Cambridge A Level Mathematics - Further 9231 — 2020 May/June Paper 4 · Variant 2

1 Two randomly selected groups of students, with similar ranges of abilities, take the same examination in different rooms. One group of 140 students takes the examination with background music playing. The other group of 210 students takes the examination in silence. Each student is awarded a grade for their performance in the examination and the numbers from each group gaining each grade are shown in the following table. Grade awarded A B C Background music 49 51 40 Silence 93 68 49 Test at the 10% significance level whether grades awarded are independent of whether background music is playing during the examination. [6] ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ...................................................................................................................................................................

Mark scheme: 1 49 E = 56.8 51 E = 47.6 40 E = 35.6 93 E = 85.2 68 E = 71.4 49 E = 53.4 M1A1 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 49 56.8 51 47.6 40 35.6 93 85.2 68 71.4 49 53.4 56.8 47.6 35.6 85.2 71.4 53.4 − − − − − − + + + + + M1 = 3.096 (3.10) A1 Use appropriate tabular value = 4.605 M1 3.096 < 4.605 so the grades awarded are independent of the background A1 6

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2 The times, in milliseconds, taken by a computer to perform a certain task were recorded on 10 randomly chosen occasions. The times were as follows. 6.44 6.16 5.62 5.82 6.51 6.62 6.19 6.42 6.34 6.28 It is claimed that the median time to complete the task is 6.4 milliseconds. (a) Carry out a Wilcoxon signed-rank test at the 5% significance level to test this claim. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State an underlying assumption that is made when using a Wilcoxon signed-rank test. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) H0: population median is 6.40, H1: population median ≠ 6.40 B1 Calculate differences and signed ranks 0.04 –0.24 –0.78 –0.58 0.11 0.22 –0.21 0.02 –0.06 –0.12 2 –8 –10 –9 4 7 –6 1 –3 –5 M1A1 Test statistic = 4 + 7 + 2 +1 =14 A1 Compare with correct critical value 8 M1 Accept H0: insufficient evidence to reject claim (FT their test statistic) A1 FT 6 2(b) Symmetrically distributed about the median B1 1

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3 The continuous random variable X has probability density function f given by Z 3 ] 16 2 - x 0 G x 1 1, ` j ] 3 f ( x) = [ 1 G x G 9, 16 x ]] 0 otherwise. \ (a) Find E(X). 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The random variable Y is such that Y = X . (b) Find the probability density function of Y. 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Mark scheme: 3(a) E(X) = ( ) 1 9 0 1 3 3 16 2 d d 16 − +   x x x x x x 5 3 2 2 2 3 2 3 2 16 5 16 3 x x x     − +             A1 269 80 A1 3 3(b) ( ) 3/2 1/2 3 1 , 0 1 8 3 F 3 1, 1 9 8 8   − <     =   −  x x x x x x    M1A1 ( ) 2 3 3 1 , 0 1 8 3 G 3 1, 1 3 8 8    − <      =   −  y y y y y y    M1A1 (G(y) = 0 for 0 y  and = 1 for 3) y  ( ) ( ) 2 3 2 , 0 1 8 g 3, 1 3 8  − <  =    y y y y y    = 0 otherwise A1 5

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4 A company has two different machines, X and Y, each of which fills empty cups with coffee. The manager is investigating the volumes of coffee, x and y, measured in appropriate units, in the cups filled by machines X and Y respectively. She chooses a random sample of 50 cups filled by machine X and a random sample of 40 cups filled by machine Y. The volumes are summarised as follows. / x = 15.2 / x 2 = 5.1 / y = 13.4 / y 2 = 4.8 The manager claims that there is no difference between the mean volume of coffee in cups filled by machine X and the mean volume of coffee in cups filled by machine Y. Test the manager’s claim at the 10% significance level. 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Mark scheme: 4 H0: μ μ = x y H1: μ μ ≠ x y B1 sx2 = 2 1 15.2 5.1 49 50   −     = 0.0097796: sy2 = 2 1 13.4 4.8 0.007974 39 40   − =     M1A1 2 0.00977959 0.007974 0.0003949 50 40 = + = s M1A1 0.304 0.335 0.0003949 − = z = ( )1.56 − M1A1 Compare with 1.645 M1 Accept H0: insufficient evidence to reject manager’s claim A1 9

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5 A large number of children are competing in a throwing competition. The distances, in metres, thrown by a random sample of 8 children are as follows. 19.8 22.1 24.4 21.5 20.8 26.3 23.7 25.0 (a) Assuming that distances are normally distributed, test, at the 5% significance level, whether the population mean distance thrown is more than 22.0 metres. 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(b) Find a 95% confidence interval for the population mean distance thrown. 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Mark scheme: 5(a) 2 183.6, 4249.08,  =  = x x 22.95 = x B1 2 2 1 183.6 4249.08 5.066 7 8   = − =     s M1 H0: 22.0, μ = H1: 22.0 μ > B1 2 22.95 22.0 1.194 8 − = = t s M1A1 Compare t with correct tabular value 1.895 M1 Accept H0: mean distance thrown is not more than 22.0 m A1 7 5(b) 2 22.95 8 ± s t M1 With t = 2.365 B1 [21.1, 24.8] A1 3

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6 A bag contains 4 red balls and 6 blue balls. Rassa selects two balls at random, without replacement, from the bag. The number of red balls selected by Rassa is denoted by X. (a) Find the probability generating function, G X ( t) , of X. 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Rassa also tosses two coins. One coin is biased so that the probability of a head is 2.3 The other coin is biased so that the probability of a head is p. The probability generating function of Y, the number of heads obtained by Rassa, is G Y ( t) . The coefficient of t in G Y ( t) is 127 . (b) Find G Y ( t) . 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The random variable Z is the sum of the number of red balls selected and the number of heads obtained by Rassa. (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). 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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(BB) = 6 5 1 10 9 3 × = P(RB, BR) = 6 4 8 10 9 15 × = P(RR) = 4 3 2 10 9 15 × = M1 2 1 8 2 3 15 15 + + t t (FT their probabilities) A1 FT 2 6(b) P(1 H) = ( ) 2 1 7 1 3 3 12 − + = p p M1 1 4 = p A1 PGF of Y = 2 1 7 1 4 12 6 + + t t A1 3 6(c) PGF of Z = 2 2 1 8 2 1 7 1 3 15 15 4 12 6    + + + +       t t t t B1M1 = ( ) 2 3 4 1 15 59 72 30 4 180 + + + + t t t t AEF A1 3 6(d) Attempt to differentiate their GZ(t) and evaluate G’Z(1) M1 E(Z) = (59 + 144 + 90 + 16)/180 = 309 1.72 180 = A1 2

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