Cambridge A Level Mathematics - Further 9231 — 2023 Oct/Nov Paper 4 · Variant 1

1 Maya is an athlete who competes in 1500-metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, x minutes, this summer are summarised as follows. / x = 42.05 / x 2 = 176.83 Maya’s new practice run times are normally distributed. She believes that on average her times have improved as a result of her training. Test, at the 5% significance level, whether Maya’s belief is supported by the data. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 H0: = 4.22 H1:  4.22 B1 42.05 2 1  42.052  13  M1 = = 0.0010833 x =  = 4.205 , s =  176.83 −   10 9  10   12000  4.205 − 4.22 M1 Their 4.205, their 2s . t = s 2 10 t = −1.44 A1 Condone sign. Tabular value = 1.833: ‘1.44’ < 1.833, accept H0 M1 Compare their value with correct tabular value, signs consistent, allow ‘not significant’. There is insufficient evidence to support Maya’s belief. A1 Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 6

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2 A town council has published its plans for redeveloping the town centre and residents are being asked whether they approve or disapprove. A random sample of 250 responses has been selected from residents in the four main streets in the town: North, East, South and West Streets. The results are shown in the table. North Street East Street South Street West Street Approve 33 54 42 26 Disapprove 19 39 28 9 Test, at the 5% significance level, whether the opinions of the residents are independent of the streets on which they live. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 H0: opinion is independent of street H1: opinion is not independent of street B1 M1 At least 2 correct values or expressions. 33 32.24 54 57.66 42 43.4 26 21.7 155 A1 6 correct values or expressions. 19 19.76 39 35.34 28 26.6 9 13.3 95 52 93 70 35 250 Calculate chi-squared values: M1 Correct to 3 decimal places. 0.0179 + 0.2323 + 0.0452 + 0.8521 + 0.0292 + 0.3790 + 0.0737 + 1.3902 At least 2 correct values or expressions. 3.02 A1 SC B1 3.02 following M1M0 SC B2 3.02 following M0M0 Tabular value, 3 degrees of freedom 7.815: ‘3.02’ < 7.815, accept H0 M1 Allow ‘not significant’. Insufficient evidence to suggest that opinion depends on street. A1 CWO. Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 7

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3 Scientists are studying the effects of exercise on LDL blood cholesterol levels. Over a three-month period, a large group of people exercised for 20 minutes each day. For a randomly chosen sample of 10 of these people, the LDL blood cholesterol levels were measured at the beginning and the end of the three-month period. The results, measured in suitable units, are as follows. Person A B C D E F G H I J Beginning 72 84 120 90 102 135 64 75 80 88 Cholesterol level End 64 76 105 92 105 115 67 75 75 84 (a) Test, at the 2.5% significance level, whether there is evidence that the population mean LDL blood cholesterol level has reduced by more than 2 units after the three-month period. [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State any assumption that you have made in part (a). [1] ............................................................................................................................................................

Mark scheme: 3(a) H0: B − E = 2 H1: B − E  2 B1 May use d , but must be consistent with direction of differences found. M1 Differences, at most 2 errors. 8 8 15 –2 –3 20 –3 0 5 4 2 2 1  52 2  2728  M1 Their values but must see 9 and 10 used correctly.  d = 52,  d = 816 , s =  816 −  = = 60.62  9  10   45  52 M1 − 2 10 t = s 2 10 t = 1.29967, 1.30 A1 Tabular value = 2.262. ‘1.30’ < 2.262, accept H0 M1 Allow ‘not significant’. Insufficient evidence to suggest that cholesterol level has reduced by more A1 CWO. Correct conclusion in context, following correct than 2. work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. 7 3(b) Population differences are normally distributed B1 1

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4 f(x) x 0 2 6 As shown in the diagram, the continuous random variable X has probability density function f given by mx 0 G x G 2 , k f ( x) = + c 2 G x G 6, 2 x * 0 otherwise, where m, k and c are constants. = (a) Given that P ( X G 2 ) = 13 , show that m 1 6 and find the values of k and c. 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(b) Find the exact numerical value of the interquartile range of X. 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Mark scheme: 4(a) 2 2 B1 AG  1 P(X ⩽ 2) = , so  mx = 1 Reasoning required. 3  2  0 3  4 1 1 m = , m =    2  3 6 6 6 M1 Attempt to integrate and correct use of correct limits to  k  2  k  1 2 + c so − + cx = 4c + k = or k + 12c = 2 form linear equation in k and c.   dx =    3 x 3 3  x 2    2 2 k 1 k M1 Matching at x = 2 . Also, 2 m = + c or = + c or 3k + 12c = 4 4 3 4 1 A1 Solve, k = 1, c = 12 4 4(b) L B1 1 1 LQ: xdx = , L = 3  6 4 0 U M1 Or find and use CDF. May be in terms of m. 1  k  3 UQ: dx =  + c  +  3 4  x 2  2 k k 5 M1 Attempt at integral and correct use of correct limits. − + Uc + − 2c = Simplify to quadratic equation, may be in terms of k and U 2 12 2 c. U − U − 12 = 0 U = 4 A1 IQR = 4 − 3 A1 FT FT their UQ – their LQ, exact values only. 5

