Cambridge A Level Mathematics - Further 9231 — 2023 May/June Paper 4 · Variant 3

1 The continuous random variable X has probability density function f given by 1 - 31 - 23 6 x - x 1 G x G 27, f ( x) = ` j * 0 otherwise . (a) Find the cumulative distribution function of X. 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The random variable Y is defined by Y = X 1 (b) Find the probability density function of Y. 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(c) Find the exact value of the median of Y. 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Mark scheme: 1(a) F( 2 1 3 3 1 3 ) 3 6 2          x x x c 2 1 3 3 1 1 1 4 2 4   x x A1 AEF Correct ranges including 1 27 x  associated with their F( )x F( ) x = 0 for 1  x and F( ) x = 1 for 27  x A1 No gaps. 3 1(b) G   2 1 ( ) 2 1 4    y y y M1 Using 1 3  y x in their F(x). g   1 ( ) 1 2   y y A1 Correct, AEF, 0 otherwise not required. For 1 3 y   B1 Seen anywhere, correct variable. 3 1(c)     2 2 1 1 2 1 , 1 2 4 2      y y y OE M1 Equate their G(y) to ½ and solve to find y.   1 2   y A1 Only. 2

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2 Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, x m, for each kangaroo. He then calculates a 90% confidence interval for the population mean tail length, n m, of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that 1.11 G n G 1. 14 . Find the values of / x and / x2 for this sample. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 2 1.771 1.14 14   s x or 2 1.771 1.11 14   s x M1 SOI Allow incorrect t-value, not z-value. [Add:   1 1.14 1.11 2   x 1.125  ] 15.75   x B1 Does not depend on use of a t-value. Subtract or substitute: 2 1 1.14 1.11 14 2 1.771         s M1 Allow incorrect t-value, but not a z-value.  2 0.00100 4  s or s = 0.0316[9] A1 450 448063 , implied by correct final answer.   2 2 2 13 14     x x s M1 OE 2 17.7   x (3) A1 CWO 6

More questions on Inference using normal and t-distributions

3 A large number of students took two test papers in mathematics. The teacher believes that the marks obtained in Paper 1 will be higher than the marks obtained in Paper 2. She chooses a random sample of 9 students and compares their marks. The marks are shown in the table. Student A B C D E F G H I Paper 1 46 73 55 64 86 42 66 68 60 Paper 2 41 66 61 63 90 40 58 42 70 (a) Carry out a Wilcoxon matched-pairs signed-rank test, at the 5% significance level, to test whether the data supports the teacher’s belief. 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(b) State an assumption that you have made in carrying out the test in part (a). 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Mark scheme: 3(a) 5 7 -6 1 –4 2 8 26 –10 4 6 -5 1 –3 2 7 9 –8 M1 Differences, allow at most 2 errors. A1 Correct rank order, ignore signs. [P = 29] Q = 16 A1 Condone P not excluded. H0: population medians equal or 1 2  m m H1: population median X > population median Y or 1 2  m m B1 ‘Population’ required. Accept use of m, not . Do not accept ‘difference between population medians > 0’ without X, Y OE specified. Critical value = 8 B1 16 > 8 and accept Ho / not significant M1 Compare their ‘16’ with their ‘8’ and conclusion. Their ‘16’ must be less than 23. Ignore their hypotheses. Condone ‘reject H1’. Insufficient evidence to support teacher’s belief or insufficient evidence that the marks/median in Paper 1 are/is higher than the marks/median in Paper 2 A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. e.g. Proves that the teacher is incorrect scores A0. A0 if hypotheses wrong way round. 7 3(b) The population differences are symmetrical (about the median difference) B1 Words in bold, or their equivalent, are required. 1

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4 An inspector is checking the lengths of metal rods produced by two machines, X and Y. These rods should be of the same length, but the inspector suspects that those made by machine X are shorter, on average, than those made by machine Y. The inspector chooses a random sample of 80 rods made by machine X and a random sample of 60 rods made by machine Y. The lengths of these rods are x cm and y cm respectively. Her results are summarised as follows. / x = 164.0 / x 2 = 338.1 / y = 124 .8 / y 2 = 261 .1 (a) Test at the 10% significance level whether the data supports the inspector’s suspicion. 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(b) Give a reason why it is not necessary to make any assumption about the distributions of the lengths of the rods. 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Mark scheme: 4(a) x x y B1   2 2 1 164 338.1 0.02405 79 80          xs and   2 2 1 124.8 261.1 0.02569 59 60          ys B1 Both. Implied by 19 379 , 790 14750 or 3sf. 2 0.02405 0.02569 80 60   s M1 2 0.0007289  s or 0.026998 or 0.0270  s A1 Implied by 1.11  z 164.0 124.8 80 60   z s M1 FT their value for s. 1.11  A1 ‘1.11’ < 1.282 Accept H0/ not significant M1 Compare with correct z-value 1.282 and consistent signs. Condone ‘reject H1’. Using probabilities, P(Z >1.11) = 0.1333 > 0.1. Insufficient evidence to support the inspector’s suspicion/ Insufficient evidence that the (mean) lengths of rods from machine X are shorter than the (mean) lengths of rods from machine Y A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. e.g. proves that the inspector is incorrect scores A0 A0 if hypotheses wrong way round. 8 Question Answer Marks Guidance 4(b) Large sample sizes OR central limit theorem applies. B1 1

