Cambridge A Level Mathematics - Further 9231 — 2025 May/June Paper 4 · Variant 4

1 A random sample of 12 observations of a normal random variable is taken. The results give unbiased estimates for the population mean and variance as 10.24 and 0.52 respectively. Test, at the 10% significance level, the null hypothesis that the population mean is 10.6 against the alternative hypothesis that the population mean is less than 10.6. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 10.24 − 10.6 M1 0.52  t =   Must be in correct form, allow in denominator. 0.52 12 12 (  )1.729 A1 CWO, AWRT (  )1.73 . '− 1.729 ' −1.363 (or 1.729  1.363) , reject H 0 (significant). M1 Compare their t with 1.363 (consistent signs) and appropriate conclusion (may be in terms of H1). Reject H0 can be implied by an attempt at an appropriate conclusion in context. Allow M1 for comparison of their t with 1.356 , 1.796, or 1.782 (consistent signs) and appropriate conclusion. Sufficient evidence to support/suggest that the [population] mean A1 Correct conclusion in context from correct working ignoring is less than 10.6. hypotheses. Level of uncertainty in language used (for example, not ‘prove’). Allow  or mean for population mean. 4

More questions on Inference using normal and t-distributions

2 The level of sound produced by a particular type of machine was measured for a random sample of 11 such machines. The results, in suitable units, are shown below. Machine A B C D E F G H I J K Sound level 7.66 8.48 8.21 7.98 8.01 7.77 8.25 8.11 8.03 8.16 7.92 (a) Use a Wilcoxon signed-rank test to test whether the average sound level produced by this type of machine is more than 8.00. Use a 5% significance level. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Give a reason why a Wilcoxon signed-rank test may be more appropriate than a t-test in this case. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) H0 : population median = 8.00 B1 Allow m , but not . If m defined, must see population. H1 : population median  8.00 –0.34, 0.48, 0.21, –0.02, 0.01, –0.23, 0.25, 0.11, 0.03, 0.16, –0.08 M1 Attempt at (signed) differences. –10, 11, 7, –2, 1, –8, 9, 5, 3, 6, –4 M1 Attempt at (signed) ranks, ranks may be reversed. ( P = 42, Q = 24) T = 24 A1 '24' > 13, accept H 0 (not significant). M1 Compare their 24 (from rank sum only) with 13 and appropriate conclusion (may be in terms of H1). Accept H0 can be implied by an attempt at an appropriate conclusion in context. Allow M1 for comparison of their 24 (from rank sum only) with 10 and appropriate conclusion. Insufficient evidence to support/suggest that the average sound A1 Correct conclusion in context from correct working ignoring level is more than 8.00. hypotheses. Level of uncertainty in language used (for example, not ‘prove’). Must refer to average/median sound (level). 6 2(b) It is not known if the population is normal. B1 Must refer to population and/or distribution. It is not known if the (underlying) distribution is normal. B0 for data is normal. B0 for (population) differences are normal. 1

