Cambridge A Level Mathematics - Further 9231 — 2022 May/June Paper 4 · Variant 3

1 The times taken by members of a large quiz club to complete a challenge have a normal distribution with mean n minutes. The times, x minutes, are recorded for a random sample of 8 members of the club. The results are summarised as follows, where x is the sample mean. x = 33 .8 / ( x - x ) 2 = 94. 5 Find a 95% confidence interval for n. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1   2 94.5 13.5 7   s CI: 2 33.8 2.365 8  s M1 Correct formula with a t value, must be 33.8. B1 2.365 used.   33.8 3.072 30.7, 36.9   A1 Either form. 4

More questions on Inference using normal and t-distributions

2 A scientist is investigating the size of shells at various beach locations. She selects four beach locations and takes a random sample of shells from each of these beaches. She classifies each shell as large or small. Her results are summarised in the following table. Beach location A B C D Total Large 68 69 96 81 314 Size of shell Small 28 55 64 39 186 Total 96 124 160 120 500 Test, at the 10% significance level, whether the size of shell is independent of the beach location. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 H0 : the size of shell is independent of the location H1 : the size of shell is not independent of the location B1 O 68 69 96 81 E 60.29 77.87 100.48 75.36 O 28 55 64 39 E 35.71 46.13 59.52 44.64 M1 A1 Attempt at calculation of E values. Calculate values of   2 O E E  : 0.9859 1.0104 0.1997 0.4221 1.6646 1.7055 0.3372 0.7126 M1 Attempt at Chi-squared contributions. Accuracy of 0.1  . Test statistic = 7.04 A1 Correct total to 3sf. Tabular value for 3df = 6.251 7.04 > 6.251 Reject Ho M1 Compare with 6.251 and correct FT conclusion. May be implied by FT correct conclusion in words. Insufficient evidence to suggest size of shell is independent of location/Evidence suggests that size of shell is dependent on (beach) location A1 Correct conclusion, in context. Level of uncertainty in language is used. 7

More questions on χ²-tests

3 George throws two coins, A and B, at the same time. Coin A is biased so that the probability of obtaining a head is a. Coin B is biased so that the probability of obtaining a head is b, where b 1 a . The probability generating function of X, the number of heads obtained by George, is G X ( t) . The coefficients of t and t2 in G X ( t) are 125 and 121 respectively. (a) Find the value of a. 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The random variable Y is the sum of two independent observations of X. (b) Find the probability generating function of Y, giving your answer as a polynomial in t. 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(c) Find Var(Y ). 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Mark scheme: 3(a) GX(t) = 2 6 5 1 12 12 12   t t P(0 heads) =    6 1 1 12    a b P(1 head) =    5 1 1 12     a b a b P(2 heads) = 1 12  ab 7 1 , 12 12          a b ab Solve to give 1 1 3 4 a b         A1 Correct value for a . 2 3(b) GY (t) = 2 2 6 5 1 12 12 12         t t M1 Square their GX (t).   2 3 4 1 36 60 37 10 144     t t t t M1 Obtain quartic polynomial from 3-term GX (t). A1 3 Question Answer Marks Guidance 3(c) GY '(t) =   2 3 1 60 74 30 4 144    t t t GY "(t) =   2 1 74 60 12 144   t t M1 Differentiate their GY (t) twice. Var(Y) = GY " (1) + GY ' (1) – (GY ' (1))2   1 7 49 59 146 0.819 144 6 36 72     M1 Use correct formula and attempt at Var(Y). 0.819 A1 3

