12 – M11/5/MATME/SP1/ENG/TZ1/XX/M !!!!!!! 1)! SECTION B 8 .

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!!!!!!!!Stats and Prob test Answers!!!!– 12 –!M11/5/MATME/SP1/ENG/TZ1/XX/MSECTION B8.(a)(i)s 1(ii)evidence of appropriate approache.g. 21 16 , 12 8 q 15A1q 5(iii)p 7, r3N1(M1)A1N2A1A1N2[5 marks](b)(i)P (art music)(ii)METHOD 1P (art)581216A234N2A1evidence of correct reasoning3 5e.g.4 8R1the events are not independentAGN0METHOD 2P (art) P (music)9625638A1evidence of correct reasoning12 85e.g.16 16 16R1the events are not independentAGN0[4 marks](c)3(seen anywhere)P (first takes only music)16A17(seen anywhere)15A1P (second takes only art)evidence of valid approach3 7e.g.16 15P (music and art)21240(M1)780A1N2[4 marks]Total [13 marks]

!!!!!!!!Stats and Prob test Answers– 15 –9.(a)M12/5/MATME/SP1/ENG/TZ2/XX/M(i)4 33, and6 66(ii)2 11, and3 22A1A1A1multiplying along the correct branches (may be seen on diagram)3 2e.g.7 664217N3(A1)A1N2[5 marks](b)P (bag A)261, P (bag B)3appropriate approache.g. P(WW A) P(WWcorrect calculation1 1 2 2 2,e.g.3 7 3 7 42P(2W )602525214623(seen anywhere)(A1)(A1)(M1)B)A1842A1N3[5 marks]continued  

!!!Stats and Prob test Answers– 16 –Question 9 /XX/MSPEC/5/MATME/SP1/ENG/TZ0/XX/Mrecognizing conditional probabilityP ( A B)P (WWA) ASECTIONe.g., P ( A WW )P ( B)P (WW )1AE ! ADcorrect2numerator6 2 1attemptto finde.g.,P AWWAD42 6 21e.g. AB ! BD , u ! v(M1)correct denominator11 !" 1AE #60(u, !5v ) " # u ! v #e.g.2 21 22 %252(A1)841probability 1ED # AE # 420(u ! v )52A1(A1)M1A1[3 marks]A1A1N3[4 marks]DC # 3vA1attempt to find EC1e.g. ED ! DC, (u ! v ) ! 3v2M117 " 1!EC # u ! v " # (u ! 7v ) #22 2%N2A1N2[4 marks]Total [7 marks]2.(a)min value of r is &1 , max value of r is 1(b)C(c)linear, strong negativeA1A1N2[2 marks]A1N1[1 mark]A1A1N2[2 marks]Total [5 marks]

!!!!!!4)!!!!!!!!!!!!!!!!!!!!!!!Stats and Prob test Answers!!!!!!– 11 –M12/5/MATME/SP2/ENG/TZ1/XX/MSECTION B8.(a)(b)(i)p 17, q 11(ii)75 T85evidence of valid approache.g. adding frequencies760.81720439376P (T 95)0.81793A1A1N2A1N1[3 marks](M1)A1N2[2 marks](c)(d)(i)10A1N1(ii)50A1N1[2 marks](i)evidence of approach using mid-interval values (may be seen in part (ii)) (M1)79.1397849A2x 79.1N316.438606116.4N1(ii)!A1[4 marks](e)evidence of valid approache.g. standardizing, z 0.9648.0.8326812P (T 95) 0.833(M1)A1N2[2 marks]Total [13 marks]!!!!!!!!!!!!!

!!!!!5)!!!!!!!!!!!!!!!!!!!!!!!Stats and Prob test Answers– 11 –SPEC/5/MATME/SP2/ENG/TZ0/XX/MSECTION B8.A1A1N2[2 marks]additional cost per box (unit cost)A1N1fixed costsA1N1[2 marks](a)y ! 10.7 x ! 121(b)(i)(ii)(c)(d)attempt to substitute into regression equatione.g. y " 10.7 # 60 ! 121, y " 760.12 M1cost " 760 (accept 763 from 3 s.f. values)A1setting up inequality (accept equation)e.g. 19.99 x % 10.7 x ! 121M1x % 12.94 A113 boxes (accept 14 from x % 13.02, using 3 s.f. values)A1N2[2 marks]N2Note: Exception to the FT rule: if working shown, award the final A1 for acorrect integer solution for their value of x.[3 marks](e)(i)this would be extrapolation, not appropriateR1R1N2(ii)this would be interpolation, appropriateR1R1N2[4 marks]Total [13 marks]

q 5 A1 N2 (iii) p 7, r 3 A1A1 N2 [5 marks] (b) (i) 5 P(art music) 8 A2 N2 (ii) METHOD 1 12 3 P(art) 16 4 §· ¹ A1 evidence of correct reasoning R1 e.g. 35 48 z the events are not independent AG N0 METHOD 2 96 3 P(art) P(music) 256 8 §· u ¹ A1 evidence of correct reasoning R1 e.g. 12 8 5 16 16 16 uz the events are not .

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