : and A B A B H H μ μ μ μ = > B1 2 21.2 47.92  =  = x x 2 2 1 21.2 47.92 0.33067 9 10   = − =     xs 124 375 = 2 2 1 24.48 53.75 0.34644 11 12   = − =     ys 9527 27500 = M1 A1 2 9 0.33067 11 0.34644 10 12 2 × + × = + − s = 0.3393 M1 A1 2.12 2.04 0.321 1 1 10 12 − = = + t s M1 A1 Accept 0.321 – 0.322. Critical value: 1.325 0.321 < 1.325 accept 0 H . M1 Compare their value with 1.325 and correct FT conclusion. Insufficient evidence that mean of A is greater than mean of B A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. EITHER: Distributions are normal and equal variances OR: Distributions are normal B1 Assumptions consistent with method used. Accept ‘population is normal’. ‘Distribution of difference of means is normal’. Alternative method for question 6 0 1 : : and A B A B H H μ μ μ μ = > B1 Question Answer Marks Guidance 2 21.2 47.92  =  = x x 2 2 1 21.2 47.92 0.33067 9 10   = − =     xs 124 375 = 2 2 1 24.48 53.75 0.34644 11 12   = − =     ys 9527 27500 = M1 For one correct unsimplified. A1 For both correct. 2 0.33067 0.34644 10 12 = + s = 0.061936 M1 A1 ( ) 2.12 2.04 0.321 5 − = = t s M1 A1 Accept 0.321 – 0.322. Critical value: 1.325 0.321 < 1.325 accept 0 H . M1 Compare their value with 1.325 and correct FT conclusion. Insufficient evidence that mean of A is greater than mean of B. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. EITHER: Distributions are normal and equal variances OR: Distributions are normal. B1 Assumptions consistent with method used. Accept ‘population is normal’. ‘Distribution of difference of means is normal’. 10

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