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S1 Jan 2012 Edexcel Post Exam Discussion Thread - Solutions in the first post

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are u sure about question 2 with the probablility question doesn't p(a) include the intersection so 2/3 - 1/4 should give you p(A'nb) then you just fill in the ven diagram from there

Reply 61

alz333

the answer to p(b) was 7/12 with P(a) 1/4 p(ANB) 1/6 IT works now adding this up gives you the union of 2/3 1/4 + 7/12 - 1/6

Reply 62

Arsey

Original post by alz333

are u sure about question 2 with the probablility question doesn't p(a) include the intersection so 2/3 - 1/4 should give you p(A'nb) then you just fill in the ven diagram from there

A'nB means not A AND B this is the 15 and only the 15 region in the Venn

Reply 63

sdfgs

is anyone posting a mark scheme?

Reply 64

sanjeevp

Original post by alz333

the answer to p(b) was 7/12 with P(a) 1/4 p(ANB) 1/6 IT works now adding this up gives you the union of 2/3 1/4 + 7/12 - 1/6

Because A and B are independent the intersection is P(A) x P(B) therefore if you substitute this into the conditional rule you get:
P(A) x P(B) = P(A) + P(B) - P(AuB)

You don't know P(B) so you input values and rearrange to find P(B) as ou have eveything else. So Arsey is correct :/

Reply 65

Shafski

Original post by sdfgs

is anyone posting a mark scheme?

Seriously?

Reply 66

MushyMorshy

I'm confused with Arsey's answer for Q5f.
I thought coding does not affect the product moment correlation?

Reply 67

That Was So Funny!

Original post by Arsey

A'nB means not A AND B this is the 15 and only the 15 region in the Venn

For the last question, if I worked out the probability of w> 234.4 using tables then multiplied that by 0.8 to get a 4 decimal place answer which rounds to 0.16 but then forgot to multiply it by two would I get any method marks? I know I should have just used the simpler values but surely the values I got from the table are more accurate ?

Reply 68

Arsey

Original post by That Was So Funny!

I am not sure what you mean, why did you work out P(w>234.4) in the first place?

Reply 69

Arsey

Original post by MushyMorshy

I'm confused with Arsey's answer for Q5f.
I thought coding does not affect the product moment correlation?

5f has nothing to do with coding

Reply 70

Arsey

Original post by vishalb

But that means that the question wasn't phrased very well :frown:

It could've been interpreted in 2 different ways?

I completely agree with you.

This question is very naughty.

From the calculations it would appear that the fake 5 year old coin weighing 20g is an outlier. Statistically, removing an outlier from a data set should increase the correlation between the two sets of data (It should make any formula generated more reliable).

So we would say PMCC increases, hence my solution.

However, given that the value of PMCC was -0.908, it would get closer to -1. So the value of the negative number should decrease. That said, you can understand why someone would say it is getting bigger, the negative number is getting bigger....

I have no idea how they will mark this one.

Furthermore, the data given is not possible. I appreciate it states 'without further calculation' but if you were to remove the fake coin, reduce your data set to 9 coins, calculate new values of Stt, Sww, Stw.

Stw changes to a positive value, meaning that there is now evidence of positive correlation, so PMCC WOULD INCREASE Numerically (if not statistically)

Sww is negative, which is impossible.

With such a small data set, 1 extreme piece of data will have a massive impact, so having an original PMCC so close to -1 doesn't make any sense in the first place.

Reply 71

That Was So Funny!

Original post by Arsey

I am not sure what you mean, why did you work out P(w>234.4) in the first place?

I think the question was to work out the probability of the 2 items being picked and one being within the given range. So the prob of the given range was 0.2 and the prob of before the given range was 0.5 as it was half the diagram which left 0.3 which is the value I should have used. 0.3 was the value of p(w>234.4) but I used the tables instead being an idiot and got a more accurate value for that to 4 decimal places. Then I multiplied it by 0.8 to get a 4 decimal place answer which rounded to 0.16 but didn't multiply it by 2

Reply 72

Sarah456

Arsey, will i lose all 9 marks even if a attempted to do number 2? I got mixed up between mutually exclusive and independent..I assumed they were separate. Did i just lose all 9 marks?

Reply 73

That Was So Funny!

And if i rounded my 0.026 to 0.03 for the regression line ?

Reply 74

master_blaster66

Original post by Sarah456

Arsey, will i lose all 9 marks even if a attempted to do number 2? I got mixed up between mutually exclusive and independent..I assumed they were separate. Did i just lose all 9 marks?

i think so :frown:

, as all your other answers will be wrong too.
sorry

Reply 75

Sarah456

Thats just being too mean =/. Damnit...I wish i could get marks for attempting..

Reply 76

destruction

Original post by Sarah456

Thats just being too mean =/. Damnit...I wish i could get marks for attempting..

You would get method marks depending on what the mark scheme says, and in some cases there is something where you get a mark even if your wrong because of the question before as long as it makes some kind of sense.

You most prob get some of the 9 marks but depends on generosity of the markers

Reply 77

destruction

Original post by That Was So Funny!

i one of the previous question they said area= 0.2 if within range
So prob not in range is 1-0.2=0.8

Then it was just
In range x Not in range + Not in Rangex In range Or one of those multiplied by 2

I did it but forgot the x2 :angry:

Reply 78

Arsey

Original post by Sarah456

Arsey, will i lose all 9 marks even if a attempted to do number 2? I got mixed up between mutually exclusive and independent..I assumed they were separate. Did i just lose all 9 marks?