Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 4 · Variant 3

1 Ellie is investigating the heights of two types of beech tree, A and B, in a certain region. She has chosen a random sample of 60 beech trees of type A in the region, recorded their heights, x m, and calculated unbiased estimates for the population mean and population variance as 35.6 m and 4.95 m 2 respectively. Ellie also chooses a random sample of 50 beech trees of type B in the region and records their heights, y m. Her results are summarised as follows. / y = 1654 / y 2 = 54 850 Find a 95% confidence interval for the difference between the population mean heights of type A and type B beech trees in the region. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 2 1  1654 2  B1 3392 xs =  54850 −   = 2.769  49 50 1225   2 4.95 2.769 M1 A1 s = +  = 0.1379  60 50 1654 M1 Correct formula with a z value. CI: 35.6 −  1.96 s 50 B1 1.96 seen. 1.79, 3.25  A1 Accept inequality. 6

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2 A school with a large number of students is updating its logo. Each student has designed a new logo and two teachers have each awarded a mark out of 50 for each logo. The marks awarded to a random sample of 12 students are shown in the following table. Student A B C D E F G H I J K L Teacher 1 36 38 40 36 22 34 45 44 48 35 28 30 Teacher 2 38 42 32 41 32 41 42 50 36 44 42 41 One of the students claims that Teacher 2 is awarding higher marks than Teacher 1. (a) Carry out a Wilcoxon matched-pairs signed-rank test, at the 5% significance level, to test whether the data supports the claim. 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It was later discovered that Teacher 1 had entered her mark for student C incorrectly. Her intended mark was 24 not 40. This was corrected. (b) Determine whether this correction affects the conclusion of the test carried out in part (a). 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Mark scheme: 2(a) 2 4 −8 5 10 7 −3 6 −12 9 14 11 M1 Signed differences, allow one error. 1 3 −7 4 9 6 −2 5 −11 8 12 10 M1 Attempt at ranking. (Sum of + ranks = 58) Test statistic = 20 A1 H0: population median for teacher 2 = population median for teacher 1 B1 H1: population median for teacher 2 > population median for teacher 1 Allow use of m for population median. Critical value, from tables, is 17 B1 ‘20’ > 17, accept H0 M1 Insufficient evidence to support the claim A1 Correct work only except possibly hypotheses, in Insufficient evidence that the scores of Teacher 2 are higher than those of Teacher 1 context, level of uncertainty in language. 7 2(b) New test statistic is ‘13’ B1 FT Their 20 minus 7. 13 < 17, so conclusion is now ‘reject H0’, (sufficient evidence to support claim) B1 Must be 13 and 17. 2

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3 A statistician believes that the number of telephone calls received by an advice centre in a 10-minute interval can be modelled by the Poisson distribution Po(1.9). The number of calls received in a randomly chosen 10-minute interval was recorded on each of 100 days. The results are summarised in the table, together with some of the expected frequencies corresponding to the distribution Po(1.9). 6 or Number of calls 0 1 2 3 4 5 more Observed 10 18 35 21 11 4 1 frequency Expected 14.957 28.418 26.997 1.322 frequency (a) Complete the table. 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(b) Carry out a goodness of fit test, at the 10% significance level, to determine whether the statistician’s belief is reasonable. 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Mark scheme: 3(a) 17.098 8.122 3.086 B1 One correct. B1 All correct. 2 3(b) M1 0 1 2 3 4 or more Last two or three columns combined. 10 18 35 21 16 14.957 28.418 26.997 17.098 12.53 Contributions to test statistic are: M1 1.6428 3.8192 2.3724 0.8905 0.9609(7) May be implied by awrt 9.69 Test statistic is 9.69 A1 9.686 H0: Po(1.9) is a good fit for the data B1 H1: Po(1.9) is not a good fit for the data Critical value is 7.779, compare ‘9.69’ > 7.779 reject H0 M1 4 degrees of freedom Sufficient evidence to suggest that Po(1.9) is a not a good fit for the data/ A1 Correct work only, including hypotheses, in Sufficient evidence to reject/not support the statistician’s claim context, level of uncertainty in language. 6

