Cambridge A Level Mathematics - Further 9231 — 2023 May/June Paper 4 · Variant 1

1 The lengths of the leaves of a particular type of tree are normally distributed with mean μcm. The lengths, x cm, of a random sample of 12 leaves of this type are recorded. The results are summarised as follows. / x = 91.2 / x 2 = 695.8 Find a 95% confidence interval for μ. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1   2 2 91.2 1 91.2 7.6 695.8 12 11 12 x x s           67 0.2436 275              M1 Both. CI:   11 0.2436 7.6 0.975 12 t   M1 With a t-value. 2.201 seen B1 7.6 0.3136  =  7.29, 7.91 A1 4

More questions on Inference using normal and t-distributions

2 The children at two large schools, P and Q, are all given the same puzzle to solve. A random sample of size 10 is taken from the children at school P. Their individual times to complete the puzzle give a sample mean of 9.12 minutes and an unbiased variance estimate of 2.16 minutes2. A random sample of size 12 is taken from the children at school Q. Their individual times, x minutes, to complete the puzzle are summarised by / x = 99. 6 / ( x - x ) 2 = 21.5 , where x is the sample mean. Times to complete the puzzle are assumed to be normally distributed with the same population variance. Test at the 5% significance level whether the population mean time taken to complete the puzzle by children at school P is greater than the population mean time taken to complete the puzzle by children at school Q. [8] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2   2 99.6 21.5 43 8.3 1.9545 or 12 11 22 x x s           H0: P Q    H1: P Q    B1 Pooled estimate = 9 2.16 11 1.9545 10 12 2      M1 FT FT their 1.9545 only 2.047 (2.05) A1 May be implied, allow 2.04 9.12 8.3 1 1 2.047 10 12 t     M1 FT FT their pooled estimate 1.339 (or 1.338) A1 Accept 1.34 '1.339' 1.725  Accept H0 / not significant M1 Compare with correct tabular value 1.725 and conclusion without context. Condone ‘reject H1’. Insufficient evidence to suggest that the (mean) time taken at P is greater than the (mean) time taken at Q. A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if H0 and H1 reversed. 8

More questions on Inference using normal and t-distributions

3 A random sample of 50 values of the continuous random variable X was taken. These values are summarised in the following table. Interval 1 G x 1 1 .5 1.5 G x 1 2 2 G x 1 2.5 2.5 G x 1 3 3 G x 1 3.5 3.5 G x G 4 Observed frequency 3 3 8 11 13 12 It is required to test the goodness of fit of the distribution with probability density function f given by 1 4 2 + x 1 G x G 4, 2 f ( x) = * 024e x o otherwise . The expected frequencies, correct to 4 decimal places, are given in the following table. Interval 1 G x 1 1.5 1.5 G x 1 2 2 G x 1 2.5 2.5 G x 1 3 3 G x 1 3.5 3.5 G x G 4 Expected frequency 4.4271 a 6.1285 8.4549 b 14.9678 (a) Show that a = 4.6007 and find the value of b. 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(b) Carry out a goodness of fit test, at the 10% significance level, to test whether f is a satisfactory model for the data. 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Mark scheme: 3(a) 3 2 2 2 1.5 1 4 1 4 d 24 24 3 x x x x x                  M1 Integration with correct powers and correct limits seen. 50 a  0.092014 = 4.6007 A1 AG 53 576 or 1325 288 or 0.092014 or 4.60069 seen 11.4211 b  B1 Or 11.421(0) 3 3(b) Combine first two columns: 6, 9.0278 M1 Must be seen, or implied by 1.0155. Calculate value of chi-squared: 1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885 M1 At least 2 correct values (at least 3sf) or expressions seen. Allow columns not combined or three columns combined. 3.16 A1 CWO. Dependent on M1M1 scored. SCB1 for 3.16 with no working H0: f is a good fit for the data H1: f is not a good fit for the data B1 Tabular value of chi-squared: 7.78 ‘3.16’ < 7.78 and accept H0 M1 Correct tabular value: allow correct FT value if columns not combined (9.236) or three columns combined (6.251). Insufficient evidence to suggest that f is not a good fit for the data Condone: sufficient evidence to suggest that f is a good fit for the data. A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if hypotheses reversed 6

