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    Page 3 in edexcel book, top irght.

    If X1 + X2 + ... + X9 + Y1 +Y2 + ... + Y5 distrubuted normally (9u1 + 5u2, 9(p1^2) + 5(p2)^6

    Where u = mu, (mean) and p(sigma or delta?) the standard deviation squared.

    My question is why is it not 9^2 and 5^2 I thought that you had to square the coefficients?

    Thanks anyone?
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    (Original post by Moordland)
    Page 3 in edexcel book, top irght.

    If X1 + 2 + ... + X9 + Y1 +Y2 + ... + Y5 distrubuted normally (9u1 + 5u2, 9(p1^2) + 5(p2)^6

    Where u = mu, (mean) and p(sigma or delta?) the standard deviation squared.

    My question is why is it not 9^2 and 5^2 I thought that you had to square the coefficients?

    Thanks anyone?
    You only square the coefficients when it's a multiple of the same random variable.

    So 2X1 would have a var of 4 times var(X1) but X1 + X2 has variance var(X1) + var(X2)
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    (Original post by davros)
    You only square the coefficients when it's a multiple of the same random variable.

    So 2X1 would have a var of 4 times var(X1) but X1 + X2 has variance var(X1) + var(X2)
    So would it have to be 5X1 + 5X2 + ... 5X9 for the variance to be 25(Varx1) like that?

    Or actually just being multiplied instead of added?

    Thanks for the reply
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    (Original post by Moordland)
    So would it have to be 5X1 + 5X2 + ... 5X9 for the variance to be 25(Varx1) like that?

    Or actually just being multiplied instead of added?

    Thanks for the reply
    The variance of that would be 25Var(X1) + 25Var(X2) + ... + 25Var(X9)

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    (Original post by davros)
    The variance of that would be 25Var(X1) + 25Var(X2) + ... + 25Var(X9)

    Yes that makes sense thanks.

    What I still don't get is that X and Y are of different distributions from the OP. So under what conditions would it have to be squared.

    Sorry to keep asking self-teaching with this ambivalent book is annoying at times!
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    (Original post by Moordland)
    Yes that makes sense thanks.

    What I still don't get is that X and Y are of different distributions from the OP. So under what conditions would it have to be squared.

    Sorry to keep asking self-teaching with this ambivalent book is annoying at times!
    If I'm understanding your question correctly, it's only squared when you're measuring a random variable that is defined as a multiple of another one, so if Y = aX where a is a constant, then

    Var(Y) = Var(aX) = (a^2)Var(X)

    However, if you take 2 measurements X and Y, then Var(X + Y) = Var X + Var Y
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    (Original post by davros)
    If I'm understanding your question correctly, it's only squared when you're measuring a random variable that is defined as a multiple of another one, so if Y = aX where a is a constant, then

    Var(Y) = Var(aX) = (a^2)Var(X)

    However, if you take 2 measurements X and Y, then Var(X + Y) = Var X + Var Y
    Understood thank you for your perfect explanation.
 
 
 
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