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4 The continuous random variable X has probability density function f given by kx 3 0 G x 1 1 , f ( x) = *k ( 5 - x) 1 G x G 5 , 0 otherwise, where k is a constant. (a) Sketch the graph of f. [1] (b) Show that k = 4 . 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(c) Find the cumulative distribution function of X. 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(d) Find the median value of X. 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Mark scheme: 4(a) Correct sketch. B1 Label 1 and 5 on x-axis, ignore labels, if any, on y-axis. 1 4(b) kx 4 k B1 For either part correct. with limits 0 and 1 gives 4 4  x 2  k  5 x −  with limits 1 and 5 gives 8k  2  k 4 B1 AG + 8k = 1, k = 4 33 2 4(c)  M1 Integrate both parts. 0 x  0   A1 Middle two parts correct AEF.  1 4 x 0 ≤ x  1 A1 0 and 1 correct.  33 F( x ) =   4  1 2 17  5 x − x − 1 ≤≤x 5    33  2 4    1 x  5.  3 4(d) 1 3 m 1 M1* Or integrate m to 5. Or use F(x) = 0.5.  0kx dx +  1 k ( 5 − x ) dx = 2 4  1 1 2 1  1 4  25 1 2  1 M1 Must be a quadratic in m. + 5m − m − 5 + = or − 5m + m =     33  4 2 2  2 33  2 2  2 4  1 2 17  1 or 5 m − m − =   33  2 4  2 4 m 2 − 40 m + 67 = 0 *DM1 1 A1 m = 10 − 33 = 2.13 ( ) 2 4

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5 Nikita has three coins. One coin is fair, one coin is biased so that the probability of obtaining a head is 1 and the third coin is biased so that the probability of obtaining a head is 15. The random variable X is 3 the number of heads that Nikita obtains when he throws all three coins at the same time. (a) Find the probability generating function of X. 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Rajesh has two fair six-sided dice with faces labelled 1, 2, 3, 4, 5, 6. The random variable Y is the number of 4s that Rajesh obtains when he throws the two dice. The random variable Z is the sum of the number of heads obtained by Nikita and the number of 4s obtained by Rajesh. (b) Find the probability generating function of Z, expressing your answer as a polynomial. 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(c) Use your answer to part (b) to find E(Z). 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Mark scheme: 5(a) 8 14 7 1 B1 All correct. P(0 heads) = , P(1 head) = , P(2 heads) = , P(3 heads) = 30 30 30 30 8 14 7 2 1 3 M1 Cubic polynomial. GX (t) = + t + t + t 30 30 30 30 A1 FT FT their probabilities that sum to one. 3 5(b) 25 10 1 2 B1 GY (t) = + t + t 36 36 36 1  8 14 7 2 1 3  2 M1 With attempt to multiply. 25 + 10t + t GZ (t) =  + t + t + t  ( ) 30  36  30 30 30 30  1 2 3 4 5 M1 A1 Obtain quintic polynomial. 200 + 430t + 323t + 109t + 17t + t ( ) 1080 4 5(c) 1 M1 Differentiate and substitute t = 1. GZ’ (1) = ( 430 + 646 + 327 + 68 + 5 ) 1080 1476 41 A1 = = 1.37 1080 30 2

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6 Ansal is investigating the wingspans of Monarch butterflies in two different regions, X and Y. He takes a random sample of 8 Monarch butterflies from region X and records their wingspans, x cm. His results are as follows. 8.2 7.0 7.3 8.8 7.8 8.5 9.2 7.4 Ansal also takes a random sample of 9 Monarch butterflies from region Y and records their wingspans, y cm. His results are summarised as follows. / y = 71.10 / y 2 = 567.13 Ansal suspects that the mean wingspan of Monarch butterflies from region X is greater than the mean wingspan of Monarch butterflies from region Y. It is known that the wingspans of Monarch butterflies in regions X and Y are normally distributed with equal population variances. Test, at the 10% significance level, whether Ansal’s suspicion is supported by the data. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 6 2 1  64.2 2  M1 851 xs =  519.46 −   = 0.607857  7 8 1400   17 2 1  71.12  , both. 25 ys =  567.13 −   = 0.680  8  9  2 8  0.680 + 7  0.607857 M1 A1 1939 s =  = 0.64633 Correct formula for pooled variance used, . 9 + 8 − 2 3000 H0: X − Y = 0 B1 H1: X − Y  0 8.025 − 7.9 0.125 M1 A1 Dependent on pooled variance being used t = = = 0.320 1 1 0.3906 s + 8 9 ‘0.320’ < 1.341 accept H0 M1 Compare with 1.341 and conclusion Insufficient evidence to support Ansal’s suspicion A1 Correct work only, except possibly B1, in Insufficient evidence to suggest that wingspan from region X is greater than context, level of uncertainty in language wingspan from region Y. 8

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