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January 2019

Mark Scheme Mock Paper (Set1) Pearson Edexcel GCE A Level Mathematics Statistics (9MA0/31)

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PEARSON EDEXCEL GCE MATHEMATICS

General Instructions for Marking

1. The total number of marks for the paper is 100
2. These mark schemes use the following types of marks:  M marks: Method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.  A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.  B marks are unconditional accuracy marks (independent of M marks)  Marks should not be subdivided.
3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes.  bod – benefit of doubt  ft – follow through  the symbol will be used for correct ft  cao – correct answer only  cso - correct solution only. There must be no errors in this part of the question to obtain this mark  isw – ignore subsequent working  awrt – answers which round to  SC : special case  o.e. – or equivalent (and appropriate)  d or dep – dependent  indep – independent  dp decimal places  sf significant figures   The answer is printed on the paper or ag- answer given
4. All M marks are follow through. A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but answers that don’t logically make sense e.g. if an answer given for a probability is >1 or <0, should never be awarded A marks.

5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected.
6. Where a candidate has made multiple responses and indicates which response they wish to submit, examiners should mark this response. If there are several attempts at a question which have not been crossed out, examiners should mark the final answer which is the answer that is the most complete.
7. Ignore wrong working or incorrect statements following a correct answer.
8. Mark schemes will firstly show the solution judged to be the most common response expected from candidates. Where appropriate, alternatives answers are provided in the notes. If examiners are not sure if an answer is acceptable, they will check the mark scheme to see if an alternative answer is given for the method used. If no such alternative answer is provided but the response is deemed to be valid, examiners must escalate the response for a senior examiner to review. Question Scheme Marks AOs 1(a)

[ ´ x =¿ ]

B1 cao 1.1b

Question Scheme Marks AOs 2(a)(i) Extrapolation is making predictions outside the original data range.

B1 1.

(a)(ii) This is unreliable as the trend may not continue. B1 2. (2) (b) The product moment correlation coefficient cannot be greater than 1

B1 1.

(c) r =0.76279 … awrt 0.763 B1 1.1b

(d) H 0 :^ ρ =^0 H 1 :^ ρ >^0 B1 2.

Critical value 0.7155 M1 1.1a

Reject H 0

There is evidence that the product moment correlation coefficient is greater than 0 A1f 2.2b (3) (e) This suggests that on average (female hook-billed) kites with longer tails have longer wings. B1 3.2a (1) (8 marks) Notes: (a) B1: for a correct definition of extrapolation B1: for a correct statement of the dangers of extrapolation (b) B1: for a correct statement (c) B1: for awrt 0. (d)

B1: for both hypotheses in terms of ρ

M1: for selecting a suitable 1% critical value compatible with their H 1

A1: for correct conclusion stated f their (c) provided − 1 ≤ r ≤ 1

(e)

B1: for correct interpretation in context f their (d) provided − 1 ≤ r ≤ 1

Question Scheme Marks AOs

3(a) 82

× 100 (¿12.8125)

M1 1.1b 13 A1 1.1b (2) (b)(i) [ F = faulty , T = tests positive ] P^ (^ F | T^ )=^

P ( F ∩T )

P ( T )

M1 3.1b

P ( F ∩T ) =0.02 × 0.7 [¿ 0.014 ] M1^ 1.1b

P ( T )=0.02 × 0.7+0.98 × 0.1 [¿ 0.112 ] M1 1.1b

P ( F | T )=0.125 A1 1.1b (4) b(ii) Most machines that test positive do not have faults therefore the company’s test is not very useful oe B1 3.2a (1)

(c) P ( A ∩ B ) =0.18 M1 2.

e.g. P ( A ) × P ( B ) =0.35 × 0.55=0.1925 ≠ P ( A ∩B )=0.18 A1 1.1b

(d) P ( A ∨ B not both )=0.35+0.55− 2 × 0.18 oe M1 3.1b

=0.54 A1 1.1b (2) (10 marks) Notes: (a) M1: for a correct calculation for the strata size A1: for 13 (b) M1: for identifying correct calculation

M1: for method for finding P ( F ∩T )

M1: for method for finding P ( T )

A1: a correct answer (c) M1: for correctly finding

A ∩B

P ¿

) oe A1: for a fully correct explanation: correct probabilities and correct comparisons (d) M1: for a correct expression A1: cao

Question Scheme Marks AOs 5(a) W = number of scratch cards out of 20 that win,

W B (20,0.45)

B1 3.

S =number of stores with at least 12 winning cards

S B ( 8, p )

M1 3.1b

p = P ( W ≥ 12 ) =0. 130765 A1^ 3.

1 −[ P ( S = 1 )+ P ( S = 0 ) ] M1 3.

So P ( S ≥ 2 )=0.2818 … A1 1.1b

(b) Number of trials is large and probability of success is close to

B1 1.

(c) X N (135,74.25) B1, B1 1.1b,1.

b

P ( X <122.5)= P

Z <

√74.25 )^

M1 3.

¿ 0.0734 … A1 1.1b

(d) The probability is greater than 0.025 therefore there is insufficient evidence at the 5% significance level to suggest that the proportion is different from 45% B1 2.2b (1) (11 marks) Notes: (a) B1 may be implied by subsequent working 1 st^ M1: for selection of appropriate model for S 1 st^ A1: for a correct values of the parameter p 2 nd^ A1: for awrt 0. (b) B1: both correct conditions

Accept n is large, np > 5 and n ( 1 − p )> 5

(c) B1: for correct mean B1: for correct variance M1: for continuity correction A1 awrt 0. (d) B1: for correct statement