Cambridge A Level Mathematics - Further 9231 — 2021 May/June Paper 4 · Variant 2

1 A random sample of 7 observations of a variable X are as follows. 8.26 7.78 7.92 8.04 8.27 7.95 8.34 The population mean of X is n. (a) Test, at the 10% significance level, the null hypothesis n = .822 against the alternative hypothesis n 1 8.22. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State an assumption necessary for the test in part (a) to be valid. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) 56.56 8.08 7 x = = B1 Accept unsimplified. 2 2 1 56.56 457.275 6 7 s   = −     = 0.045033 1351 30000       = M1 Accept unsimplified. 2 8.08 8.22 7 t s − = = − 1.745 M1 A1 1.74 – 1.75 Tabular value 1.440 1.745 > 1.440 , so reject 0 H M1 Comparison with 1.440 and correct FT conclusion. There is insufficient evidence to support the hypothesis that the (population) mean is 8.22 OR sufficient evidence to suggest mean is less than 8.22 OR sufficient evidence to suggest 8.22 μ < OE A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6 1(b) Underlying distribution is normal/ population is normal B1 X is normal distributed. 1

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2 A driving school employs four instructors to prepare people for their driving test. The allocation of people to instructors is random. For each of the instructors, the following table gives the number of people who passed and the number who failed their driving test last year. Instructor A Instructor B Instructor C Instructor D Total Pass 72 42 52 68 234 Fail 33 34 41 58 166 Total 105 76 93 126 400 Test at the 10% significance level whether success in the driving test is independent of the instructor. 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Mark scheme: 2 72 (61.425) 42 (44.46) 52 (54.405) 68 (73.71) 33 (43.575) 34 (31.54) 41 (38.595) 58 (52.29) M1 At least 3 correct to 4sf. Calculates E values (in brackets above) A1 All correct to 4sf (if 6.037 not found). Test value: ( ) 2 O E E −  = 1.8206 + 0.1361 + 0.1063 +0.4423 + 2.5664 + 0.1919 + 0.1499 + 0.6235 M1 Correct formula used. = 6.037 A1 Correct to 3 sf. H0: (driving) test success is independent of instructor B1 Tabular value: 3 df 10% = 6.251 6.04 < 6.251 Accept 0 H M1 Compare with correct tabular value and conclusion. Sufficient evidence to suggest that (driving) test success is independent of instructor A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Allow ±1 in the third significant figure of test value. 7

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3 The continuous random variable X has cumulative distribution function F given by 0 x 1 0, G x G 9, F ( x) = 811 x 2 0 * 1 x 2 9. (a) Find X . 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(b) Find Var X . 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(c) The random variable Y is given by Y 3 = X . Find the probability density function of Y. 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Mark scheme: 3(a) 9 1.5 0 2 81 x dx  B1 ( ) 2 f , 81 x x = correct expression as integrand. 2.5 2 2 81 5 x × M1 Integrate, FT only on their PDF expression. 12 2.4 5 = A1 3 3(b) ( ) 9 2 2 0 2 d 81 x x their a −  M1 3 2 2 9 6 2.4 81 3 25 × − = A1 2 3(c) G(y) = 6 1 81 y M1 g(y) = 5 2 27 y M1 Fully correct including ‘for 3 0 9 y   and 0 otherwise’ A1 Condone 2.08 3

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4 A scientist is investigating the lengths of the leaves of birch trees in different regions. He takes a random sample of 50 leaves from birch trees in region A and a random sample of 60 leaves from birch trees in region B. He records their lengths in cm, x and y, respectively. His results are summarised as follows. / x = 282 / x 2 = 1596 / y = 328 / y 2 = 1808 n The population mean lengths of leaves from birch trees in regions A and B are n A cm and B cm respectively. Carry out a test at the 5% significance level to test the null hypothesis n A = n B against the alternative hypothesis n A ! n B . 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Mark scheme: 4 2 2 1 282 1596 49 50 xs   = −     = 0.11265 138 1225       = M1 One correct, unsimplified. 2 2 1 328 1808 59 60 ys   = −     = 0.25311 224 885   =     A1 Both correct to 3sf. 2 0.11265 0.25312 0.006471 50 60 s = + = … M1 A1 May be implied. 282 328 50 60 their z s − = M1 [z] = 2.155 A1 Tolerance 2.15 – 2.16 Comparison with tabular value 1.96 OE 2.155 > 1.96 Reject 0 H M1 Comparison with 1.96 OE and conclusion. Sufficient evidence to reject population means are equal OR sufficient evidence to accept population means are not equal OR sufficient evidence to suggest A B μ μ ≠ OE A1 Correct conclusion following correct work. Level of uncertainty in language is used. 8

