Cambridge A Level Mathematics - Further 9231 — 2025 Oct/Nov Paper 4 · Variant 3

1 A group of 10 school children are asked to estimate the size of an angle i° in a given acute angled triangle. These estimates, in degrees, are as follows. 84 85 77 85 84 87 86 88 83 85 (a) Stating any assumptions you make, calculate a 95% confidence interval for i. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Give a reason why the assumptions made in part (a) may not be appropriate in this case. [1] ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a)   = 13415 = 8.933 M1 Correct expression, implied by AWRT 8.93. s 2 = 19 ( '71314'− '844'10 2 ) 2.262 B1 2.262 or 2.26 seen. '844' 10  '2.262' '8.933'10 M1 Correct form, must be a t-value. [82.3, 86.5] A1 Accept with inequality signs or open brackets. Condone [86.5, 82.3]. Do not accept 84.4  2.1 . The distribution of estimates of angles is normal. B1 OR The estimates are a random sample from some population. OR The estimates are independent. OR Underlying distribution is normal. 5 1(b) Population unlikely to be normal as  is close to right-angle / data B1 Must refer to the context. is skewed. OR Estimates may not be independent, for example due to collusion. OR No indication that sample is random. 1

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2 The manager of a car park claims that the number of cars entering the car park follows a Poisson distribution with mean 2.8. The numbers of cars entering the car park are recorded on a working day during successive 5-minute periods. The following table contains the observed frequencies, together with most of the expected frequencies and their contributions to the | 2 -test statistic. Number of cars 0 1 2 3 4 5 H 6 Observed frequency 2 15 31 29 13 3 7 Expected frequency 6.081 17.03 23.84 p 15.57 8.721 6.511 | 2 -test statistic 2.739 0.241 2.152 q 0.425 3.753 0.037 (a) Find the value of p and the value of q. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Carry out a goodness of fit test at the 5% significance level to investigate the manager’s claim. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) e −2.8 (2.8) 3 B1 Or by subtraction. p =  100 = 22.248 Allow 22.2 or 22.3. 3! (29 − 22.25) 2 B1 AWRT 2.05. q = = 2.05 22.25 2 2(b) H0: Po(2.8) fits the data. B1 Must mention distribution and data/number of cars. H1: Po(2.8) does not fit the data. OR H0: Po(2.8) is a satisfactory model for the number of cars (entering the car park). H1: Po(2.8) is not a satisfactory model for the number of cars (entering the car park). 2.739 + 0.241 + 2.152 + '2.049'+ 0.425 + 3.753 + 0.037  = 11.396  M1 Sum of chi-square contributions using their q. May be implied by 11.4. '11.4'  12.59, accept H0 / do not reject H0 / not significant. M1 Compare their 11.4 with 12.59 and appropriate result (may be in terms of H1). Test result may be implied by an attempt at an appropriate conclusion in context. Insufficient evidence to suggest that Po(2.8) is not a good fit to the A1 Correct conclusion from correct working ignoring hypotheses. data. In context. OR Level of uncertainty in language is used (for example, not ‘prove’). Insufficient evidence to suggest that the manager’s claim is false. Allow ‘Not enough evidence that / to show / conclude that…’ Do not accept statements such as ‘there is sufficient evidence to suggest…’ or ‘no evidence…..’ 4

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3 A random sample of 10 newborn baby boys is taken and their masses in kg are recorded. From this sample, the population standard deviation of all newborn baby boys is estimated as 0.6 kg. A random sample of 5 newborn baby girls is taken and their masses in kg are recorded as follows. 3.9 3.1 2.9 3.1 3.6 It is assumed that the masses of newborn baby boys and girls have the same population standard deviation, v kg. By pooling the two samples, calculate an estimate of v. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3  ( x − x )2 = 9  0.36 = 3.24 B1 43 86 or 0.688 . ) ) '16.6'2 M1 Can be implied by 0.172 ( = 250 ( = 125 '55.8'− 5 3.24 + 0.688 M1 Correct use of formula for pooled variance using their values. ( = 0.302 ) 10 + 5 − 2 s = 0.550 A1 CAO 4

