Cambridge A Level Mathematics - Further 9231 — 2021 Oct/Nov Paper 4 · Variant 3

1 The times taken for students at a college to run 200 m have a normal distribution with mean n s. The times, x s, are recorded for a random sample of 10 students from the college. The results are summarised as follows, where x is the sample mean. x = 25. 6 / ( x - x ) 2 = 78. 5 (a) Find a 90% confidence interval for n. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ A test of the null hypothesis n = k is carried out on this sample, using a 10% significance level. The test does not support the alternative hypothesis n 1 k . (b) Find the greatest possible value of k. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) 2 78.5 8.7222 9 s = = 157 18 CI: 2 25.6 10 s t ± M1 Correct expression with a t value. With t = 1.833 A1 With correct t value. ( ) 25.6 1.71 2 ± or [23.9, 27.3] A1 Accept in either form, ISW. Accept inequality form. 4 1(b) 2 25.6 10 k t s −  M1 With t = −1.383 A1 Allow 1.383. 26.9 A1 CAO 3

More questions on Inference using normal and t-distributions

2 The continuous random variable X has cumulative distribution function F given by Z ] 0 x 1 - 1, ]] 1 2 2 ( 1 + x) - 1 G x G 0 , F ( x) = [ 1 2 1 - 2 ( 1 - x) 0 1 x G 1, ] ] 1 x 2 1. \ (a) Find the probability density function of X. 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(b) Find P - 12 1 G X G 2 [2] b l. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find E ( X 2 ) . 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(d) Find Var ( X 2 ) . 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Mark scheme: 2(a) , f ( ) otherwise. 1 1 0 1 1 0 ì + - ïïïï = - < íïïïïî x x x x x ≤ ≤ 0 ≤, M1 Differentiation attempted. A1 All correct, including 0 otherwise. 2 2(b) P 1 1 2 2 X   −       = 1 1 7 1 F F 2 2 8 8     − − = −         M1 Can do by integration with correct limits. 3 4 A1 2 2(c) ( ) ( ) 0 1 2 2 1 0 1 1 x x dx x x dx − + + −   3 4 3 4 1 1 1 1 3 4 3 4 x x x x     + + −         M1 1 1 1 12 12 6 + = A1 2 Question Answer Marks Guidance 2(d) ( ) ( ) 0 1 2 4 4 1 0 1 1 1 6 x x dx x x dx −   + + − −      5 6 5 6 1 1 1 1 1 5 6 5 6 36 x x x x     + + − −         M1 Complete method with their mean squared explicit. 1 1 7 15 36 180 − = A1 2

More questions on Continuous random variables

3 A supermarket sells pears in packs of 8. Some of the pears in a pack may not be ripe, and the supermarket manager claims that the number of unripe pears in a pack can be modelled by the distribution B(8, 0.15). A random sample of 150 packs was selected and the number of unripe pears in each pack was recorded. The following table shows the observed frequencies together with some of the expected frequencies using the manager’s binomial distribution. Number of unripe pears per pack 0 1 2 3 4 5 H6 Observed frequency 35 48 43 15 6 3 0 Expected frequency 40.874 p 35.641 12.579 2.775 0.392 q (a) Find the values of p and q. 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(b) Carry out a goodness of fit test, at the 5% significance level, to test whether the manager’s claim is justified. 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Mark scheme: 3(a) x 0 1 2 3 4 5 6 or more O 35 48 43 15 6 3 0 E 40.874 57.704 35.641 12.579 2.775 0.392 0.0352 p = 57.704 B1 3 dp or better. q = 0.035 or 0.036 B1 Accept numbers which round to 0.035 or 0.036. 2 Question Answer Marks Guidance 3(b) Combine frequencies less than 5: last 4 columns give 24 / 15.7812 M1 Allow last 2 or 3 columns combined for this M1. ( ) 2 0.8444 1.6319 1.5199 4.2803 O E E −  = + + + M1 8.28 A1 Accept 8.27 – 8.28. SC: 25.60, if values not combined, scores M1A0. SC: 14.96 if last 3 combined, scores M1A0. SC: if last 2 combined, scores M1A0. H0: B(8, 0.15) fits the data B1 Must mention distribution and data. 3 degrees of freedom, tabular value = 7.815. 8.28 > 7.815 Reject 0 H . M1 Compare their value with 7.815 and correct FT conclusion 3 columns combined compared with 9.488 and correct conclusion Insufficient evidence to support manager’s claim. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 6