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5 The random variable X has the geometric distribution Geo(p). pt (a) Show that the probability generating function of X is , where q = 1 - p . [3] 1- qt ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ q (b) Use the probability generating function of X to show that Var ( X ) = 2 . 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Kenny throws an ordinary fair 6-sided dice repeatedly. The random variable X is the number of throws that Kenny takes in order to obtain a 6. The random variable Z denotes the sum of two independent values of X. (c) Find the probability generating function of Z. 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Mark scheme: 5(a) r −1 B1 Implied by G(t) =  pq r −1 t r . P(X = r) = p (1 − p ) r −1 = pq r −1 M1 (Infinite summation) with common ratio (qt) and first G(t) =  pt ( qt ) term or common factor pt indicated or GP identified.  1 pt A1 AG G(t) = pt  = Convincingly obtained.  1 − qt  1 − qt 3 5(b) −2 −1 −2 M1 Attempt to differentiate as a product/quotient. G'(t) = pqt (1 − qt ) + p (1 − qt ) [= p (1 − qt ) ] 2 −3 −2 −3 M1 Attempt to differentiate their G'(t). G''(t) = 2 pq t (1 − qt ) + 2 pq (1 − qt ) [= 2qp (1 − qt ) ] −2 1 2𝑞 M1 Substitute t = 1 into their G'(t) and G''(t) and use formula G'(1) = p (1 − q ) = , G''(1) = 𝑃2 for Var(X) (at some point). p 2 2𝑞 1 1 M1 (1−q ) replaced by p in denominator throughout, and Var(X) = + 𝑃2 𝑝−(𝑝) some cancellation seen (at some point). q A1 AG 2 Convincingly obtained. p 5 5(c) 1 B1 Implied by correct final answer. t  t  6 GX(t) = = 5   1 − t  6 − 5t  6 2 B1 1 2  t   t  GZ(t) = (GX(t))2 =   Allow  6  .  6 − 5t   1 − 5 t   6  2

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6 A school is conducting an experiment to see whether the distance that children can throw a ball increases in hot weather. On a cold day, all the children at the school were asked to throw a ball as far as possible. The distances thrown were measured and recorded. The median distance thrown by a random sample of 25 of the children was 22.0 m. The children were asked to throw the ball again on a hot day. The distances thrown by the same 25 children were measured and recorded and these distances, in m, are shown below. 21.2 23.5 22.9 18.6 19.4 22.1 26.5 20.2 25.7 20.6 22.3 17.4 22.2 27.0 23.9 28.2 22.6 27.2 23.0 23.7 19.8 22.7 23.3 21.5 24.3 The teacher claims that on average the distances thrown will be further when it is hot. Carry out a Wilcoxon signed-rank test, at the 5% significance level, to test whether the data supports the teacher’s claim. 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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 H0: population medians are equal B1 Do not accept ‘differences between population medians < H1: population median after > population median before 0’ unless difference defined. Allow H 0 : m = 22, H1 : m  22.0 M1 Signed differences, allow at most 4 errors –0.8 -7 1.5 12 0.9 8 –3.4 –19 –2.6 –18 0.1 1 4.5 21 –1.8 –14 3.7 20 –1.4 –11 M1 Attempt at ranks (ignore signs). 0.3 3 –4.6 –22 0.2 2 5 23 1.9 15 6.2 25 0.6 5 5.2 24 1 9 1.7 13 –2.2 -16 0.7 6 1.3 10 –0.5 –4 2.3 17 (W+ = 214) W− = 111 A1 Cao identified, or used, as test statistic. 1 1 B1 Normal: mean = n ( n + 1) =  25  26  = 162.5 4 4 1 1 B1 Variance = n ( n + 1)( 2n + 1) =  25  26  51 = 1381.25 24 24 111.5 − 162.5 M1 Allow missing or incorrect continuity correction. z- value: Their 111 must come from ranks. 1381.25 −1.37 A1 CAO Tabular value is −1.645 : ‘ −1.37' −1.645 , or 0.915<0.95 oe, accept H0, M1 Consistent signs. Allow ‘not significant’. Insufficient evidence to support the teacher’s claim A1 All correct. Correct conclusion in context, following Insufficient evidence to suggest that the distances thrown are further when it correct work, level of uncertainty in language. is hot A0 if hypotheses wrong way round or missing 10

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