Cambridge A Level Mathematics - Further 9231 — 2021 May/June Paper 4 · Variant 3

1 Farmer A grows apples of a certain variety. Each tree produces 14.8 kg of apples, on average, per year. Farmer B grows apples of the same variety and claims that his apple trees produce a higher mass of apples per year than Farmer A’s trees. The masses of apples from Farmer B’s trees may be assumed to be normally distributed. A random sample of 10 trees from Farmer B is chosen. The masses, x kg, of apples produced in a year are summarised as follows. / x = 152.0 / x 2 = 2313.0 Test, at the 5% significance level, whether Farmer B’s claim is justified. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 2 2 1 152 2313 0.28889 9 10 s   = − =     M1 13 45 , accept unsimplified. H0 : μ = 14.8 H1 : μ > 14.8 B1 If μ not used, ‘population mean’ required. Must see 14.8, must be = and >. [ ] 152 14.8 10 / 10 t s − = M1 Using unbiased estimate. p-value is 0.0215. [ ] t = 2.35 A1 Rounds to 2.35. Tabular value = 1.833 ‘2.35’ > 1.833 Reject 0 H M1 Comparison with 1.833 and correct FT conclusion. Sufficient evidence to accept Farmer B’s claim oe A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Allow ±1 difference in third significant figure of their t for this mark. 6

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2 A company is developing a new flavour of chocolate by varying the quantities of the ingredients. A random selection of 9 flavours of chocolate are judged by two tasters who each give marks out of 100 to each flavour of chocolate. Chocolate A B C D E F G H I Taster 1 72 86 75 92 98 79 87 60 62 Taster 2 84 72 74 95 85 87 82 75 68 Carry out a Wilcoxon matched-pairs signed-rank test at the 10% significance level to investigate whether, on average, there is a difference between marks awarded by the two tasters. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 H0: difference in population medians = 0 (ma = mb) B1 Correct hypotheses, allow md . Do not allow μ or mean. H1: difference in population medians ≠ 0 (ma ≠ mb) B1 ‘population’ included. Diff: 12 −14 −1 3 −13 8 −5 15 6 M1 Allow one error. Rank: 6 −8 −1 2 −7 5 −3 9 4 M1 Allow one error. Smaller sum 19 A1 (Other sum is 26.) Critical tabular value = 8 ‘19’ > 8 so accept H0 M1 Comparison with 8 and correct FT conclusion. There is insufficient evidence that marks differ A1 Correct conclusion, in context, following correct work, except possibly the hypotheses. Level of uncertainty in language is used. 7

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3 The heights, x m, of a random sample of 50 adult males from country A were recorded. The heights, y m, of a random sample of 40 adult males from country B were also recorded. The results are summarised as follows. / x = 89.0 / x 2 = 159.4 / y = 67.2 / y 2 = 113. 1 Find a 95% confidence interval for the difference between the mean heights of adult males from country A and adult males from country B. 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Mark scheme: 3 89 67.2 1.78 1.68 50 40 x y = = = = B1 Implied by 0.1 in the CI formula. 2 2 1 89 159.4 49 50 x σ   = −     = 0.02 1 50   =     M1 One variance correct unsimplified. 2 2 1 67.2 113.1 39 40 y σ   = −     = 0.0052308 17 3250   =     A1 Both correctly calculated. 2 0.02 0.0052308 50 40 σ = + = 69 130000       M1 Unsimplified expression must be seen, 40 and 50 in correct places. [ 2 σ =] 0.0005308 A1 May be implied by correct final answer. CI : 1.78 – 1.68 ± z × 0.0005308 M1 Correct form for CI with a z-value. with z = 1.96 A1 0.10 ± 0.0452 or [0.0548, 0.145] A1 Allow either form. 8 SC Assuming equal variances award B1M0A0, M1A1 (for use of pooled variance formula, gives 0.0135), M1A1A0.

More questions on Inference using normal and t-distributions

4 X is a discrete random variable which takes the values 0, 2, 4, … . The probability generating function of X is given by 1 GX ( )t = 2 . 3 - 2t (a) Find E ( X ) and Var ( X ). 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(b) Find P ( X = 4). 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Mark scheme: 4(a) G’(t) = ( ) 2 2 4 3 2 t t − M1 Differentiate to obtain ( ) 2 2 3 2 kt t − OE. so E(X) = G’(1) = 4 A1 CAO WWW G’’(t) = ( ) 2 3 2 12 24 3 2 t t + − or ( ) ( ) 2 3 2 2 2 4 3 2 32 3 2 t t t − − − + − M1 OE. Differentiate, allow only numerical or sign slips. Var(X) = G ' '(1) + G '(1) – ( ) ( ) 2 G' 1 = 36 + 4 – 16 M1 Substitute their values into correct formula, dependent on attempt at G ' '(t). [Var(X)] = 24 A1 CAO WWW 5 4(b) ( ) ( ) 1 2 2 1 3 2 3 2 X G t t t − = = − − = 2 4 1 2 4 1 . 3 3 9 t t   + + +…     M1 Expand given expression or give expression for the term in t4 . P(X = 4) = their coefficient of t4 M1 Found from legitimate method. [P(X = 4) =] 4 27 A1 WWW 3