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4 A researcher believes that the median m of a population has changed from its known previous value m0. The researcher collects a random sample of size 28. She ranks the data and calculates a test statistic T using the Wilcoxon signed-rank test. The conclusion of the test carried out at a 1% significance level is that there is not sufficient evidence to support her belief. Using a normal approximation, find the least possible value of T. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 1 4  28  ( 28 + 1)  = 203 B1 Both expressions correct. Implied by 203 and 1928.5 seen. 1  28  ( 28 + 1)  ( 56 + 1)  = 1928.5  24 T + 0.5 − '203' M1 Correct form or expression for CI with any z-value.  z =  Condone T or T − 0.5 instead of T + 0.5 for M1. '1928.5' T + 0.5 − '203' M1 Compares their z with  2.576 . −2.576 '1928.5' Condone T or T − 0.5 instead of T + 0.5 for M1. T  89.4 A1 Allow T = 89.4 . T = 90 A1 All correct with no errors seen, including correct continuity correction. 5

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5 A continuous random variable X has probability density function f given by Z ] 1 x 0 G x 1 4 , ]] 16 f ( x) = [ 1 4 G x G ,9 ] k x ] 0 otherwise, \ where k is a constant. (a) Show that k = 3 . 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(b) Find the median value of X. 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The random variable Y is defined by Y = X . (c) Find the probability density function of Y. 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Mark scheme: 45(a) M1 Equation in terms of k formed following attempt to integrate with 9 2 1 1  2   x 3 + 2 x  = 1 correct limits.     4 16 3 k   0 1 2 ( 8 − 0 ) + ( 3 − 2 ) = 1 24 k OR 1 2 + = 1 3 k 2 A1 AG, no errors seen. = 2, k = 3 k 3 2 15(b) 1 4 1 m − 2 1 *M1 Use of 4  m  9 to form equation equal to 0.5, OE. 16 0 x d x + 3 4 x dx = 2 Equation may be in terms of k. OR 1 9 − 12 3 m x d x = 0.5 1 m − 2 = 12 DM1 Integrate and form an equation in m . 3 + 32 ( ) OR 2 3 − m = 12 3 ( ) m = 1681 [ = 5.0625] A1 Accept AWRT 5.06. 3 M1 Attempt to integrate both parts, limits not required.5(c)  2 0 ≤ x  4  241 x 3 F ( x ) =  1 May be seen in part 5(a). ≤ x ≤ 9  23 x 2 − 1 4 B1 For constant –1 obtained in expression for 4 ≤ x ≤ 9 . May be seen in part 5(a). ≤ y  2 M1 For changing to y .  241 y 3 0 G ( y ) =  Limits not required. ≤ y ≤ 3  23 y − 1 2  1 ≤ y  2 M1 For differentiation, limits not required. 8 y 2 0  2 ≤ y ≤ 3 g ( y ) =  3 2 A1 Fully correct with correct domain covering all reals.  0 otherwise  Alternative method for question 5(c) Using chain rule (or inverse function): M1 M1 for changing to y . 2 d 2 2 y . y = 2 yf y G ( y ) = F ( ) so g ( y ) = F ( ) ( ) dy A1 M2  y 2 twice, limits not required. 2 y 1 y 2 0 ≤ y  2 Use of g ( y ) = 2 yf ( ) ) ( 16  M1 for one expression. g ( y ) =    1  2 y   2 ≤ y ≤ 3 2    3 y     1 ≤ y  2 A1 Fully correct with correct domain covering all reals. 8 y 2 0  2 ≤ y ≤ 3 g ( y ) =  3 2  0 otherwise  5

