It’s been a looooooong day. And yeah the A Level Maths papers have been a little tricky this year.

While I recognize that many of have had a tough time working out the papers for the past two mornings, do spare a thought for me who have been forced by my sis to working out three A Level Maths papers in the past two days (plus tuition time) as you browse through my suggested answers to today’s H2 Maths Paper 2 which you may access here by clicking the button (if you haven’t yet done so).

Latest version:

H2 Maths Paper 2

1.1: Q2(ii) Modified rejection reason to 2x < n (as kindly pointed out by Chen Jing) Q3(iii): x = -0.4847 instead of -0.4877 (misread due to pen no ink :P) Q8(iv): 13.5 – corrected rounding error (as kindly pointed out by Ah Wei) Q12(iii) – removed lower limit of 0 since there’s nothing stopping the number of people joining the queue to be greater than the number leaving the queue (as pointed out by sam)

1.2: Q2(ii) – Simplified the final answer further. Q8(i) – Removed extra point from scatter diagram which should contain only 5 points (I swear I plotted only 5 points that night!!!) Q12(iii) – removed the continuity correction since the approximated aggregated Normal distribution is already continuous. Therefore no continuity correction is needed (apologies for this oversight and I hope this will bring closure to this part). Q12(iv) – added missing answer (too tired that night and missed this!)

GCE A Level H2 Maths 9740 Paper 2 (left) H1 Maths 8864 Paper 1 (right) suggested solutions & answers

As usual, please leave a comment should you spot any mistake in this set of suggested solutions – as mentioned it’s been rather tiring doing the many, many A Level Maths papers these two days *faints*

Last but not least, if you haven’t yet done so, do say hi to me when you drop by in Facebook 😉

@sam: Yes there shouldn't be a lower limit of 0 since there is a possibility that the number of people joining the queue is greater than the number leaving the queue.

Hi! I think the rejection reason for the rectangle should be 2x<n if u looked at the left side! Cause one of the x value is ard 0.2n the other 0.7n so Reject the 0.7n

Hi, For 12iii, there should not be a lower limit of 0, as the number of people entering the queue can be more than number of people leaving. In addition, continuity correction should not be used, as the distribution of Y15-X15~N(9,45) is not based on a poisson distribution. seems like it's more tricky than you thought 😀

@yellowfishx: Sorry I've been away in the past days.

Yes a continuity correction is used only when the distribution modeled is discrete. In this case, the aggregated normal distribution is already continuous (though it is approximated) so a further continuity correction is unnecessary.

Therefore no continuity correction is needed and I've amended the solutions to reflect this in v1.2.

Apologies for this oversight and I hope this will bring closure to everyone on this part.

P.S. The lower limit issue has be corrected since v1.1.

@Curious: I've removed the continuity correction (cc) step for Q12(iii) in version 1.2.

However, the reason has more to do with the fact that the final approximated normal distribution is continuous (and not a discrete Poisson distribution anymore) that leads to cc being unnecessary.

CC would be required if the final modeled distribution is discrete, regardless of whether it is the result of any 'subtraction' or 'addition' between distributions.

@Anonymous: The scatter diagram is y on x so that you can show clearly the non-linear relationship between y and x in your scatter diagram.

Please note that you should simply sketch using arbitrary points as required by the instructions in Q9(i) i.e. a is positive ⇒ positive y-intercept b is negative ⇒ plots negatively sloped equally-spaced positive x- and y-values ⇒ all points in first quadrant

You should not waste time plotting using the data in the table (or worse use all 7 points in your scatter diagram when you are instructed to include only 5) which is actually meant for subsequent parts.

@clarionx: 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.

@Mac: 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 (^{18}C_{4} × ^{12}C_{6}), out of every possible combination of choosing a committee of 10 from 30 people (^{30}C_{10}).

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).

my math teacher once told me that there is no bell curve for math. because too many people do well..he and i may be wrong. oh well, this is def more difficult than last year's. I'm between 70-75 hope i get an A. I NEED THE A...like i have the right to dismiss everyone else because i got a conditional offer from UK. HAHA, everybody, good luck!

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

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## 36 Comments

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12 part three. why is there a bottom limit of zero. shouldnt it be just P(Y-X <11.5) ?

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@sam: Coz you're supposed to "assume that the queue does not become empty during this period".

