Author Topic: Exam 2 Suggested Solutions Finished (Read 27301 times) Tweet Share

gabrielle__ · « **Reply #45 on:** November 10, 2008, 03:50:44 pm »

Quote from: coblin on November 10, 2008, 03:49:54 pm

Quote from: Glockmeister on November 10, 2008, 03:46:50 pm
with that kind of mark, I think you could do it

really depends how the state took to this paper now...

enwiabe dropped 3 marks on Exam 2 (full on Exam 1) and got 44. (97% on SACs).

Quote from: enwiabe on November 10, 2008, 08:15:04 am
I got 44 in methods in 2006 with 40/40 in exam 1, 77/80 in exam 2 and 97% for my SACs... I thought it was a friggen' joke.

Hmm...

aw shit..

Collin Li · « **Reply #46 on:** November 10, 2008, 03:52:03 pm »

2006 was considered easy though. I also did 2006. For comparison, I dropped 1 mark in Exam 1 (and 0 in Exam 2), and got 99% in SACs and got a 49.

chris · « **Reply #47 on:** November 10, 2008, 03:53:07 pm »

I cant believe i dropped 4 marks in mchoice. 18/22

. I am dying to see how i went on s.answer, but i cbf redoing all of them to check. Has anyone worked all the s.answer out? -- im so tired and dont think i can start spec study yet.

gabrielle__ · « **Reply #48 on:** November 10, 2008, 03:54:15 pm »

i didnt find these exams difficult though
so ss's will probably be similar to 2006?

mskwmskw · « **Reply #49 on:** November 10, 2008, 03:54:55 pm »

what do i need for my exams to get a 30 raw?

Slayer · « **Reply #50 on:** November 10, 2008, 04:01:16 pm »

What r the final MC answers cause there are alot of variations? cause at the start i dropped like 9 den i dropped like 2 ... :S

Collin Li · « **Reply #51 on:** November 10, 2008, 04:05:02 pm »

Quote from: Slayer on November 10, 2008, 04:01:16 pm

What r the final MC answers cause there are alot of variations? cause at the start i dropped like 9 den i dropped like 2 ... :S

They're only suggested solutions, so they could continue changing as trinon finds flaws in his work.

Glockmeister · « **Reply #52 on:** November 10, 2008, 04:08:25 pm »

Quote from: gabrielle__ on November 10, 2008, 03:54:15 pm

i didnt find these exams difficult though
so ss's will probably be similar to 2006?

i wouldnt say that acutally... remember I said how the state went

and enwiabe I think got unlucky cause 2006 papers were regarded as too easy... which means you couldn't lose much marks before losing study scores.

Synesthetic · « **Reply #53 on:** November 10, 2008, 04:10:44 pm »

Quote from: chris on November 10, 2008, 03:53:07 pm

I cant believe i dropped 4 marks in mchoice. 18/22 . I am dying to see how i went on s.answer, but i cbf redoing all of them to check. Has anyone worked all the s.answer out? -- im so tired and dont think i can start spec study yet.

http://vcenotes.com/forum/index.php/topic,7408.0.html

iamdan08 · « **Reply #54 on:** November 10, 2008, 04:11:17 pm »

Quote from: Slayer on November 10, 2008, 04:01:16 pm

What r the final MC answers cause there are alot of variations? cause at the start i dropped like 9 den i dropped like 2 ... :S

Do you do Cas or Non Cas? The solutions are different between the two!

Slayer · « **Reply #55 on:** November 10, 2008, 04:12:37 pm »

Yeh cud ppl put the sugessted sols for MC for CAS pls

pinchies · « **Reply #56 on:** November 10, 2008, 04:16:29 pm »

I reckon, since last year 62.5 got you an A+, that if you got 58+ you could be pretty confident with an A+ this year, as I found it slightly harder. (Damn <-AC-> question stumped me!!)

hanhanchampion · « **Reply #57 on:** November 10, 2008, 04:19:43 pm »

15 is A... they are independent which means Pr(a/b)=Pr(a) I checked the book. So A must be contained or contain B. Only A is right. I think it's A{1,2,4} B{1,2} or something like that..

Collin Li · « **Reply #58 on:** November 10, 2008, 04:22:32 pm »

Quote from: hanhanchampion on November 10, 2008, 04:19:43 pm

15 is A... they are independent which means Pr(a/b)=Pr(a) I checked the book. So A must be contained or contain B. Only A is right. I think it's A{1,2,4} B{1,2} or something like that..

Yes, independent means: $\mbox{Pr}(A|B) = \mbox{Pr}(A)$ , which implies:

$\frac{\mbox{Pr}(A \cap B)}{\mbox{Pr}(B)} = \mbox{Pr}(A)$

$\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \times \mbox{Pr}(B)$

The answer is E. See this post. I have modified it so the first case is E, and the last case is A:

Quote from: coblin on November 10, 2008, 03:27:07 pm

Quote from: Nelle91 on November 10, 2008, 03:24:28 pm
I got 15 B aswell but it seemed like the most logical choice, had no idea lol

You work out Q15 by verifying this:

$\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \times \mbox{Pr}(B)$

For example (option E):

If $A = \{ 1,2\}$ , $B = \{ 2,4,6\}$ , then: $A \cap B = \{2 \}$

$\mbox{Pr}(A) = \frac{1}{3}$

$\mbox{Pr}(B) = \frac{1}{2}$

$\mbox{Pr}(A \cap B) = \frac{1}{6}$

Since $\mbox{Pr}(A) \times \mbox{Pr}(B) = \frac{1}{6} = \mbox{Pr}(A \cap B)$ , then events $A$ and $B$ are independent.

Another example (option A):

If $A = \{ 1,2,3\}$ , $B = \{ 1,2\}$ , then: $A \cap B = \{1,2 \}$

$\mbox{Pr}(A) = \frac{1}{2}$

$\mbox{Pr}(B) = \frac{1}{3}$

$\mbox{Pr}(A \cap B) = \frac{1}{3}$

Since $\mbox{Pr}(A) \times \mbox{Pr}(B) = \frac{1}{6} \neq \mbox{Pr}(A \cap B)$ , then events $A$ and $B$ are not independent!

pinchies · « **Reply #59 on:** November 10, 2008, 04:23:41 pm »

Talking to my teacher he said E?

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