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4 A random sample of 13 technology companies is chosen and the numbers of employees in 2018 and in 2022 are recorded. Company A B C D E F G H I J K L M Number in 2018 104 19 126 234 970 514 35 149 429 12 86 304 1104 Number in 2022 106 24 127 228 1012 525 32 156 449 24 78 294 1154 A researcher claims that there has been an increase in the median number of employees at technology companies between 2018 and 2022. (a) Carry out a Wilcoxon matched-pairs signed-rank test, at the 5% significance level, to test whether the data supports this claim. 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The researcher notices that the figures for company G have been recorded incorrectly. In fact, the number of employees in 2018 was 32 and the number of employees in 2022 was 35. (b) Explain, with numerical justification, whether or not the conclusion of the test in part (a) remains the same. 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Mark scheme: 4(a) 2 5 1 –6 42 11 –3 7 20 12 –8 –10 50 2 4 1 –5 12 9 –3 6 11 10 –7 –8 13 M1 Differences (allow up to 3 errors) A1 Correct rank order, ignore signs   68, 23 P Q   A1 CWO H0: difference in population medians = 0 H1: population median in 2022 > population median in 2018 B1 ‘Population’ required. Accept use of m, not . Do not accept ‘difference between population medians > 0’ without 2022, 2018 OE specified Critical value 21 B1 ‘23’ > ‘21’ and accept H0 M1 Compare their 23 with their 21 and FT conclusion. Their 23 must be less than 46. Insufficient evidence to support researcher’s belief / insufficient evidence that the (median) number of employees in 2022 is greater than the (median) number of employees in 2018 A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if hypotheses reversed 7 4(b) Rank for G will now be +3, giving Q = 20 which is < 21 M1 Must include numbers, ft their 21 from part (a) and ft their 20 (23 – 3) Change in conclusion A1 CWO Condone ‘reject H0’ OE. 2

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5 Harry has three coins. • One coin is biased so that, when it is thrown, the probability of obtaining a head is 1.3 • The second coin is biased so that, when it is thrown, the probability of obtaining a head is 1.4 • The third coin is biased so that, when it is thrown, the probability of obtaining a head is 1.5 The random variable X is the number of heads that Harry obtains when he throws all three coins together. (a) Find the probability generating function of X. 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Isaac has two fair coins. The random variable Y is the number of heads that Isaac obtains when he throws both of his coins together. The random variable Z is the total number of heads obtained when Harry throws his three coins and Isaac throws his two coins. (b) Find the probability generating function of Z, expressing your answer as a polynomial in t. 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(c) Use the probability generating function of Z to find E(Z). 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Mark scheme: 5(a) P(3H) = 1 60 P(2H) = 9 60 P(1H) = 26 60 P(0H) = 24 60 B1 Two probabilities correct, seen anywhere. GX (t) = 2 3 24 26 9 1 60 60 60 60 t t t    M1 Cubic polynomial with 4 probabilities as coefficients with the correct powers of t from their working. Equivalent forms are acceptable. A1 Correct, AEF. 3 5(b) GY (t) = 2 1 1 1 4 2 4 t t   B1 GZ(t) = 2 3 2 24 26 9 1 1 1 1 60 60 60 60 4 2 4 t t t t t               M1 Attempt to multiply their two PGFs.   2 3 4 5 1 24 74 85 45 11 240 t t t t t      M1 Obtain quintic expression and collect terms A1 Correct, AEF. 4 5(c) Differentiate: G'Z (t) =   2 3 4 1 74 170 135 44 5 240 t t t t     M1 Differentiate their G. G'Z (1) = 428 107 1.78 240 60   A1 Any correct form. 2

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6 The continuous random variable X has probability density function f given by 3 2 1 x - 21 x 28 e + 4 e 0 G x G 2 ln 3 , f ( x) = ` j * 0 otherwise . (a) Find the cumulative distribution function of X. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 2 The random variable Y is defined by Y = e 1 ( X ) . (b) Find the probability density function of Y. 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(c) Find the 30th percentile of Y. 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(d) Find E ( Y 4 ) . 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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(a) F(   1 1 2 2 3 ) 2e 8e 28 x x x c             F( 1 1 2 2 3 9 ) 2e 8e 28 14 x x x             A1 Correct ranges including 0 2ln3 x   associated with their F( ) x F( x ) = 0 for 0 x  and 1 for 2ln3 x  B1 No gaps in range. 3 6(b) G( 3 8 ) 2 6 28 y y y          M1 y substituted into their F. 1 3 y   B1 Seen anywhere, condone ln3 1 e y   . g( 2 3 8 ) 2 28 y y         A1 Correct expression, not containing logs. 3 6(c) 3 8 3 2 6 28 10 y y          M1 Their G(y) 3 10  2 5 8 20 0 y y    M1 Obtain from G(y) and solve quadratic equation to find y. 1.35 y  A1 Single correct answer. 3 Question Answer Marks Guidance 6(d) E(Y 4) =   4 4 2 2 3 3 1 1 3 8 3 2 d 2 8 d 28 28 y y y y y y            = 5 3 3 2 8 28 5 3 y y        1 Integrate 4 y their g(y). Limits must be 1 and 3. 17.8 A1 89 5 2

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