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    (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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    (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.
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    (Original post by Glavien)
    I still don't seem to be getting the correct answer of 0.3844.
    I get it.

    0.1^2 + 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
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    (Original post by Zacken)
    I get it.

    0.1^2 + 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!!
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    (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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    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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    (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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    I thought because it's a uniform distribution you could use that formula. Name:  ImageUploadedByStudent Room1465497031.659425.jpg
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    (Original post by Glavien)
    I thought because it's a uniform distribution you could use that formula. Name:  ImageUploadedByStudent Room1465497031.659425.jpg
Views: 86
Size:  150.8 KB


    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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    (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.
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    No worries.
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    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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    (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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    (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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    (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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    (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!
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    (Original post by Glavien)
    Thank you so much!
    No worries.
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    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

    Name:  ImageUploadedByStudent Room1465924620.325705.jpg
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    (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

    Name:  ImageUploadedByStudent Room1465924620.325705.jpg
Views: 77
Size:  167.0 KB
    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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    (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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