More questions on χ²-tests

4 Manet has developed a new training course to help athletes improve their time taken to run 800 m. Manet claims that his course will decrease an athlete’s time by more than 2 s on average. For a random sample of 10 athletes the times taken, in seconds, before and after the course are given in the following table. Athlete A B C D E F G H I J Before 150 146 131 135 126 142 130 129 137 134 After 145 138 129 135 122 135 132 128 127 137 Use a t-test, at the 5% significance level, to test whether Manet’s claim is justified, stating any assumption that you make. 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Mark scheme: 4 0 1 : 2 and : 2 B A B A H H μ μ μ μ − = − > B1 Or use of d μ . Differences: 5 8 2 0 4 7 −2 1 10 −3 M1 Allow one error. 2 32, 272 d d  =  = 2 2 1 32 3.2, 272 18.84 9 10 d s   = = − =     848 45   =     M1 Sample mean and variance. 2 3.2 2 0.874 10 t s − = = M1 A1 Compare with tabular value 1.833: 0.874 < 1.833. Accept 0 H . M1 Compare their value with 1.833 and conclusion. Insufficient evidence to support claim. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. Assumption: population differences are normally distributed B1 8

More questions on Inference using normal and t-distributions

5 Nine balls labelled 1, 2, 3, 4, 5, 6, 7, 8, 9 are placed in a bag. Kai selects three balls at random from the bag, without replacement. The random variable X is the number of balls selected by Kai that are labelled with a multiple of 3. (a) Find the probability generating function G X (t ) of X. 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The balls are replaced in the bag. Jacob now selects two balls at random from the bag, without replacement. The random variable Y is the number of balls selected by Jacob that are labelled with an even number. (b) Find the probability generating function G Y (t ) of Y. 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The random variable Z is the sum of the number of balls that are labelled with a multiple of 3 selected by Kai and the number of balls that are labelled with an even number selected by Jacob. (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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Mark scheme: 5(a) P(3, 6, 9) = 3 2 1 6 9 8 7 504 × × = P(Two of 3, 6, 9) = 3 2 6 108 3 9 8 7 504 × × × = P(one of 3, 6, 9) = 3 6 5 270 3 9 8 7 504 × × × = P(none of 3, 6, 9) = 6 5 4 120 9 8 7 504 × × = B1 At least 2 probabilities correct. ( ) 2 3 20 45 18 1 G 84 84 84 84 X t t t t = + + + M1 A1 Attempt with at least 3 probabilities in a polynomial, CAO. 3 5(b) P(both even ) = 12 72 P(one even) = 40 72 P(no even) = 20 72 M1 ( ) 2 5 10 3 G 18 18 18 Y t t t = + + A1 CAO 2 Question Answer Marks Guidance 5(c) 2 3 1 20 45 18 1 1512 84 84 84 84 t t t   + + +     2 5 10 3 18 18 18 t t      +  + M1 Method and attempt to multiply. M1 Multiplication to obtain single quintic polynomial. 2 3 4 5 1 (100 425 600 320 64 3 1512 t t t t t + + + + + ) A1 CAO 3 5(d) G'(1) = ( ) 1 425 1200 960 256 15 1512 + + + + M1 17 9 A1 CWO 2

More questions on Probability generating functions

6 The blood cholesterol levels, measured in suitable units, of a random sample of 11 women and a random sample of 12 men are shown below. Women 51 55 242 167 152 256 75 137 98 238 235 Men 311 262 170 302 175 320 220 260 72 351 86 333 Carry out a Wilcoxon rank-sum test, at the 5% significance level, to test whether, on average, there is a difference in cholesterol levels between women and men. 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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 51 1 72 3 55 2 86 5 75 4 170 10 98 6 175 11 137 7 220 12 152 8 260 17 167 9 262 18 235 13 302 19 238 14 311 20 242 15 320 21 256 16 333 22 351 23 M1 Attempt at ranking. Total ranks: 95 A1 0 1 : and : x y x y H m m H m m = ≠ B1 Allow words, must include ‘population’ median. Use normal approximation with attempts at mean and variance. M1 Mean =132, variance = 264 A1 95.5 132 264 − M1 Allow no or wrong continuity correction for M1 only. Question Answer Marks Guidance 6 – 2.246 A1 CAO Critical value is – 1.96. – 2.246 < – 1.96 reject 0 H . M1 Compare their value with – 1.96. Or area comparison 0.0123 or 0.0124 with 0.025 and FT conclusion. There is sufficient evidence of a difference in levels. A1 Correct conclusion, in context, following correct work. Level of uncertainty in language is used. 9

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