More questions on Non-parametric tests

3 A shop selling electrical goods has a team of three salespeople: Avril, Ben and Charlie. The manager wishes to investigate whether the salespeople are equally successful at selling particular types of items. The following table gives a record of a random sample of 250 sales of laptops, cameras and televisions, with the number sold by each of the three salespeople. Type of item Laptop Camera Television Total Avril 31 40 24 95 Ben 23 45 29 97 Charlie 21 25 12 58 Total 75 110 65 250 Test, at the 10% significance level, whether there is independence between the type of item sold and the salesperson. 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Mark scheme: 3 H 0 : type of item sold is independent of salesperson. B1 Allow dependent, association or relationship for not independent. For example: H0: No association/relationship between… H1 : type of item sold is not independent of salesperson. H1: Association/relationship between… Laptop Camera Television M1 Calculate expected values (must be seen). M1 for at least 4 expected values correct to 3sf or better. Avril 31 (28.5) 40 (41.80) 24 (24.70) Ben 23 (29.1) 45 (42.68) 29 (25.22) A1 All expected values correct to 3sf or better. Charlie 21 (17.4) 25 (25.52) 12 (15.08) ( 31 − 28.5 ) 2 ( 40 − 41.8 ) 2 (15.08 − 12 ) 2 M1 At least 2 terms (not including 250) of correct form seen. + ++ May see individual terms not summed. 28.5 41.8 15.08 May be implied by AWRT 3.67 or 3.68. OR 312 40 2 12 2 + ++ − 250 28.5 41.8 15.08 Test statistic = 3.67 (or 3.68) A1 AWRT 3.67 or 3.68. '3.67' < 7.779, accept H 0 (not significant). M1 Compare their 3.67 (or 3.68) with 7.779 and appropriate conclusion (may be in terms of H1). Accept H0 can be implied by an attempt at an appropriate conclusion in context. Allow M1 for comparison of their 3.67 (or 3.68) with 9.488 and appropriate conclusion. Insufficient evidence to suggest that type of item sold and A1 Correct conclusion in context from correct working ignoring salesperson are not independent. hypotheses. Condone ‘no sufficient evidence’. OR Level of uncertainty in language used (for example, not ‘prove’). Insufficient evidence to suggest that there is dependence between Must refer to both item and salesperson. type of item sold and salesperson. Do not accept statements such as ‘there is sufficient evidence to suggest…’. 7

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4 The continuous random variable X has probability density function f given by kx 0 G x 1 1, f ( x) = *kx 2 1 G x G 2 , 0 otherwise. (a) Show that k = 6 . 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(b) Find the cumulative distribution function of X. 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(c) Find the median value of X. 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Mark scheme:  kx dx = 1 kx  +  kx  = 1  2  3  0 14(a)  10 2 2  1 2 1 1 3 2  M1 Linear equation in terms of k formed following attempt to kx d x +  1   integrate (all powers correct) and use of correct limits. 1 1 for first region. 2 1 k ) 2 k + 13 k ( 8 − 1) = 1 May consider area of triangle ( 17 6 k = 1, k = 176 A1 AG, shown convincingly. No errors seen. 2 4(b)  0 x  0 M1 Attempt to integrate both parts, all powers correct, condone  3 2 missing c . x 0 ≤ x  1  Sight of expressions for 0 ≤ x  1 and 1 ≤ x ≤ 2 sufficient. F ( x ) =  17 2 3 1 ≤ x ≤ 2  17 x + 17 1  1 x  2 A1 Both parts correct with c = 171 . Sight of expressions for 0 ≤ x  1 and 1 ≤ x ≤ 2 sufficient. A1 All correct, fully defined including domain. Domain must cover all reals, may overlap at boundaries. 3 1 ≤ m ≤ 2 and their F ( x ) (or areas) to form equation with 0.5.4(c) F ( m ) = 2 so 172 m 3 + 171 = 12 M1 Use 1 3 15 3 15 A1 FT FT their non-zero c , to obtain m with 1 ≤ m ≤ 2 . m = 4 , m = 4  = 1.55 2 2 M1 Correct method and attempt to integrate both parts. All powers4(d)   1    1 6  1   2 6 2  1  6 1 6 1 2 x correct, limits not required.  E   =   17 x   dx +  17 x   dx = 17  0 + 17  2 x 1   X   0  x  1  x  6 15 1  = 0.882  A1 CWO, AWRT 0.882. (1 + 2 − 17 2 ) = 17 2