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5 Chai packs china mugs into cardboard boxes. Chai’s manager suspects that breakages occur at random times and that the number of breakages may follow a Poisson distribution. He takes a small sample of observations and finds that the number of breakages in a one-hour period has a mean of 2.4 and a standard deviation of 1.5. (a) Explain how this information tends to support the manager’s suspicion. 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The manager now takes a larger sample and claims that the numbers of breakages in a one-hour period follow a Poisson distribution. The numbers of breakages in a random sample of 180 one-hour periods are summarised in the following table. Number of breakages 0 1 2 3 4 5 6 7 or more Frequency 21 33 46 31 23 16 10 0 The mean number of breakages calculated from this sample is 2.5. (b) Use the data from this larger sample to carry out a goodness of fit test, at the 10% significance level, to test the claim. 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Mark scheme: 5(a) (Mean of distribution = 2.4) Variance = 2.25 Mean approx. equal to variance (so Poisson might be suitable) A1 Must have ‘approximately’ oe, not equality. Must have 2.25. 2 5(b) Po(2.5) leads to frequencies: M1 At least 3 correct to 3sf. 14.77[5], 36.938, 46.173, 38.477, 24.048, 12.024, 5.010, 2.554 A1 All correct to 4sf. Combine last two values: 7.564 M1 FT their table values if final figure is less than 5. ( ) 2 2.6227 0.4198 0.00065 1.4529 0.04570 O E E −  = + + + + 1.3148 0.7845 + + M1 Apply correct formula to their frequencies. 6.64 A1 Correct to 3sf. 5(b) H0: distribution fits the data B1 Must mention data and distribution . Tabular value, 5 degrees of freedom, is 9.236 ‘6.64’ < 9.236, so accept H0 M1 Compare with correct tabular value and conclusion. Insufficient evidence to show that data does not follow a Poisson distribution A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Allow ±1 difference in third significant figure. 8

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6 The continuous random variable X has probability density function f given by 1 8 0 G x 1 1, f ()x = 1 (8 - x) 1 G x G 8, 28 *0 otherwise. (a) Find the cumulative distribution function of X. 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(b) Find the value of the constant a such that P ( X G a) = 57 . 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The random variable Y is given by Y = 3 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) 2 1 0 1, 8 F( ) 1 1 1 8 1 8. 28 2 7 x x x x x x  <  =     − −         A1 no constant term. Both parts correct, with domains. Any equivalent form, for example: 2 2 1 1 7 56 7 x x − − . F(x)= 0 for x < 0 and F(x) = 1 for x > 8. A1 Alternative method for question 6(a) ( ) 2 1 0 1, 8 F( ) 1 1 8 1 8. 56 x x x x x  <  =  − −     M1 A1 Integrate both parts (all powers of x increase by 1), allow M1 if no constant term. Both parts correct, with domains. Any equivalent form, for example: ( ) 2 1 16 8 56 x x − − . F(x)= 0 for x < 0 and F(x) = 1 for x > 8. A1 3 6(b) F(a) = 5 7 leading to 2 1 1 1 5 8 28 2 7 7 a a   − − =     *M1 Correct method and attempt to simplify, ft their part (a). 2 16 48 0 a a − + = [ ] 4 12 a a = = DM1 Obtain quadratic and solve. a = 4 A1 Correct single answer, WWW. 3 Question Answer Marks Guidance 6(c) ( ) 3 6 3 1 8 G 1 0 1, 1 16 1 2 ( 56 7 ) . y y y y y y   =    <  − −    M1 A1 Substitute x = y3 into both parts of their F(x) from part (a). Correct functions in terms of y, any equivalent form. 0  y < 1 and 1  y  2seen in CDF or PDF A1 Condone  instead of <. ( ) 2 2 5 3 0 1, 8 3 g( ) 8 1 2, 28 0 otherwise. y y y y y y  <   = − <       M1 A1 Differentiate their G(y), all powers of y decrease by 1. Must include ‘otherwise’. 5 Question Answer Marks Guidance 6(c) Alternative method for question 6(c) 1 2 2 3 3 1 , d d , d 3 d 3 y x y x x x y y − = = = M1 ( ) 2 1 3 0 1, f d d d 8 8 x x x x y y < = =  M1 ( ) 2 3 g 8 y y = A1 ( ) ( ) ( ) 3 2 1 1 1 8, f d 8 d 8 3 d 28 28 x x x x x y y y = − = −  ( ) ( ) 2 5 3 8 28 g y y y = − A1 0  y < 1 and 1  y  2 seen in CDF or PDF A1 5

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