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5 The random variable X has probability generating function GX ( t) given by = k ( 1 + 3t + 4t ), GX ( t) 2 where k is a constant. (a) Show that E ( X ) = 118 . 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The random variable Y has probability generating function GY ( t) given by = 3 t ( 1 + 2 t). GY ( t) 1 2 The random variables X and Y are independent and Z = X + Y . (b) Find the probability generating function of Z, expressing your answer as a polynomial in t. 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(c) Use your answer to part (b) to find the value of Var(Z ). 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(d) Write down the most probable value of Z. 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Mark scheme: 5(a)   2 G ( ) 1 3 4 X t k t t          1 1 3 4 1, 8    k k G'X (t) =   3 8  k t , M1 Or 1 3 4 11 0 1 2 8 8 8 8              px E(X) = G'X (1) = 11 8 A1 AG Evidence of using 1  t or px required. CWO 3 5(b) GZ(t) =     2 2 1 1 1 3 4 1 2 8 3     t t t t M1 Multiply the two PGFs to obtain a single polynomial of degree 5.   2 3 4 5 1 5 10 8 24    t t t t A1 May have 2t as a factor. 2 Question Answer Marks Guidance 5(c) G'Z(t) =   2 3 4 1 2 15 40 40 24    t t t t G''Z(t) =   2 3 1 2 30 120 160 24    t t t M1 Differentiate twice, allow one slip. Var(X) = G''Z(1) + G'Z(1) – (G'Z(1))2 =   2 1 97 97 312 24 24 24        M1 Use correct formula. 0.707 A1 407 576 3 5(d)   4  Z B1 FT FT their final polynomial in part (b) 1

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6 A scientist is investigating whether the ability to remember depends on age. A random sample of 150 students in different age groups is chosen. Each student is shown a set of 20 objects for thirty seconds and then asked to list as many as they can remember. The students are graded A or B according to how many objects they remembered correctly: grade A for 16 or more correct and grade B for fewer than 16 correct. The results are shown in the table. Age of students 11-12 years 13-14 years 15-16 years Grade A 25 16 19 Grade B 28 45 17 (a) Carry out a | 2 -test at the 2.5% significance level to test whether grade is independent of age of student. 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The scientist decides instead to use three grades: grade A for 16 or more correct, grade B for 10 to 15 correct and grade C for fewer than 10 correct. The results are shown in the following table. Age of students 11-12 years 13-14 years 15-16 years Grade A 25 16 19 Grade B 12 27 11 Grade C 16 18 6 With this second set of data, the test statistic is calculated as 10.91. (b) Complete the | 2 -test at the 2.5% significance level for this second set of data. 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(c) State, with a reason, whether you would prefer to use the result from part (a) or part (b) to investigate whether the ability to remember depends on age. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 6(a) H0: grade is independent of age H1: grade is not independent of age B1 Condone ‘Ability to remember’ instead of ‘grade’. Calculate expected values, shown in table in bold 25 21.2 16 24.4 19 14.4 60 28 31.8 45 36.6 17 21.6 90 53 61 36 150 M1 At least 2 correct values or expressions. A1 6 correct values or expressions. Calculate chi-squared values: 0.6811 + 2.8918 + 1.4694 + 0.4541 + 1.9279 + 0.9796 M1 At least 2 correct values (at least 3sf) or expressions seen. 8.40[4] A1 Tabular value, 2 degrees of freedom = 7.378 ‘8.404’ > 7.378 and reject H0/significant M1 Compare their value with 7.378 and conclusion without context. Condone ‘accept H1’. Sufficient evidence to suggest that grade is not independent of age. A1 Correct conclusion in context, following correct work, level of uncertainty in language (not ‘prove’, not ‘there is no evidence’), no contradictions. A0 if hypotheses wrong way round. 7 6(b) Value, 4 degrees of freedom = 11.14 10.91 < 11.14 Accept H0/not significant M1 Condone ‘reject H1’. Insufficient evidence to suggest that grade is not independent of age. A1 CAO 2 Question Answer Marks Guidance 6(c) For example, result in part (b) because the table contains more information e.g. More degrees of freedom, more groups, more detail B1 Any appropriate comment to support part (a) or part (b). Allow ‘more specific’. Not ‘more data’ or ‘more accurate’ on its own. 1

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