Cambridge A Level Mathematics - Further 9231 — 2024 May/June Paper 4 · Variant 2

1 The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean n minutes. The times taken, x minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where x is the sample mean. x = 42.8 / ( x - x ) 2 = 941.5 Find a 95% confidence interval for n. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 941.5 2 13 s  CI: 2 42.8 2.160 14 s  M1 A1 Correct formula with a t value. 2.160 seen in a CI formula for A1. [37.9, 47.7] A1 (37.9, 47.7) A1 37.9 ' ' 47.7 x   A1 (ignore symbol used inside the inequality) Condone [47.7, 37.9] A1 Final answer 42.8 4.91  A0 4

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2 A large number of students are taking a Physics course. They are assessed by a practical examination and a written examination. The marks out of 100 obtained by a random sample of 15 students in each of the examinations are as follows. Student A B C D E F G H I J K L M N O Practical 66 63 24 52 59 76 88 51 48 36 91 72 68 67 60 examination Written 63 57 39 50 47 71 87 65 56 39 78 70 61 62 70 examination Use a sign test, at the 10% significance level, to test whether, on average, the practical examination marks are higher than the written examination marks. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 H0: practical results and written results are equal H1: practical results are greater than written results B1 Or H0 : 0 P W m m , H1 : 0 P W m m , oe (m is population median). There are 10 + signs; B(15, 0.5) soi M1 (P(   10) 1 P 9 X X    =) 1 – 0.849 = 0.151 A1 P(X ⩽ 5) = 0.151 (implies M1) Compare with 0.1: ‘0.151’ > 0.1 accept H0 M1 ‘0.151’ must come from a valid binomial calculation Correct ft conclusion for their 0.151 and 0.1 Condone Reject H1 ‘Accept H0’ can be implied by a conclusion that is consistent with their 0.151 and 0.1 Insufficient evidence to suggest that practical results are greater than written results A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not ‘prove’. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’. 5

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3 A factory produces metal discs. The manager claims that the diameters of these discs have a median of 22.0 mm. The diameters, in mm, of a random sample of 12 discs produced by this factory are as follows. 22.4 20.9 22.8 21.5 23.2 22.9 23.9 21.7 19.8 23.6 22.6 23.0 (a) Carry out a Wilcoxon signed-rank test, at the 10% significance level, to test whether there is any evidence against the manager’s claim. [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State an assumption that is necessary for this test to be valid. [1] ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Signed differences 0.4 1.1 0.8 0.5 1 .2 0.9 1 .9 0.3 2.2 1 .6 0.6 1 .0     Ranks: 2 8 5 3 9 6 1 1 1 12 1 0 4 7     M1 Attempt at ranking. (P+ = 54) P- = 24 A1 All ranks must be correct. H0: population median = 22.0 H1: population median  22.0 B1 Accept m used as ‘population median’ Critical value from table = 17 B1 ‘24’ > ‘17’ Accept H0 M1 ‘24’ must come from a ranking of signed differences, ‘17’ must be critical value from tables (7, 9, 10, 13, 17 or 21). Correct ft conclusion for their 24 and their 17. Condone Reject H1. ‘Accept H0’ can be implied by a conclusion that is consistent with their 24 and their 17. Insufficient evidence to support (population) median not being 22.0 Insufficient evidence against the manager’s claim A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not ‘prove’. Condone ‘no sufficient’ ‘not enough’. Do not accept statements such as ‘there is sufficient evidence to suggest…..’. 7 Question Answer Marks Guidance 3(b) Distribution is symmetric about the population median Underlying/population distribution is symmetric about the median B1 Need to see population or underlying distribution mentioned. Underlying distribution is symmetric B0. Population distribution is symmetric B0. ‘data’ implies B0 but condone ‘population data’ ‘mean’ implies B0. 1 Question Answer Marks Guidance G’Y (t) =     

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4 The random variable Y is the sum of two independent observations of the random variable X. The probability generating function G Y ( t) of Y is given by t 2 G Y ( t) = 4 . ( 4 - 3t) (a) Find E ( Y ) . 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(b) Write down an expression for the probability generating function of X. 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(c) Find P ( X = 4) . 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Mark scheme: 4(a) GY (t) =   2 4 , 4 3 t t    4 3 2 8 4 3 2 12 4 3 4 3 t t t t t     or     2 4 5 2 12 4 3 4 3 t t t t    M1 ‘Two’ terms obtained, correct denominator(s). E(Y) = G’Y (1) = 14 M1 A1 t = 1 in their expression. CWO 3 4(b) GX (t) =   2 4 3 t t  B1 1 Question Answer Marks Guidance 4(c) GX (t) =   2 2 2 3 3 3 9 27 4 3 1 1 3 4 16 4 16 2 16 64 t t t t t t t t                        M1 Expansion as far as 3t OR term in 4t calculated. Use P(X = 4) = their coefficient of 4t M1 Find the numerical value of their coefficient of 4t . 27 256 A1 Accept 0.105. 3

