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S1 Help Please.

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Reply 40
Avatar for Zacken
Zacken
22
Original post by Glavien


You're missing P(1 and 1) and P(3 and 3) and P(5 and 5).
Reply 41
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Glavien
OP
11
Original post by Zacken
You're missing P(1 and 1) and P(3 and 3) and P(5 and 5).


I still don't seem to be getting the correct answer of 0.3844.:s-smilie:
Reply 42
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Zacken
22
Original post by Glavien
I still don't seem to be getting the correct answer of 0.3844.:s-smilie:


I get it.

0.12+0.282+0.242+2(0.1×0.28)+2(0.1×0.24)+2(0.28×0.24)=0.38440.1^2 + 0.28^2 + 0.24^2 + 2(0.1 \times 0.28) + 2(0.1 \times 0.24) + 2(0.28 \times 0.24) = 0.3844
Reply 43
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Glavien
OP
11
Original post by Zacken
I get it.

0.12+0.282+0.242+2(0.1×0.28)+2(0.1×0.24)+2(0.28×0.24)=0.38440.1^2 + 0.28^2 + 0.24^2 + 2(0.1 \times 0.28) + 2(0.1 \times 0.24) + 2(0.28 \times 0.24) = 0.3844


Ohhh, sorry about that, I was doing P(1 and 1) P(3 and 3) P(5 and 5) twice by mistake. Thank you!! :smile:
Reply 44
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Zacken
22
Original post by Glavien
Ohhh, sorry about that, I was doing P(1 and 1) P(3 and 3) P(5 and 5) twice by mistake. Thank you!! :smile:


No problem.
Reply 45
Avatar for Glavien
Glavien
OP
11
For question 6(c) of this paper, why can't you use 1/2n(n+1) to work out E(B).

Paper:https://1a388b28f5ce9318f837a64962fc3ceab9c61da8.googledrive.com/host/0B1ZiqBksUHNYU2s1SVY1VDN2Q0U/January%202013%20QP%20-%20S1%20Edexcel.pdf
Reply 46
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Zacken
22
Original post by Glavien


Huh? 1/2 n(n+1) only sums 1 + 2 + ... + n
Reply 47
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Glavien
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I thought because it's a uniform distribution you could use that formula. ImageUploadedByStudent Room1465497031.659425.jpg


Posted from TSR Mobile
Reply 48
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Zacken
22
Original post by Glavien
I thought because it's a uniform distribution you could use that formula. ImageUploadedByStudent Room1465497031.659425.jpg


Posted from TSR Mobile


Keyword "defined on the set {1, 2,...,n}" The one in your question not defined on this sort of set.

In any case, ignore all these stupid formulae and just work out the expectation using normal arithmetic all the time.
Reply 49
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Glavien
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11
Original post by Zacken
Keyword "defined on the set {1, 2,...,n}" The one in your question not defined on this sort of set.

In any case, ignore all these stupid formulae and just work out the expectation using normal arithmetic all the time.


Cool, thanks a lot. :smile:
Reply 50
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Zacken
22
No worries.
Reply 51
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Glavien
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11
How do you work out the standard deviation for question 2 of this paper please?
I looked at the exam solutions video and he just confused me. There must be a simpler way of going about it.

Paper: https://1a388b28f5ce9318f837a64962fc3ceab9c61da8.googledrive.com/host/0B1ZiqBksUHNYU2s1SVY1VDN2Q0U/June%202014%20MA%20-%20S1%20Edexcel.pdf
Reply 52
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Zacken
22
Original post by Glavien
How do you work out the standard deviation for question 2 of this paper please?
I looked at the exam solutions video and he just confused me. There must be a simpler way of going about it.

Paper: https://1a388b28f5ce9318f837a64962fc3ceab9c61da8.googledrive.com/host/0B1ZiqBksUHNYU2s1SVY1VDN2Q0U/June%202014%20MA%20-%20S1%20Edexcel.pdf



Have you learnt stuff like E(ax) = aE(X) and Var(aX) = a^2Var(X), and Var(X + a) = Var(X). You should have, at least.

So, in this case, you have Var(Y) = Var(1.4X - 20) = Var(1.4X) = 1.4^2Var(X).

So if we solve for Var(X):

Var(X) = Var(Y) / 1.4^2

Then standard deviation is square root of variance:

So sqrt(Var(X)) = sqrt(Var(Y) / 1.4^2) means s.d(X) = s.d(Y) / 1.4
Reply 53
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Glavien
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11
Original post by Zacken
Have you learnt stuff like E(ax) = aE(X) and Var(aX) = a^2Var(X), and Var(X + a) = Var(X). You should have, at least.

So, in this case, you have Var(Y) = Var(1.4X - 20) = Var(1.4X) = 1.4^2Var(X).

So if we solve for Var(X):

Var(X) = Var(Y) / 1.4^2

Then standard deviation is square root of variance:

So sqrt(Var(X)) = sqrt(Var(Y) / 1.4^2) means s.d(X) = s.d(Y) / 1.4


Hi, thanks the standard deviation makes sense now. :smile:

For the mean, I know now how to use E(X) to work the mean out, but why would you even think about using it in this case? I thought it was to do with probability and the working about expected value. I don't know why but it just seemed like a random thing to do. Sorry, to keep bothering you.
Reply 54
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Zacken
22
Original post by Glavien
Hi, thanks the standard deviation makes sense now. :smile:

For the mean, I know now how to use E(X) to work the mean out, but why would you even think about using it in this case? I thought it was to do with probability and the working about expected value. I don't know why but it just seemed like a random thing to do. Sorry, to keep bothering you.


*shrugs* It's sort of a standard thing in my mind, hopefully you'll know when to use it now. It comes up when you're given a function of a random variable (which is still a random variable).
Reply 55
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Glavien
OP
11
Original post by Zacken
*shrugs* It's sort of a standard thing in my mind, hopefully you'll know when to use it now. It comes up when you're given a function of a random variable (which is still a random variable).


Thank you so much! :smile:
Reply 56
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Zacken
22
Original post by Glavien
Thank you so much! :smile:


No worries.
Reply 57
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Glavien
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11
How do you do part f of this question from the Jan 2016 IAL paper please?

I thought it would be 0.75*(0.25)^2=3/64

ImageUploadedByStudent Room1465924620.325705.jpg
Reply 58
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Zacken
22
Original post by Glavien
How do you do part f of this question from the Jan 2016 IAL paper please?

I thought it would be 0.75*(0.25)^2=3/64

ImageUploadedByStudent Room1465924620.325705.jpg


Heh, this tripped everybody up in the exam - I managed it though. :tongue:

0.75 includes the probability that the third person got a non-merit grade, i.e: gets a distinction or w/e it is.
Reply 59
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Glavien
OP
11
Original post by Zacken
Heh, this tripped everybody up in the exam - I managed it though. :tongue:

0.75 includes the probability that the third person got a non-merit grade, i.e: gets a distinction or w/e it is.


Ohhh, that took me a while too understand. Thanks for that.

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