Cambridge A Level Mathematics - Further 9231 — 2020 Oct/Nov Paper 4 · Variant 2

1 The heights of the members of a large sports club are normally distributed. A random sample of 11 members of the club is chosen and their heights, x cm, are measured. The results are summarised as follows, where x denotes the sample mean of x. x = 176.2 / ( x - x ) 2 = 313. 1 Test, at the 5% significance level, the null hypothesis that the population mean height for members of this club is equal to 172.5 cm against the alternative hypothesis that the mean differs from 172.5 cm. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 2 313.1 31.3 10 s = = B1 Can be implied 176.2 172.5 / 11 t s − = M1 = 2.19 A1 ‘2.19’ < 2.228 M1 Comparison of their t value with 2.228 accept H0: insufficient evidence to reject mean height is 172.5 OR sufficient evidence to accept mean height is 172.5 A1 FT Correct conclusion in context, level of uncertainty in language used. No contradictions. CWO Do not accept: ‘mean height is 172.5’ Do not accept use of symbols that are undefined e.g. insufficient evidence to reject μ = 172.5 5

More questions on Inference using normal and t-distributions

2 A large school is holding an essay competition and each student has submitted an essay. To ensure fairness, each essay is given a mark out of 100 by two different judges. The marks awarded to the essays submitted by a random sample of 12 students are shown in the following table. Student A B C D E F G H I J K L Judge 1 62 74 52 48 68 55 56 64 37 70 81 59 Judge 2 65 70 47 49 76 74 67 54 50 77 72 75 (a) One of the students claims that Judge 2 is awarding higher marks than Judge 1. Carry out a Wilcoxon matched-pairs signed-rank test at the 5% significance level to test whether the data supports the student’s claim. [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ It is discovered later that the marks awarded to student A have been entered incorrectly. In fact, Judge 1 awarded 65 marks and Judge 2 awarded 62 marks. (b) By considering how this change affects the test statistic, explain why the conclusion of the test carried out in part (a) remains the same. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) H0: difference in (population) medians = 0 H1: difference in (population) medians < 0 (or >0) B1 B1 Correct hypotheses (median or m) All notation identified e.g. m = population median if used Differences: 3 4 5 1 8 19 11 1 0 13 7 9 16 − − − − − − − − M1 At most 3 errors Ranks: 2 3 4 1 6 12 9 8 10 5 7 11 − − − − − − − − A1 Award for correct rank order, ignore signs Sum T = 3 + 4 + 8 + 7 = 22 A1 cwo Compare with critical value 17: ‘22’ > 17 M1 Compare their T with 17 Accept H0 Data does not support student’s claim A1 In context, all correct, except possibly second B1 Level of uncertainty in language used. No contradictions. 7 2(b) Rank for A becomes + 2, M1 T = 24 Changing sign of difference can only reduce evidence in favour of the claim. A1 still > 17 and test result unchanged 2

More questions on Non-parametric tests

3 A random sample of 200 observations of the continuous random variable X was taken and the values are summarised in the following table. Interval 0 G x 1 0.5 0.5 G x 1 1 1 G x 1 1.5 1.5 G x 1 2 2 G x 1 2.5 2.5 G x 1 3 Observed frequency 5 23 40 41 46 45 It is required to test the goodness of fit of the distribution with probability density function f given by 1 9 x ( 4 - x) 0 G x G 3, f ( x) = *0 otherwise. Most of the relevant expected frequencies, correct to 2 decimal places, are given in the following table. Interval 0 G x 1 0 .5 0.5 G x 1 1 1 G x 1 1.5 1.5 G x 1 2 2 G x 1 2.5 2.5 G x 1 3 Expected frequency p q 37.96 43.52 43.52 37.96 (a) Show that p = 10.19 and find the value of q. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Carry out a goodness of fit test, at the 5% significance level, to test whether f is a satisfactory model for the data. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 0.5 2 0 4 1 d [ 9 9 x x x   − =      2 3 1 1 2 0.5 0.5 9 3   × −     ] (= 0.0509(26) or 11 216 ) M1 Statement of integration with correct limits Freq = p = 0.0509(26) × 200 = 10.19 AG A1 Requires sight of 0.0509 or 11 216 By addition to 200, q = 26.85 725 27       B1 Or by integration 3 3(b) O 5 23 40 41 46 45 E 10.19 26.85 37.96 43.52 43.52 37.96 Test statistic = 2.6433 + 0.5520 + 0.1096 + 0.1459 + 0.1413 + 1.3056 M1 At least 4 correct 4.89 or 4.90 A1 ‘4.90’ < 11.07: M1 Compare their value with 11.07 PDF is a satisfactory model for the data. A1 FT Correct conclusion in context, ft only their 4.90 4

