# S1 Help Please.Watch

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3 years ago
#41
(Original post by Glavien)
For question 1(d), why is it incorrect to do this:

2*P(1and3) + 2*P(3and5) + 2*P(5and1)

Paper: https://1a388b28f5ce9318f837a64962fc...%20Edexcel.pdf
You're missing P(1 and 1) and P(3 and 3) and P(5 and 5).
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Thread starter 3 years ago
#42
(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. 0
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3 years ago
#43
(Original post by Glavien)
I still don't seem to be getting the correct answer of 0.3844. I get it. 0
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Thread starter 3 years ago
#44
Ohhh, sorry about that, I was doing P(1 and 1) P(3 and 3) P(5 and 5) twice by mistake. Thank you!! 0
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3 years ago
#45
(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!! No problem.
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Thread starter 3 years ago
#46
For question 6(c) of this paper, why can't you use 1/2n(n+1) to work out E(B).

Paper:https://1a388b28f5ce9318f837a64962fc...%20Edexcel.pdf
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3 years ago
#47
(Original post by Glavien)
For question 6(c) of this paper, why can't you use 1/2n(n+1) to work out E(B).

Paper:https://1a388b28f5ce9318f837a64962fc...%20Edexcel.pdf
Huh? 1/2 n(n+1) only sums 1 + 2 + ... + n
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Thread starter 3 years ago
#48
I thought because it's a uniform distribution you could use that formula. Posted from TSR Mobile
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3 years ago
#49
(Original post by Glavien)
I thought because it's a uniform distribution you could use that formula. 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.
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Thread starter 3 years ago
#50
(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. 0
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3 years ago
#51
No worries.
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Thread starter 3 years ago
#52
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://1a388b28f5ce9318f837a64962fc...%20Edexcel.pdf
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3 years ago
#53
(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://1a388b28f5ce9318f837a64962fc...%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
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Thread starter 3 years ago
#54
(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. 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.
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3 years ago
#55
(Original post by Glavien)
Hi, thanks the standard deviation makes sense now. 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).
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Thread starter 3 years ago
#56
(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! 0
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3 years ago
#57
(Original post by Glavien)
Thank you so much! No worries.
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Thread starter 3 years ago
#58
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 0
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3 years ago
#59
(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 Heh, this tripped everybody up in the exam - I managed it though. 0.75 includes the probability that the third person got a non-merit grade, i.e: gets a distinction or w/e it is.
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Thread starter 3 years ago
#60
(Original post by Zacken)
Heh, this tripped everybody up in the exam - I managed it though. 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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