Edexcel S3 Urgent question :o Watch

RosesAreBlue
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S3 June 2015, 5c and 5d... Someone please explain, Im beyond confused


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Zacken
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(Original post by RosesAreBlue)
S3 June 2015, 5c and 5d... Someone please explain, Im beyond confused


(Question Paper)
https://drive.google.com/file/d/0B3t...s4QW5aWEE/view

(Mark Scheme)
https://drive.google.com/file/d/0B3t...NFTEVkVXc/view
You'll need to be more specific. 5(c) is just using Var(X+Y) = Var(X) + Var(Y).
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RosesAreBlue
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(Original post by Zacken)
You'll need to be more specific. 5(c) is just using Var(X+Y) = Var(X) + Var(Y).
5c is an explain question, I dont understand why its not independent.
5d, what was the reason for bringing all U1-U5 all into one fraction?

Is there something in the S3 book I haven't learnt? :confused:
This is coming from someone whos only relied on past papers and a variety of online videos to self-teach.
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Zacken
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(Original post by RosesAreBlue)
5c is an explain question, I dont understand why its not independent.
5d, what was the reason for bringing all U1-U5 all into one fraction?

Is there something in the S3 book I haven't learnt? :confused:
This is coming from someone whos only relied on past papers and a variety of online videos to self-teach.
1. I'll talk about independence first:

Well, think of it, two things being independent means knowing something about one thing gives you no information about the other. And the other way around for dependent.

So, if we know something about U1, that gives us a bit of information about Ubar since we calculate Ubar using U1 (and other stuff), but the key thing is that knowing U1 or U2 or U3 or U4 or U5 gives us information about Ubar.

So they are not independent.

However, knowing something about U1 gives you no information whatsoever about U2 or U3 or anything, hence U1 and U2 are independent, same for U2 and U3 and U1 and U3 and etc...

Remember that just because the things are all independent to one another (U1 independent to U2 to U3 to U4, etc...) doesn't mean that the sum of those things (Ubar) is also independent of them.

2. So since U1, U2, etc... are all independent it is not the case that Ubar and U1 or U2 or... are independent, so you need to convert Ubar into its definition as the sum of all 5 U's and dividing by 5 which are then all indepdenent of one another so you can use the Var(X + Y) = Var(X) + Var(Y) formula.

Hope that helped, I've self-taught all my modules as well, so I know how tough it can be.
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RosesAreBlue
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(Original post by Zacken)
1. I'll talk about independence first:

Well, think of it, two things being independent means knowing something about one thing gives you no information about the other. And the other way around for dependent.

So, if we know something about U1, that gives us a bit of information about Ubar since we calculate Ubar using U1 (and other stuff), but the key thing is that knowing U1 or U2 or U3 or U4 or U5 gives us information about Ubar.

So they are not independent.

However, knowing something about U1 gives you no information whatsoever about U2 or U3 or anything, hence U1 and U2 are independent, same for U2 and U3 and U1 and U3 and etc...

Remember that just because the things are all independent to one another (U1 independent to U2 to U3 to U4, etc...) doesn't mean that the sum of those things (Ubar) is also independent of them.

2. So since U1, U2, etc... are all independent it is not the case that Ubar and U1 or U2 or... are independent, so you need to convert Ubar into its definition as the sum of all 5 U's and dividing by 5 which are then all indepdenent of one another so you can use the Var(X + Y) = Var(X) + Var(Y) formula.

Hope that helped, I've self-taught all my modules as well, so I know how tough it can be.
Thanks a ton I couldn't thank you more.
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Zacken
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(Original post by RosesAreBlue)
Thanks a ton I couldn't thank you more.
No problem at all.

Good luck for tomorrow! :woo:
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