Statistics Question

[dooyeon1998]

Hello This is a question from Neap 2016 Trial exam

The weights of the raspberries produced by a farm may be assumed to be normally distributed with mean 10 grams and standard deviation 1.5 grams. 12 of these raspberries are selected at random and put into a container. The probability that the combined weight of the 12 raspberries is greater than 130 grams, correct to four decimal places, is

{Answers}
Let W be the total weight of the 12 raspberries.E(W) = 12 × 10 = 120 and var(W) = 12 × 1.52 So W ~ N(120, 12 × 1.52) and Pr(W > 130) = 0.0271


What I was wondering is that if we let X=weight of a raspberry. Then W = 12X. Therefore E(w)=12x10 and Var (W)= Var(12X) = 12^2Var(x) =  (12^2) x (1.5^2).  So I am keep on getting wrong variance. Where did I go wrong with my working out?

Thank you

[shlblk]

You ended up choosing E for this question right?
I've done this paper and I'm pretty sure that Neap has made an error.

sd(X) = 1.5
Hence Var(X) = 1.5²
Then...
Var(12X) = 12²Var(X)...which leads to the answer we got.

It seems Neap forgot to square the 12.

[adnauseam]

Yo
I think the answers are correct.

Since we're dealing with the combined weights of 12 things, not 12 times the weight of one thing, wouldn't Variance = 12 x 1.5^2?
So i think making W=12X would be wrong in this case.
Please let me know if im wrong though, i'm not the best in this area.
Cheers

[reachfortheatars]

Yeah, I asked my teacher about this - as adnauseum said, it's not 12X, but rather (X1 + X2 + X3... + X12), with each variable having the same mean (10) and variance (1.5), so the variance would be (12*1)Var(X) = 12Var(X) = 12*(1.5^2)