More questions on Inference using normal and t-distributions

5 Georgio has designed two new uniforms X and Y for the employees of an airline company. A random sample of 11 employees are each asked to assess each of the two uniforms for practicality and appearance, and to give a total score out of 100. The scores are given in the table. Employee A B C D E F G H I J K Uniform X 82 74 42 59 60 73 94 98 62 36 50 Uniform Y 78 75 63 56 67 82 99 90 72 48 61 (a) Give a reason why a Wilcoxon signed-rank test may be more appropriate than a t-test for investigating whether there is any evidence of a preference for one of the uniforms. 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(b) Carry out a Wilcoxon matched-pairs signed-rank test at the 10% significance level. 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Mark scheme: 5(a) Underlying distribution or population of differences is unknown Underlying distributions or population of scores for both X and Y are unknown B1 Not known to be normal. Condone ‘scores for X and Y cannot be assumed to be normally distributed’. 1 5(b) 0 :difference of population medians 0 H = B1 Correct hypotheses, allow m but not μ or mean. 1 :difference of population median 0 H ≠ B1 ‘population’ included. Diff: 4 −1 −21 3 −7 −9 −5 8 −10 −12 −11 M1 Differences, allow one error. Rank: 3 −1 −11 2 −5 −7 −4 6 −8 −10 −9 M1 Signed ranks, allow one sign error. [ ] 55, 11 Q P =− = A1 P = 11 Critical tabular value = 13 ‘11’ < 13 so reject 0 H M1 Comparison with 13 and correct ft conclusion. Sufficient evidence of a preference for one of the uniforms OE A1 Correct conclusion, in context, following correct work, except possibly second B1. Level of uncertainty in language is used. 7

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6 Tanji has a bag containing 4 red balls and 2 blue balls. He selects 3 balls at random from the bag, without replacement. The number of red balls selected by Tanji is denoted by X. (a) Find the probability generating function GX ( )t of X. 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Tanji also has two coins, each biased so that the probability of obtaining a head when it is thrown is 1.4 He throws the two coins at the same time. The number of heads obtained is denoted by Y. (b) Find the probability generating function GY ( )t of Y. 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The random variable Z is the sum of the number of red balls selected by Tanji and the number of heads obtained. (c) Find the probability generating function of Z, expressing your answer as a polynomial. 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(d) Use the probability generating function of Z to find E ( Z ) and Var ( Z ). 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 6(a) P(3R, 2R, 1R, 0R) = 1 3 1 , , , 0 5 5 5 M1 At least 2 correct probabilities used in a polynomial. ( ) 2 3 1 3 1 G 5 5 5 X t t t t = + + A1 2 6(b) P(2H, 1H, 0H) = 1 6 9 , , 16 16 16 M1 One correct probability and all three adding to 1 used in a polynomial. ( ) 2 9 6 1 G 16 16 16 Y t t t = + + A1 2 6(c) Attempt to multiply results from part (a) and part (b) M1 Obtain polynomial M1 2 3 4 5 9 33 28 9 1 80 80 80 80 80 t t t t t + + + + A1 Accept exact equivalent decimals. 3 6(d) E(Z) = ( ) ( ) ' 1 1 9 66 84 36 5 80 Z G = + + + + M1 Differentiate and put t = 1. = 200 2.5 80 = A1 ( ) ( ) ' ' 1 362 1 66 168 108 20 4.525 80 80 Z G   = + + + = =     M1 Question Answer Marks Guidance 6(d) Var (Z) = ( ) ( ) ( ) ( ) 2 1 1 ' 1 G G G ′ + − ′′ M1 Use result. 2 362 200 200 80 80 80   + −    = 31 0.775 40   =     A1 5

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