Cambridge A Level Mathematics - Further 9231 — 2023 Oct/Nov Paper 4 · Variant 2

1 A factory produces small bottles of natural spring water. Two different machines, X and Y, are used to fill empty bottles with the water. A quality control engineer checks the volumes of water in the bottles filled by each of the machines. He chooses a random sample of 60 bottles filled by machine X and a random sample of 75 bottles filled by machine Y. The volumes of water, x and y respectively, in millilitres, are summarised as follows. / x = 6345 / ( x - x ) 2 = 243.8 / y = 7614 /( y - y 2) = 384.9 x and y are the sample means of the volume of water in the bottles filled by machines X and Y respectively. Find a 95% confidence interval for the difference between the mean volume of water in bottles filled by machine X and the mean volume of water in bottles filled by machine Y. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 2 243.8  1219  2 384.9  3849  B1 Implied by correct s or pooled estimate s x =  = = 4.132  , s y =  = = 5.201 243.8 + 384.9 59  295  74  740  = 4.727 60 + 75 − 2 2 4.132 5.201 M1 Using their sample variances. s = +  = 0.1382  or s = 0.3718 Pooled estimate M0. 60 75 A1 6345 7614 M1 With a z-value CI: −  1.96  '0.3718' 60 75 A1 With 1.96 (with their s ) 6345 7614 1 1 Pooled −  1.96  2.174 + 60 75 60 75 4.96, 3.5  etc.  0 , 4.96  3.5, 4.96 ) A1 4.23 0.729 is A0, condone  3.5  or ( 6

More questions on Inference using normal and t-distributions

2 The number of breakdowns on a particular section of road is recorded each day over a period of 90 days. It is suggested that the number of breakdowns follows a Poisson distribution with mean 3.5. The data is summarised in the table, together with some of the expected frequencies resulting from the suggested Poisson distribution. Number of 8 or 0 1 2 3 4 5 6 7 breakdowns per day more Observed frequency 0 5 13 17 21 16 9 5 4 Expected frequency 2.718 9.512 16.646 16.993 11.895 3.469 2.407 (a) Complete the table. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Carry out a goodness of fit test, at the 10% significance level, to determine whether or not Po(3.5) is a good fit to the data. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) B1 Each. 0 5 13 17 21 16 9 5 4 B1 2.718 9.512 16.646 19.421 16.993 11.895 6.939 3.469 2.407 2 2(b) H0: Po(3.5) is a good fit to the data B1 H1: Po(3.5) is not a good fit to the data Combine first 2 columns: 5, 12.23 M1 Both. And last 2 columns: 9, 5.876 Chi-squared values: M1 Allow if no or incorrect number of columns 4.274 + 0.799 + 0.302 + 0.945 + 1.417 + 0.612 + 1.661 added. At least two ‘correct’ values (3 sf) or expressions seen from their grouping (or lack of). 2.718 + 2.140 + … + 0.6757 + 1.054 10.0 A1 AWRT 10.0 If M0 awarded then SC B1 for 10.0. Tabular value: 10.64 ‘10.0’ < 10.64, accept H0 /not significant M1 Allow equivalent to 10.64 if columns not combined or only one pair combined (12.02 one pair combined, 13.36 none combined). Insufficient evidence to suggest that Po(3.5) is not a good fit to the data A1 Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses the wrong way round or missing. 6

More questions on χ²-tests

3 Toby has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 3 marbles from the bag, without replacement. The random variable X is the number of red marbles that Toby obtains. (a) Find the probability generating function of X. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Ling also has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 2 marbles from his bag, without replacement. The random variable Y is the number of red marbles that Ling + 6 t + 5 t ) . obtains. It is given that the probability generating function of Y is 121 ( 1 2 The random variable Z is the total number of red marbles that Toby and Ling obtain. (b) Find the probability generating function of Z, expressing your answer as a polynomial in t. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Use the probability generating function of Z to find Var(Z ). [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 120 270 108 6 B1 2 correct probabilities. P(3R) = P(2R) = P(1R) = P(0R) = 504 504 504 504 1 2 3 1 2 3 M1 Cubic polynomial with their probabilities. GX(t) = 6 + 108t + 270t + 120t = 1 + 18t + 45t + 20t ( ) ( ) 504 84 1 3 15 2 5 3 A1 Correct. + t + t + t 84 14 28 21 3 3(b) 1 2 3 1 2 M1 Attempt to multiply out the brackets. GZ(t) = 1 + 18t + 45t + 20t  1 + 6t + 5t ( ) ( ) 84 12 1 2 3 4 5 M1 Obtain quintic polynomial (may not be 1 + 24t + 158t + 380t + 345t + 100t ( ) simplified). 1008 1 1 79 2 95 3 115 4 25 5 A1 Correct. + t + t + t + t + t 1008 42 504 252 336 252 3 3(c) 1 2 3 4 M1 Differentiate twice. May not see derivatives in G' = 24 + 316t + 1140t + 1380t + 500t ( ) terms of t. 1008 1 2 3 G'' = 316 + 2280t + 4140t + 2000t ( ) 1008  3360  10 B1 Seen or implied, NFWW (e.g. sampling with E( Z ) =   replacement).  1008  3 1) and G' (1) . 8736 10  10 2 26 10 100  M1 Use formula with their G  ( Var(Z) = + − = − −    1008 3  3  3 3 9  8 A1 Or 0.889 9 4

