Cambridge A Level Mathematics - Further 9231 — 2022 May/June Paper 4 · Variant 1

1 A manager is investigating the times taken by employees to complete a particular task as a result of the introduction of new technology. He claims that the mean time taken to complete the task is reduced by more than 0.4 minutes. He chooses a random sample of 10 employees. The times taken, in minutes, before and after the introduction of the new technology are recorded in the table. Employee A B C D E F G H I J Time before 10.2 9.8 12.4 11.6 10.8 11.2 14.6 10.6 12.3 11.0 new technology Time after 9.6 8.5 12.4 10.9 10.2 10.6 12.8 10.8 12.5 10.6 new technology (a) Test at the 10% significance level whether the manager’s claim is justified. 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(b) State an assumption that is necessary for this test to be valid. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) Differences: 0.6 1.3 0.0 0.7 0.6 0.6 1.8 –0.2 –0.2 0.4 M1 Attempt at differences, allow one error. 2 5.6 6.74 d d      2 2 1 5.6 6.74 0.400 4 9 10 s          M1 H0: 0.4 B A     and H1: 0.4 B A     B1 Allow use of d  0.56 0.4 0.4004 10 t   (0.800) M1 t =0.7996 (0.800) A1 Critical value = t0.90(9) = 1.383 Compare: ‘0.7996’ < 1.383 Accept H0 M1 Compare calculated value with 1.383 and correct FT conclusion. Insufficient evidence to support manager’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 7 1(b) Distribution of population differences is normal B1 1

More questions on Inference using normal and t-distributions

2 The probability generating function, GY (t), of the random variable Y is given by = 0.04 + 0.2t + 0.37t + 0.3t + 0 .09 t . GY ( t) 2 3 4 (a) Find Var(Y). 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The random variable Y is the sum of two independent observations of the random variable X. (b) Find the probability generating function of X, giving your answer as a polynomial in t. 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Mark scheme: 2(a) GY ' (t) = 2 3 0.2 0.74 0.9 0.36 t t t    GY "(t) = 2 0.74 1.8 1.08 t t   M1 PGF differentiated twice. GY '(1) = 2.2 GY "(1) = 3.62 A1 Var(Y) = GY "(1) + GY '(1) – (GY '(1))2 3.62 + 2.2 – 2.22 M1 Correct formula used and attempt at Var(Y). 0.98 A1 4 2(b) GY (t) =   2 2 a bt ct   M1* Correct method. So 2 0.04, 0.2 a a   2 0.09, 0.3 c c   0.1 2 0.2, 0.5 0.2 ab b    depM1 Attempt to find all coefficients. GX (t) = 2 0.2 0.5 0.3 t t   A1 Note: GY =    2 2 1 1 2 3 100 t t   M1, accept without 1 100 . So GX =    1 1 2 3 10 t t   dep M1, must have 1 10 . Final answer A1. 3

More questions on Probability generating functions

3 The continuous random variable X has probability density function f given by kx ( 4 - x) 0 G x 1 2, f ( x) = k ( 6- x) 2 G x G 6 , * 0 otherwise, where k is a constant. (a) Show that k = 403 . [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that E ( X ) = 2.5 , find Var(X). 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(c) Find the median value of X. 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Mark scheme: 3(a) 2 3 2 1 1 2 6 1 3 2 k x x k x x           8 40 8 36 18 12 2 1 3 3 k k              3 40 k  1 3(b) E(X 2) =     2 6 3 4 2 3 0 2 4 6 d k x x dx k x x x      = 4 5 3 4 1 1 2 5 4 k x x k x x                =   9.6 96 7.92 k   M1* Attempt at E(X 2), correct limits, integrated. Var(X) = E(X2) – E2(X) = 2 (7.92) 2.5 their  depM1 Use of correct formula, with their E(X2). 1.67 A1 3 Question Answer Marks Guidance 3(c)     2 0 2 1 4 6 2 m kx x dx k x       M1* Correct expression and attempt to integrate.   2 0 2 4 5 kx x dx    B1 May be implied 2 2 1 1 6 5 2 2 k x x          2 1 1 6 12 2 2 10 m m k     2 68 12 0 3 m m    depM1 First integral and 1 2 may be seen combined as 1 10 . Obtain quadratic equation. 2 6 30 3 m     2 6 30 2.35 3 m    A1 Single answer. 4

