Comments on: GCE A Level 2011 Oct/Nov H2 Maths 9740 Paper 2 & H1 Maths 8864 Paper 1 Suggested Answers & Solutions

By: Miss Loi

Miss Loi — Wed, 07 Nov 2012 13:21:17 +0000

*waves waves*

By: Miss Loi

Miss Loi — Wed, 07 Nov 2012 13:20:47 +0000

*waves*

By: Joey

Joey — Wed, 07 Nov 2012 06:07:25 +0000

Hi, will you guys upload the answers for this year's H1/H2 math papers? Thanks so much!

By: Jess

Jess — Wed, 07 Nov 2012 05:41:26 +0000

Hi, when will the H2 maths 2012 A Level answers be out?

Thanks!

By: Miss Loi

Miss Loi — Tue, 12 Jun 2012 14:39:03 +0000

This diagram illustrates what will probably happen should Miss Loi commit SEAB treason by uploading the hallowed question papers

By: weng

weng — Mon, 28 May 2012 18:09:43 +0000

can the question pape r be uploaded

By: weng

weng — Mon, 28 May 2012 18:09:13 +0000

hmm, are the question paper be uploaded too

By: Mr Loi

Mr Loi — Wed, 16 Nov 2011 14:10:27 +0000

Refer to this comment last year for some heads up on continuity correction. In short, for the CC in Q7(iv), P(R < 25) becomes P(R < 24.5) ∵ 25 is not included in the interval.

By: Mr Loi

Mr Loi — Wed, 16 Nov 2011 13:15:50 +0000

For Q12(iii), P("at most 11 people left the queue overall by 0945") as described in the suggested solution vs P("at least 24 people in the queue by 0945") yields the same outcome. I suppose your workings is as follows, in which case is fine: X ~ Po(27) → X ~ N(27, 27) approximately Y ~ Po(18) → X ~ N(18, 18) approximately No. of people in queue at 0945, W ~ N(35 + 18 − 27, 18+27) W ~ N(26, 45) P(W ≥ 24) ≈ 0.617 (3sf) As for Q11(i), this is actually a question on Permutations & Combinations where P(R = 4) is the probability of getting a combination of 4 women in the committee of 10 (¹⁸C₄ × ¹²C₆), out of every possible combination of choosing a committee of 10 from 30 people (³⁰C₁₀). If you do B(10, 0.6), P(R = 4) would then refer to the probability of picking a woman 4 times out of 10 independent tries. This scenario is different from the first and will give you a different answer which will result in you having difficulty in proving part (ii).

By: Mr Loi

Mr Loi — Wed, 16 Nov 2011 12:26:05 +0000

Yes, a binomial model requires the calls to be independent of each other i.e. who I call now will have no bearing on who I call next and vice versa.