More questions on Probability generating functions

4 The continuous random variable X has probability density function f given by 3 1 1 + 2 1 G x G 3, f ( x) = 8 e x o otherwise. * 0 (a) Find X [3] E` j. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The random variable Y is given by Y = X 2 . (b) Find the probability density function of Y. 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(c) Find the 40th percentile of Y. 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Mark scheme: 4(a) E(   3 0.5 1.5 2 1 3 1 3 ) 1 d d 8 8 X x x x x x x                      1.5 0.5 3 2 2 8 3         x x A1   3 2 2 1 2 3 2 1 3 8 3 2 3            ( = 1.37) A1 3 Question Answer Marks Guidance 4(b)  3 1 8         F x x x M1 Integrate.  3 1 8           G y y y M1 Change variable throughout in their CDF.      0.5 1.5 3 , 1 9 16     g y y y y   A1 Differentiate to find g(y). 0 otherwise A1 1 9 y   and 0 otherwise, dependent on second M1. 4 4(c)  0.4  F p so 3 1 0.4 8         p p M1* Correct method. 2 15 16 15 0    p p    5 5 3 3 5 0, 3     p p p M1 dep Obtain quadratic and solve. 40th percentile of Y is 25 9 A1 3

More questions on Continuous random variables

5 A manager claims that the lengths of the rubber tubes that his company produces have a median of 5.50 cm. The lengths, in cm, of a random sample of 11 tubes produced by this company are as follows. 5.56 5.45 5.47 5.58 5.54 5.52 5.60 5.35 5.59 5.51 5.62 It is required to test at the 10% significance level the null hypothesis that the population median length is 5.50 cm against the alternative hypothesis that the population median length is not equal to 5.50 cm. Show that both a sign test and a Wilcoxon signed-rank test give the same conclusion and state this conclusion. 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Mark scheme: 5 Signs: + – – + + + + – + + + and use Binomial M1 11 signs (values not needed) allow one error. Consider B(11, 0.5) with P( 8) X  or P( 3) X  M1 Use of correct binomial. Compare 0.227 with 0.1 or 0.113 with 0.05 A1 Differences: 0.06 –0.05 –0.03 0.08 0.04 0.02 0.10 –0.15 0.09 0.01 0.12 M1 Allow one error. Signed Ranks: 6 -5 -3 7 4 2 9 -11 8 1 10 M1 Allow one swap. (P = 47) Q = 19 A1 Critical value from table = 13 B1 19 > 13 Accept H0 M1 FT their (19) compared with 13 and correct FT conclusion Insufficient evidence to support median not being 5.50 A1 BOTH conclusions correct from correct working, in context. Level of uncertainty in language. 9

More questions on Non-parametric tests

6 A company has two machines, A and B, which independently fill small bottles with a liquid. The volumes of liquid per bottle, in suitable units, filled by machines A and B are denoted by x and y respectively. A scientist at the company takes a random sample of 40 bottles filled by machine A and a random sample of 50 bottles filled by machine B. The results are summarised as follows. / x = 1120 / x 2 = 31400 / y = 1370 / y 2 = 37600 The population means of the volumes of liquid in the bottles filled by machines A and B are denoted by n A and n B . n (a) Test at the 2% significance level whether there is any difference between n A and B . 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(b) Find the set of values of a for which there would be evidence at the a% significance level that n - n A B is greater than 0.25. 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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) 2 2 1 1120 31400 1.0256 39 40          xs 2 2 1 1370 37600 1.2653 49 50          ys 40 39 62 49 Both required 2 1.0256 1.2653 40 50 s   M1 2434 47775 Condone 2 1.03 1.27 or 0.05115 40 50 s   . 2 0.050947 s  A1 May be implied by 2.66 as z value. 0 1 : 0 and : 0 a b a b H H         B1 1120 1370 0.6 40 50 0.050947    z s M1 z = 2.66 A1 Critical value of z = 2.326 2.66 > 2.326 Reject H0 M1 Compare their z value with 2.326 and correct FT conclusion. Sufficient evidence to support difference in population means oe A1 Correct conclusion, in context. Level of uncertainty in language is used. 8 Question Answer Marks Guidance 6(b) 0.6 0.25 z s   M1 Condone 1.54. 1 .55 z  A1 0.9392 – 0.9395 or 0.0605 – 0.0608 A1 6.05  A1 Condone 6.05 – 6.08, accept  4

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