Cambridge A Level Mathematics - Further 9231 — 2020 Oct/Nov Paper 4 · Variant 3

1 Kayla is investigating the lengths of the leaves of a certain type of tree found in two forests X and Y. She chooses a random sample of 40 leaves of this type from forest X and records their lengths, x cm. She also records the lengths, y cm, for a random sample of 60 leaves of this type from forest Y. Her results are summarised as follows. / x = 242.0 / x 2 = 1587.0 / y = 373.2 / y 2 = 2532.6 Find a 90% confidence interval for the difference between the population mean lengths of leaves in forests X and Y. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 2 2 1 242.0 1587.0 3.15128 39 40 xs   = − =     1229 390   =     2 2 1 373.2 2532.6 3.58129 59 60 ys   = − =     26412 7375   =     Both correct 2 3.15128 3.58129 0.13847 40 60 s = + = M1 A1 Pooled variance is M0A0 or 0.3721 s = 6.05 6.22 zs − ± M1 FT their s, must be a z value = 0.17 1.645 0.13847 ± A1 With 1.645 = [− 0.442, 0.782] or [− 0.782, 0.442] A1 7

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2 Metal rods produced by a certain factory are claimed to have a median breaking strength of 200 tonnes. For a random sample of 9 rods, the breaking strengths, measured in tonnes, were as follows. 210 186 188 208 184 191 215 198 196 A scientist believes that the median breaking strength of metal rods produced by this factory is less than 200 tonnes. (a) Use a Wilcoxon signed-rank test, at the 5% significance level, to test whether there is evidence to support the scientist’s belief. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Give a reason why a Wilcoxon signed-rank test is preferable to a sign test, when both are valid. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) H0: m = 200 H1: m < 200 B1 Allow ‘median’ in words, allow m not defined 10 14 12 8 16 9 1 5 2 4 − − − − − − M1 Signed differences (at most 3 errors) Ranks: 5 7 6 3 9 4 8 1 2 − − − − − − A1 Award for correct rank order, ignore signs Sum ranks T = 16 A1 CWO Critical value 8 and compare ‘16’ > 8 M1 Compare their T with 8 Accept H0 Insufficient evidence to support scientist’s belief A1 In context, all correct, except possibly hypotheses. Level of uncertainty in language used. No contradictions 6 2(b) Magnitude of differences from median are taken into account B1 Must mention magnitude and differences 1

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3 Apples are sold in bags of 5. Based on her previous experience, Freya claims that the probability of any apple weighing more than 100 grams is 0.35, independently of other apples in the bag. The apples in a random sample of 150 bags are checked and the number, x, in each bag weighing more than 100 grams is recorded. The results are shown in the following table. x 0 1 2 3 4 5 Frequency 12 39 46 37 12 4 Carry out a goodness of fit test at the 5% significance level and hence comment on Freya’s claim. 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Mark scheme: 3 x 0 1 2 3 4 5 Observed freq 12 39 46 37 12 4 Bin prob 0.11603 0.31236 0.33642 0.18115 0.04877 0.00525 Expected freq 17.404 46.858 50.462 27.172 7.316 0.7878 M1 A1 Attempt at E values (at least 4 correct) All correct, to 2 dp or better Add last two columns: 16, 8.104 M1 ( ) ( ) ( ) 2 2 2 17.404 12 46.858 39 50.462 46 17.404 46.858 50.462 − − − + + + ( ) ( ) 2 2 27.172 37 8.104 16 27.172 8.104 − − + = 14.64 or 14.65 M1 A1 Accept 14.6 – 14.7 ’14.64’ > 9.49 M1 Compare their value with 9.49 Freya’s claim is not supported or Data does not fit the distribution A1 FT Correct conclusion in context, FT only their 14.64 SC for Poisson M0M1M1M0 max 2 7