More questions on Probability generating functions

4 f(x) x 0 4 6 The diagram shows the continuous random variable X with probability density function f given by Z ] 1 3 128 ( 4 ax - bx ) 0 G x G 4, ]] f ( x) = [ c 4 G x G 6 , ] ] 0 otherwise, \ where a, b and c are constants. The upper quartile of X is equal to 4. (a) Show that c = 18 and find the values of a and b. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the exact value of the median of X. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find E ( X ), giving your answer correct to 2 decimal places. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 6 B1 AG. 1 1 1 UQ = 4 so , c = : c  6 − 4  = Some indication of reasoning required. cdx = 4 4 8 4 4 M1 Integrate and use correct limits to form equation 1 3 3 1  2 b 4  3 4 ax − bx dx = , 2 ax − x = , 32a − 64b = 96 in a and b. ( )    128 4 128 4 4   0 Equate at x = 4: a − 4b = 1 M1 Solve: a = 5, b = 1 A1 4 4(b) m M1* Attempt at integral and use of correct limits. 1 3 1 20 x − x dx = Could be in terms of a , b, c ( )  128 2 0 Or attempt at relevant part of CDF and set OR 1 F ( m ) = . 1  2 b 4  5 2 1 4 2 ax − x = x − x 2 F ( x ) =   128  4  64 512 Could be in terms of a , b, c . m 4 − 40m 2 + 256 = 0 M1 Simplify to quartic. Could be in terms of a , b, c . m = 2 2 A1 Positive answer only. 3 4(c) Their PDF multiplied by x with correct limits. 1 4 32 72 6 1 12 M1 X = 20 x − x dx + x dx E( ) 128   8 Could be in terms of a , b, c . 0 4 4 6 5 9 3 M1 Correct use of correct limits in their integral. 1  2 2 2  1  2  8 x − x  +  x  Could be in terms of a, b, c. 128  9  0 12   4 May be implied by correct final answer. 1  8 1 10  1 4 1 A1 CAO 2 − 2 + 6 6 − 8 = + 6 = 1.67   ( ) 1.67 following (first) M0 SC B2 128  9  12 9 2 3

More questions on Continuous random variables

5 A company is deciding which of two machines, X and Y, can make a certain type of electrical component more quickly. The times taken, in minutes, to make one component of this type are recorded for a random sample of 8 components made by machine X and a random sample of 9 components made by machine Y. These times are as follows. Machine X 4.0 4.6 4.7 4.8 5.0 5.2 5.6 5.8 Machine Y 4.5 4.9 5.1 5.3 5.4 5.7 5.9 6.3 6.4 The manager claims that on average the time taken by machine X to make one component is less than that taken by machine Y. (a) Carry out a Wilcoxon rank‑sum test at the 5% significance level to test whether the manager’s claim is supported by the data. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Assuming that the times taken to produce the components by the two machines are normally distributed with equal variances, carry out a t‑test at the 5% significance level to test whether the manager’s claim is supported by the data. [9] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Question 5(c) is printed on the next page. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) In general, would you expect the conclusions from the tests in parts (a) and (b) to be the same? Give a reason for your answer. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) H0: population medians are equal or mx = my B1 Do not accept ‘difference between population H1: population median for X < population median for Y or mx < my medians < 0’ without X or Y oe specified. M1 Rankings, allow at most 3 errors. X Y 4.0 1 4.5 2 4.6 3 4.9 6 4.7 4 5.1 8 4.8 5 5.3 10 5.0 7 5.4 11 5.2 9 5.7 13 5.6 12 5.9 15 5.8 14 6.3 16 6.4 17 Sum: 55 98 Test statistic = 55 A1 Tabular value for m = 8, n = 9 is 54 B1 ‘55’ > ‘54’, accept H0 /not significant M1 Ft their ‘55’ Must come from ranks. Ft their ‘54’, must come from table. Insufficient evidence to support manager’s claim. A1 Correct conclusion in context, following correct Insufficient evidence to suggest that the median time of machine X is less than the work, level of uncertainty in language. median time of machine Y. A0 if hypotheses the wrong way round or missing. 6 5(b) H0: 𝜇𝑥= 𝜇𝑦 H1: 𝜇𝑥< 𝜇𝑦 B1 x = 39.7 x 2 = 199.33 y = 49.5 y 2 = 275.47 B1     2 1  39.7 2  xs =  199.33 −  = 0.33125 7  8  2 1  49.5 2  B1 ys =  275.47 −  = 0.4025 8  9  2 7  0.33125 + 8  0.4025 M1 Pooled variance s = 8 + 9 − 2 0.36925 A1 39.7 − 49.5 M1 t = 8 9 s 1 + 1 8 9 t = −1.82 A1 Tabular value = 1.753: 1.82 > 1.753 M1 Reject H0, sufficient evidence that mean for machine X is less than mean for machine A1 CWO Y. Correct conclusion in context, following correct work, level of uncertainty in language. 9 5(c) t-test is assuming a normal distribution, and with equal variances. This may not be B1 Not specific to data in question. true. So, no reason to expect results to be the same. Mention of normal distribution is not enough. Outliers affect part (b) but not part (a). 1

More questions on Inference using normal and t-distributions