More questions on χ²-tests

4 The continuous random variable X has cumulative distribution function F given by 0 x 1 2 , - F ( x) = 601 x 2 1 15 2 G x G 8 , *1 x 2 8 . (a) Find P ( 3 G X G 6). [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find X . [3] E` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find X . [2] Var` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) The random variable Y is defined by Y = X3 . Find the probability density function of Y. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) F(6) – F(3) = 9 20 or 0.45 B1 1 4(b) ( ) 1 , 2 8 f 30 0, x x x otherwise  ≤ ≤  =   B1 May be implied. Only need to see ( ) 1 f 30 x x = E( X ) = 8 3 5 2 2 2 1 1 d 30 75 x x x   =        M1 Integrated = 2.34 A1 2.338… Answer of 2.34 without 5 2 1 75 x scores B1B1 (2/3) 3 4(c) Var( X ) =[ ( ) 8 2 2 3 2 1 1 d ] 30 90 x x x E X   = −      ( ) ( ) 8 8 2 2 3 2 2 1 1 Var 30 90 X x dx x E X     = = −            M1 = 5.6 – 2.3382 = 0.133 or 0.134 A1 2 Question Answer Marks Guidance 4(d) G(y) = 2 3 1 1 60 15 y − M1 CDF for Y g(y) = 1 3 1 90 y − for 8 512 y ≤ ≤ (0 otherwise) M1 A1 Differentiate to find PDF for Y Correct g(y) and correct range seen anywhere 3

More questions on Continuous random variables

5 The random variable X has the binomial distribution B(n, p). (a) Write down an expression for P ( X = r) and hence show that the probability generating function of X is ( q + pt) n , where q = 1 - p. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Use the probability generating function of X to prove that E ( X ) = np and Var ( X ) = np ( 1 - p). [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) B1 GX(t) = ( ) 0 nCr 1 n n r r r p p t − −  M1 Accept minimum of 4 terms including the last. Shown with specific value of n is M0 ( ) 0 nCr ( ) 1 n n r r pt p − −  = ( ) n q pt + A1 At least one intermediate step to be shown, with p and t grouped AG 3 5(b) G’X(t) = ( ) 1 n n q pt p − + × M1 So E(X) = G’X(1) = ( ) 1 n np q p − + and 1 q p + = so E(X) = np A1 G’’X(t) = ( )( ) 2 1 n n n q pt p p − − + × × M1 Var(X) = ( ) ( ) 2 2 1 n n p np np − + − M1 ( ) 1 np p − A1 5

More questions on Probability generating functions

6 Nassa is researching the lengths of a particular type of snake in two countries, A and B. (a) He takes a random sample of 10 snakes of this type from country A and measures the length, x m, of each snake. He then calculates a 90% confidence interval for the population mean length, n m, for snakes of this type, assuming that snake lengths have a normal distribution. This confidence interval is 3.36 G n G 4.22. Find the sample mean and an unbiased estimate for the population variance. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Nassa also measures the lengths, y m, of a random sample of 8 snakes of this type taken from country B. His results are summarised as follows. / y = 27.86 / y 2 = 98.02 Nassa claims that the mean length of snakes of this type in country B is less than the mean length of snakes of this type in country A. Nassa assumes that his sample from country B also comes from a normal distribution, with the same variance as the distribution from country A. Test at the 10% significance level whether there is evidence to support Nassa’s claim. [8] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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(a) ( ) 1 4.22 3.36 2 x = + = 3.79 B1 2 4.22 3.36 10 ts − = M1 Using z value implies M0 With t = 1.833 A1 ( 0.7418) s = variance = 0.55(0) A1 4 6(b) H0: A B μ μ = H1: A B μ μ > B1 Not x ( ) 2 2 1 27.86 98.02 0.142 507 7 8 s   = − =     M1 = ( ) 1 98.02 97.02 7 − Pooled estimate = 9 0.550316 7 0.142507 16 × + × M1 0.372 = A1 ( ) 27.86 3.79 8 1.063 1 1 0.372 10 8 t − = = − + M1A1 ‘1.063’ < 1.337 oe M1 Compare their value with 1.337 Accept H0 Nassa’s claim is not supported A1ft Correct conclusion in context, level of uncertainty in language used. No contradictions. 8

More questions on Inference using normal and t-distributions