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4 Members of the Sprints athletics club have been taking part in an intense training scheme, aimed at reducing their times taken to run 400 m. For a random sample of 9 athletes from the club, the times taken, in seconds, before and after the training scheme are given in the following table. Athlete A B C D E F G H I Time before 48.8 48.2 50.3 49.6 49.4 48.9 47.6 50.3 48.4 Time after 47.9 47.8 49.6 49.1 49.6 48.9 47.7 49.1 48.1 The organiser of the training scheme claims that on average an athlete’s time will be reduced by at least 0.3 seconds. Test at the 10% significance level whether the organiser’s claim is justified, stating any assumption that you make. 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Mark scheme: 4 Assume (population) differences are normally distributed B1 H0: 0.3 X Y μ μ − = H1: 0.3 X Y μ μ − > B1 Diff: 0.9 0.4 0.7 0.5 0.2 0 0.1 1 .2 0.3 − − M1 Signed differences 2 3.7, 3.29 d d  =  = 2 2 1 3.7 0.411, 3.29 0.2211 8 9 d s   = = − =     M1 0.4702 s = 0.411 0.3 0.2211 9 t − = 0.708 = or 0.709 M1 A1 ‘0.708’ < 1.397 M1 Compare their 0.708 with 1.397 Accept H0 Insufficient evidence to support claim A1 FT In context, except possibly hypotheses Level of uncertainty in language used. No contradictions 8

More questions on Inference using normal and t-distributions

5 Keira has two unbiased coins. She tosses both coins. The number of heads obtained by Keira is denoted by X. (a) Find the probability generating function GX ( )t of X. 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Hassan has three coins, two of which are biased so that the probability of obtaining a head when the coin is tossed is 1.3 The corresponding probability for the third coin is 1.4 The number of heads obtained by Hassan when he tosses these three coins is denoted by Y. (b) Find the probability generating function GY ( )t of Y. 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The random variable Z is the total number of heads obtained by Keira and Hassan. (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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(e) Use the probability generating function of Z to find the most probable value of Z. 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Mark scheme: 5(a) GX (t) = 2 1 1 1 4 2 4 t t + + 1 5(b) P(0H)= 12 36 P(1H) = 16 36 P(2H) = 7 36 P(3H) = 1 36 M1 A1 Attempt at probs, at least 2 correct All correct GY (t) = 2 3 12 16 7 1 36 36 36 36 t t t + + + B1 FT FT their probabilities, must be cubic with 4 non-zero terms 3 5(c) GZ (t) = ( 2 2 3 1 1 1 12 16 7 1 ) 4 2 4 36 36 36 36 t t t t t   + + + + +     M1 Attempt to multiply their two PGF = ( ) 2 3 4 5 1 12 40 51 31 9 144 t t t t t + + + + + M1 A1 Obtain quintic expression and collect terms 3 5(d) G’Z ( ) t = ( ) 2 3 4 1 40 102 93 36 5 144 t t t t + + + + M1 Differentiate E(Z) = G’Z (1) = 23 12 = (= 1.92) A1 2 5(e) 2 B1 FT FT power of term with largest coefficient in their GZ (t) 1

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6 The continuous random variable X has cumulative distribution function F given by 0 x 1 0, - x ) 0 G x G 6, F ( x) = 601 ( 16 x 2 *1 x 2 6. (a) Find the interquartile range of X. 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(b) Find E ( X 3 ) . 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The random variable Y is such that Y = X . (c) Find the probability density function of Y. 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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) UQ: F(U) = 0.75: 2 16 45 0 u u − + = M1 Obtain quadratic equation for u or l 8 19 u = − (= 3.64) A1 Value of UQ LQ: 2 16 15 0 l l − + = , 1 l = A1 Value of LQ IQR = UQ – LQ = 7 19 2.64 − = A1 FT 4 6(b) ( ) ( ) 1 8 , f 30 0, x x  −  =   0 ⩽ x ⩽ 6 otherwise B1 May be implied, only need to see ( ) ( ) 1 f 8 30 x x = − Find E(X3) by integration: ( ) 6 3 4 0 1 8 d 30 x x x −  = 6 4 5 0 1 2 5 x x   −     M1 A1 Attempt integration (with PDF not CDF) limits not required Correct integrated expression, with limits = 34.56 A1 4 6(c) ( ) ( ) 2 4 1 16 60 G y y y = − M1 CDF for Y ( ) ( ) 3 1 32 4 for 0 6 60 g y y y y = − ≤ ≤ M1A1 Differentiate to find PDF for Y Correct g(y) and correct range seen anywhere 3

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