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@Trex: Nono, i think i interpreted wrongly 🙁 Sorry.

PS: remove my comment T_T

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@sam: Yes there shouldn't be a lower limit of 0 since there is a possibility that the number of people joining the queue is greater than the number leaving the queue.

This has been amended in v1.1.

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For 8iv) must note to use 5sf and above to find value of y. If not will have rounding off errors.

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@Ah Wei: Amended in version 1.1. Thanks for pointing this out!

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Question 2 part (ii) : it should be x < 2n ,and not x < n. n/12(6 + root(12)) is still smaller than n

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@Anonymous: Amended in v1.1 as described here 🙂

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Hi! I think the rejection reason for the rectangle should be 2x<n if u looked at the left side! Cause one of the x value is ard 0.2n the other 0.7n so

Reject the 0.7n

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@Chen jing: Thanks for pointing this out. Amended in v1.1 - sorry a bit cuckoo-eyed that night 😕

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h2 math paper2, question 12 part iii) shouldnt it be X + Y instead of X-Y ??

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Hi,

For 12iii, there should not be a lower limit of 0, as the number of people entering the queue can be more than number of people leaving.

In addition, continuity correction should not be used, as the distribution of Y15-X15~N(9,45) is not based on a poisson distribution.

seems like it's more tricky than you thought 😀

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@yellowfishx: Sorry I've been away in the past days.

Yes a continuity correction is used only when the distribution modeled is

discrete. In this case, the aggregated normal distribution is already continuous (though it is approximated) so a further continuity correction is unnecessary.Therefore no continuity correction is needed and I've amended the solutions to reflect this in v1.2.

Apologies for this oversight and I hope this will bring closure to everyone on this part.

P.S. The lower limit issue has be corrected since v1.1.

曜

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omgg... got average of 70 for both papers only... that isnt an A right

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@james: same! damn sian 🙁 I screw up paper 1. wondering wat's the cut off for A this yr!

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continuity correction should be to P(R<25.5) rather than P(R<24.5)?

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@sw: 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.曜

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But mdm, isnt there is chance that the queue will get longer instead?

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@john: Yes, definitely - I overlooked this and I've since amended in v1.1.

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Hi, for 12iii, considering that there's a subtraction between both approximated distributions, isn't cc not needed? Thanks for the answers! 🙂

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@Curious: I've removed the continuity correction (cc) step for Q12(iii) in version 1.2.

However, the reason has more to do with the fact that the final approximated normal distribution is

continuous(and not a discrete Poisson distribution anymore) that leads to cc being unnecessary.CC would be required if the final modeled distribution is

discrete, regardless of whether it is the result of any 'subtraction' or 'addition' between distributions.曜

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SOrry just Wondering whether the scatter diagram is y on

X or y on x^2?

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@Anonymous: The scatter diagram is y on x so that you can show clearly the non-linear relationship between y and x in your scatter diagram.

Please note that you should simply sketch using arbitrary points as required by the instructions in Q9(i) i.e.

a is positive ⇒ positive y-intercept

b is negative ⇒ plots negatively sloped

equally-spaced positive x- and y-values ⇒ all points in first quadrant

You should not waste time plotting using the data in the table (or worse use all 7 points in your scatter diagram when you are instructed to include only 5) which is actually meant for subsequent parts.

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oh and 7i) can i say that 1 of the assumptions is to assume that the calls are independent of one another?

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@clarionx: 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.

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In my opinion it is one of the more difficult year so the A should be about 70-73 prolly:) don't worry if u get 70 and above

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For 12iii) can't you do E( 35+X-Y) where 35+X-Y is the people left in the queue? And Qn11i) is it possible to do R~B(10,0.6) then find P(R=4)?

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@Mac: 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 acombinationof 4 women in the committee of 10 (^{18}C_{4}×^{12}C_{6}), out of every possible combination of choosing a committee of 10 from 30 people (^{30}C_{10}).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).曜

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my math teacher once told me that there is no bell curve for math. because too many people do well..he and i may be wrong. oh well, this is def more difficult than last year's. I'm between 70-75 hope i get an A. I NEED THE A...like i have the right to dismiss everyone else because i got a conditional offer from UK. HAHA, everybody, good luck!

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hmm, are the question paper be uploaded too

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can the question pape r be uploaded

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

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

Thanks!

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@Jess: *waves*

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

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@Joey: *waves waves*