EDEXCEL Statistics 3 Question Random Variables

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)

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)

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