More questions on Continuous random variables

5 Eric has three identical coins, each of which is biased so that the probability of obtaining a head when it is thrown is 1. The random variable X is the number of heads obtained when Eric throws the three coins 3 at the same time. (a) Find the probability generating function GX ( )t of X. 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Eric also has two fair 6-sided dice with faces numbered 1 to 6. The random variable Y is the number of sixes obtained when Eric throws the two dice at the same time. It is given that the probability generating function of Y is 25 + 10 t + 1 t 2 . 36 36 36 Eric throws the three coins and the two dice. The random variable Z is the sum of the number of heads obtained and the number of sixes obtained. (b) Find the probability generating function GZ ( )t of Z, expressing your answer as a polynomial in t. 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(c) Use GZ ( )t to find E ( Z ) and Var ( Z) . 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Mark scheme: 5(a) Probabilities: 278 , 1227 , 276 , 271 for 0, 1, 2, 3 heads B1 27 8 + 1227 t + 276 t 2 + 271 t 3 B1 FT FT their probabilities to form a 4-term cubic. Must have  p = 1. 2 ( 3 3 + 13 t ) scores B2. 2 5(b) M1 Attempt to multiply their G X (t ) and their GY ( t ) . ( 27 8 + 1227 t + 276 t 2 + 271 t 3 )( 36 25 + 1036 t + 361 t 2 ) 1 2 3 4 5 M1 Expand to form polynomial of degree 5 (need not be simplified). 200 + 380t + 278t + 97 t + 16t + t 972 ( ) A1 CWO 3 15(c) 380 + 556t + 291t 2 + 64t 3 + 5t 4 M1 Attempt at differentiating their GZ ( t ) . G 'Z ( t ) = 972 ( ) 1296 t ) . E ( Z ) = G 'Z (1) = 972 = 34 A1 FT FT their GZ ( 1 556 + 582t + 192t 2 + 20t 3 M1 Attempt at differentiating their G'Z ( t ) . G ''Z ( t ) = 972 ( ) 1350 4 4 Use correct formula using their G 'Z (1) and their G ''Z (1) . Var ( Z ) = 972 + 34 − ( 3 = 1825 + 34 − ( 3 ) 2  ) 2  M1 17 18  = 0.944  A1 CWO 5

More questions on Probability generating functions

6 Lina and Mona are two statisticians who also write songs. The ‘time’ of a song is the number of minutes for which it lasts. For a random sample of 10 of her songs, Lina calculates a 95% confidence interval for the population mean time, n minutes. This confidence interval is 2.95 G n G 3.13 . The times, x minutes, of Lina’s songs are normally distributed. (a) Find the values of / x and / x2 for the 10 songs in Lina’s sample. 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Mona’s songs have times, y minutes, that are normally distributed. The times for a random sample of 8 of Mona’s songs are summarised as follows. / y = 24.8 / y 2 = 76.98 Mona claims that the population mean time of her songs is greater than the population mean time of Lina’s songs. (b) Assuming that the two distributions have the same population variance, test at the 5% significance level whether there is evidence to support Mona’s claim. 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Mark scheme: 6(a)  x = 30.4 B1 2.262 B1 2.26 or 2.262 seen. 10 ( 3.13 − 2.95 ) M1 Subtracting, or substituting, to find an equation involving xs . Subtract: xs = ( 2  2.262 ) Allow M1 for any of the following t values seen instead of 2.262: 1.812, 1.833, 2.228 or 3.690. xs2 = 0.0158 or xs = 0.126 A1 FT FT their 2.262 following M1. Implied by AWRT 92.6 following M1. 2 30.4 2 A1 FT FT their 2.262.  x = 9  0.0158307 + = 92.6 CWO 10 5 6(b) H0 :Y = X B1 If in words, must contain ‘population means’. d = 0, H1 :d  0 if defined or consistent with H1 :Y  X Allow H0:  working. . 24.82 24.82 M1 May be embedded, for example ys 76.98 − 18 ( 2 = 17 ( ) ) 76.98 − 8 2 9 s x2 + 7 s y2 M1 Correct form using their xs2 and their ys2 , may be embedded. s p =  = 0.015155  10 + 8 − 2 A1 SOI, can be implied by any t that rounds to 1.02 or 1.03. 3.04 − 3.1 M1 Correct form using their ps . t = = −1.028 s p 101 + 81 A1 AWRT 1.02 or 1.03 implies M1 A1. [18 degrees of freedom, so critical value is 1.746.] M1 Compare their t with 1.746 (consistent signs) and appropriate '1.028 '  1.746 , accept H 0 (not significant). conclusion (may be in terms of H1). Accept H0 can be implied by an attempt at an appropriate conclusion in context. Allow M1 for comparison of their t with 1.740 or 2.120 and appropriate conclusion. Insufficient evidence to support Mona’s claim. A1 Correct conclusion in context from correct working ignoring hypotheses. Level of uncertainty in language used (for example, not ‘prove’). 8

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