More questions on Continuous random variables

4 A scientist is investigating the numbers of a particular type of butterfly in a certain region. He claims that the numbers of these butterflies found per square metre can be modelled by a Poisson distribution with mean 2.5. He takes a random sample of 120 areas, each of one square metre, and counts the number of these butterflies in each of these areas. The following table shows the observed frequencies together with some of the expected frequencies using the scientist’s Poisson distribution. Number per square metre 0 1 2 3 4 5 6 H 7 Observed frequency 12 20 36 32 13 6 1 0 Expected frequency 9.85 24.63 30.78 25.65 p 8.02 3.34 q (a) Find the values of p and q, correct to 2 decimal places. 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(b) Carry out a goodness of fit test, at the 10% significance level, to test the scientist’s claim. 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Mark scheme: 4(a) 1.70 q  B1 Condone 1.7 2 Question Answer Marks Guidance 4(b) 0 1 2 3 4 5 6 7  12 20 36 32 13 6 1 0 9.85 24.63 30.78 25.65 16.03 8.02 3.34 1.70 Combine last two columns: O value = 1, E value = 5.04 M1 Add last 2 columns (or 3 columns: 13.06). Calculate values of   2 O E E  : 0.4693 0.8704 0.8853 1.5720 0.5727 0.5088 3.2384 M1 Calculate values. Test statistic = 8.12 A1 H0: data fits a Poisson distribution with mean 2.5 H1: data does not fit a Poisson distribution with mean 2.5 B1 Need data and distribution. e.g Data fits the given distribution. Number of butterflies per square metre fits Po(2.5). Critical value of chi-squared = 10.64 Compare: ‘8.12’ < 10.64, accept H0 M1 Compare their test statistic with 10.64. Note: allow 9.236 if 3 columns combined or 12.02 if none combined. There is sufficient evidence to support the scientist’s claim / there is sufficient evidence to suggest that the data fits a Poisson distribution with mean 2.5 A1 Correct conclusion, in context. Level of uncertainty in language is used. 6

More questions on χ²-tests

5 Raman is researching the heights of male giraffes in a particular region. Raman assumes that the heights of male giraffes in this region are normally distributed. He takes a random sample of 8 male giraffes from the region and measures the height, in metres, of each giraffe. These heights are as follows. 5.2 5.8 4.9 6.1 5.5 5.9 5.4 5.6 (a) Find a 90% confidence interval for the population mean height of male giraffes in this region. 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Raman claims that the population mean height of male giraffes in the region is less than 5.9 metres. (b) Test at the 2.5% significance level whether this sample provides sufficient evidence to support Raman’s claim. 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Mark scheme: 5(a) 2 44.4, 247.48 x x     2 2 1 44.4 53 247.48 0.1514 7 8 350 s                 M1 CI: 44.4 0.1514 1.895 8 8  M1 Correct formula with a t value. 1.895 B1 1.895 used.   5.55 0.261 5.29, 5.81   A1 Either form. 5 5(b) 0 1 : 5.9 and : 5.9 H H     B1 2 5.55 5.9 2.544 8 t s    M1 Critical value t0.975(7) = 2.365 Compare ‘– 2.544’ < – 2.365, reject H0. M1 Compare calculated value with 2.365 and correct FT conclusion. Sufficient evidence to support Raman’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 4

More questions on Inference using normal and t-distributions

6 A teacher at a large college gave a mathematical puzzle to all the students. The median time taken by a random sample of 24 students to complete the puzzle was 18.0 minutes. The students were then given practice in solving puzzles. Two weeks later, the students were given another mathematical puzzle of the same type as the first. The times, in minutes, taken by the random sample of 24 students to complete this puzzle are as follows. 18.2 17.5 16.4 15.1 20.5 26.5 19.2 23.2 17.9 18.8 25.8 19.9 17.7 16.2 17.3 16.6 17.1 20.1 20.3 12.6 16.0 21.4 22.7 18.4 The teacher claims that the practice has not made any difference to the average time taken to complete a puzzle of this type. Carry out a Wilcoxon signed-rank test, at the 10% significance level, to test whether there is sufficient evidence to reject the teacher’s claim. 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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 Differences/rank: 0.2 – 0.5 –1.6 –2.9 2.5 8.5 1.2 5.2 2 –5 –11 –18 17 24 9 21 –0.1 0.8 7.8 1.9 –0.3 –1.8 –0.7 –1.4 –1 7 23 13 –3 –12 –6 –10 –0.9 2.1 2.3 –5.4 –2.0 3.4 4.7 0.4 –8 15 16 –22 –14 19 20 4 M1 A1 Attempt at differences, allow 4 errors. Q = 110 P = 190 A1 Either sum correct. H0: population median =18.0 and H1: population median 18.0 B1 Must be ‘population’, allow m. Mean =   1 1 1 24 25 150 4 4 n n      B1 Normal approximation: mean. Variance =    1 1 1 2 1 24 25 49 1225 24 24 n n n        B1 110.5 150 1225  M1 Attempt at test statistic, allow incorrect or missing cc, FT their 110. 1.129  A1 Question Answer Marks Guidance 6 10% 2-tail, critical value is 1.645 1.129 1.645   or 0.1296 > 0.05, accept H0. M1 Compare with 1.645, correct FT conclusion. Insufficient evidence to suggest that median is not 18.0/ Insufficient evidence to reject teacher’s claim. A1 Correct conclusion, in context. Level of uncertainty in language is used. 10

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