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6 Nine athletes in a club have a new coach. The coach adopts a new training programme which he believes will reduce the race times of these athletes. Each athlete completes a 1500 m time trial before and after completing the new training programme. Their times, in seconds (s), are recorded. Athlete A B C D E F G H I Time before training (s) 250 251 252 267 276 291 310 320 335 Time after training (s) 245 251 253 261 275 293 302 313 320 (a) Carry out a paired t-test at the 5% significance level to test the coach’s belief. 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Further research suggests that the effects of the training programme tend to reduce the times of the slower athletes by more than those of the faster athletes. (b) Suggest a reason why the paired t-test used in part (a) may not have been an appropriate test in this case. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Suggest a suitable alternative test that could have been used instead of a paired t-test. 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Mark scheme: 6(a) H 0: B = A H1: B  A B1 If in words, must contain ‘population means’. Allow H 0: d = 0, H1: d 0 if defined or consistent with working. Differences: 5, 0, −1, 6, 1, −2, 8, 7, 15 M1 Attempt at signed differences. = 29.5 M1 Correct form for ds2 , note that ds = 5.43 . ds2 = 18 ( '405'− '39'9 2 )  '39' M1 Correct form. 9 = 2.393 '29.5' 9 A1 AWRT 2.39. '2.393'  1.860 , reject H0 / significant. M1 Compare their 2.393 with 1.860 and appropriate result (may be in terms of H1). Allow M1 for comparison from 2-sample test using 1.746. Test result may be implied by an attempt at an appropriate conclusion in context ignoring hypotheses. Sufficient evidence to suggest new training programme results in A1 Correct conclusion from correct working ignoring hypotheses. reduced times. In context. OR Level of uncertainty in language is used (for example, not ‘prove’). Sufficient evidence to support the coach’s belief. Do not accept statements such as “there is insufficient evidence to suggest…”. 7 6(b) The population of differences may not be normally distributed. B1 Must mention population/distribution and differences. The population of differences may not be symmetrical (and hence Accept “the sample may not be random” or “times may not be not normal). independent”. 1 6(c) A Wilcoxon matched-pairs signed-rank test. B1 Must refer to pairs / paired. OR A paired-sample sign test. Accept “paired Wilcoxon test” or “paired sign test”. 1

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7 A discrete random variable X takes values r = 0, 1, 2 with probabilities P( X = r) as given in the following table. r 0 1 2 P( X = r) a 2a b (a) Write down the probability generating function of X, and use it to find an expression for E(X ) in terms of a and b. 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(b) Show that Var ( X ) = 2b + 2 ( a + b)( 1 - 2a - 2 b) . 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The random variable Y is defined by Y = X + X + X + g + X where X , X , X , f, X are ten 1 2 3 10 1 2 3 10 independent observations of X. (c) Using the probability generating function of Y, and your answer to part (a), show that E ( Y ) = 10E( X ) . 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(d) For the case b = 0 , define fully the distribution 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 B1 G X ( t ) = a + 2at + bt ' B1 G X ( t ) = 2 a + 2bt ' E ( X ) = G X (1) = 2 a + 2b 2 7(b) '' B1 G X ( t ) = 2b 2 M1 Use correct formula with their expressions. Var ( X ) = 2b + ( 2a + 2b ) − ( 2a + 2b ) 2 A1 AG, shown convincingly. 2b + ( 2 a + 2b ) − ( 2 a + 2b ) = 2b + ( 2 a + 2b ) (1 − ( 2 a + 2b ) ) = 2b + 2( a + b )(1 − 2 a − 2b ) Alternative method for question 7(b) 2 2 2 2 B1 E X = 0  a + 1  2a + 2  b = 2a + 4b ( ) 2 M1 Use correct formula with their expressions. Var ( X ) = ( 2a + 4b ) − ( 2a + 2b ) ( 2 a + 4b ) − ( 2 a + 2b ) 2 = 2b + 2 a + 2b − (2 a + 2b ) 2 A1 AG, shown convincingly. = 2b + ( 2 a + 2b ) (1 − ( 2 a + 2b ) ) = 2b + 2( a + b )(1 − 2 a − 2b) 3 7(c) 10 G X ( t ) . a + 2 at + bt 2 G Y ( t ) = ( ) B1 FT FT their 9 GY ( t ) and evaluate at t = 1 . a + 2 at + bt 2 G Y ' ( t ) = 20 ( a + bt )( ) M1 Attempt to differentiate their E ( Y ) = 20 ( a + b )( 3a + b )9 3a + b =1 , hence E ( Y ) = 20 ( a + b ) = 10E ( X ) . A1 3a + b = 1 since total probability is 1, OE. 3 7(d) Binomial *B1 Uses PGF to identify binomial.  2  DB1 Identifies binomial with correct parameters. B  10,   3  2

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