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5 Two companies, P and Q, produce a certain type of paint brush. An independent examiner rates the quality of the brushes produced as poor, satisfactory or good. He takes a random sample of brushes from each company. The examiner’s ratings are summarised in the table. Company Poor Satisfactory Good P 18 43 64 Q 22 22 31 (a) Test, at the 5% significance level, whether quality of brushes is independent of company. 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(b) Compare the quality of the brushes produced by the two companies. 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Mark scheme: 5(a) Company Poor Satisfactory Good P (18) 25 (43) 40.625 (64) 59.375 Q (22) 15 (22) 24.375 (31) 35.625 M1 A1 Calculate E values to at least 2dp, allow one error. These must be seen. For A1, working to at least 2dp. Chi-squared contributions: Company Poor Satisfactory Good P 1.96 0.1388 0.3603 Q 3.2667 0.2314 0.6004 M1 At least 2 correct values or expressions seen. Or may be implied by AWRT 6.56 Test statistic = 6.56 A1 AWRT 6.56. H0: Quality (of items) is independent of company H1: Quality (of items) is not independent of company B1 Condone ‘no association/relationship between….’ ‘association/relationship between….’. 6.56 > 5.991 Reject H0 M1 ‘6.56’ from their attempt at chi-squared, 5.991 must be correct. Consistent signs in comparison. FT conclusion from their 6.56. Condone ‘Accept H1’. ‘Reject H0’ can be implied by a conclusion that is consistent with their 6.56. Question Answer Marks Guidance 5(a) There is sufficient evidence to suggest that the quality of brushes is not independent of company. A1 Correct work only, conclusion in context with level of uncertainty in language. Not ‘prove’. Accept ‘enough evidence’. Condone ‘some evidence’. Do not accept ‘there is insufficient…’. 7 5(b) P produces better quality brushes (than Q) or Q produces worse quality brushes (than P) P has fewer poor quality brushes than expected AND Q has more poor quality brushes than expected B1 Condone P is better (than Q). 1

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6 Jade is a swimming instructor at a sports college. She claims that, as a result of an intensive training course, the mean time taken by students to swim 50 metres has reduced by more than 1 second. She chooses a random sample of 10 students. The times taken, in seconds, before and after the training course are recorded in the table. Student A B C D E F G H I J Time before 54.2 47.4 52.1 59.0 55.3 51.0 48.9 52.2 58.4 51.4 course Time after 50.1 46.3 52.5 58.8 51.4 48.4 49.5 48.7 58.3 51.4 course (a) Test, at the 10% significance level, whether Jade’s claim is justified. 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(b) State an assumption that is necessary for this test to be valid. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) Signed differences: 4.1 1 .1 0.4 0.2 3.9 2.6 0.6 3.5 0.1 0   2 14.5, 52.81 d d     2 2 1 14.5 52.81 3.532 9 10 s          M1 H0: 1 B A     H1 : 1 B A     B1 Accept H0: 1, d  H1 : 1 d . 2 1.45 1 10 t s   M1 The – 1 must be present in the numerator. 0.757 A1 Critical value is 1.383: ‘0.757’ < 1.383 Accept H0 M1 ‘0.757’ must come from a paired sample t- calculation, 1.383 must be correct Correct ft conclusion for their 0.757 and 1.383 Condone Reject H1 ‘Accept H0’ can be implied by a conclusion that is consistent with their 0.757 and 1.383 Insufficient evidence to support Jade’s claim A1 Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not ‘prove’ Condone ‘no sufficient’ ‘not enough’ Do not accept statements such as ‘there is sufficient evidence to suggest…..’ 7 Question Answer Marks Guidance 6(b) Distribution of population differences is normal B1 Underlying distribution of differences normal B1 ‘data’ implies B0. 1 B1

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7 The continuous random variable X has probability density function f given by x 2 ( 4 - x ) 0 G x G 2 , 4 f ( x) = * 0 otherwise. (a) Find Var X [4] ` j. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The continuous random variable Y is defined by Y = X 2 . (b) Find the probability density function of Y. 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(c) Find the exact value of the median of Y. 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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: 7(a)    2 4 0 2, f 4 0 otherwise. x x x x        (E(X) =)   2 1 1 4 1 16 2 4 3 5 4 4 3 5 15 0 4 d x x x x x          ( = 1.067) (E(    2 2 1 1 8 2 32 1.5 7.2 2.5 4.5 4 4 5 9 45 0 0 ) ) 4 d 2 1.005 7 X x x x x x                M1 Integration, correct powers required, ignore limits. Var(       2 ) E X E X X   used M1 Correct formula used following attempt at integration of both terms.   2 16 32 15 45 2 0.0553         A1 112 2025 4 Question Answer Marks Guidance 7(b) F(x) =   2 4 1 8 16 x x    0 2 x  B1 G(y) =   2 1 8 16 y y  [ 0 4] y   M1 Change variable in their F(x).      1 8 2 0 4 16 g y y y     M1 Differentiate their G(y) .    1 8 2 0 4, 16 0 otherwise. y y g y        A1 Complete and correct. 4 7(c)   2 1 1 8 16 2 m m   M1 Equate their G(y) to 1 2 . 2 [ 8 8 0,] 4 2 2 m m m     A1 Single answer in an exact form. 2

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