How do you identify whether it's a sample total I'm struggling with the following question: A company packs parcels of books. Suppose the weights of hardback books are normally distributed with mean 2kg and standard deviation 0.5 kg and paper- back books are normally distributed with mean 0.8kg and standard deviation 0.3kg. Assume the weights are all independent of each other. A parcel has three hardback books and six paperback books. Find the probability that the parcel weighs more than 11kg. Here is my solution  = 3E(X) + 6E(Y) = 10.8 $ and $Var(3X+6Y) = 3^2Var(X) + 6^2Var(Y) = 5.49$ so the distribution will be $ 3X + 6Y $ ~ $ N(10.8, 5.49)$. Which is the right distribution and why? In the above picture i used the Total Sample Theorem which states that $E(T) = nu$ and $Var(T) = n*sigma^2 $ where $ T = X1 + X2 + X3 ... Xn $

Let $Z=3X+6Y$. Then $E(Z)=3E(X)+6E(Y)$

It can be proven that $Var(Z)=Var(aX+bY)=a^2Var(Y)+2abCov(X,Y)+b^2Var(Y)$.

In your case $X,Y$ are independent. Thus $Cov(X,Y)=0$.

Consequently $$Z\sim \mathcal N (3E(X)+6E(Y),a^2Var(Y)+b^2Var(Y))=\mathcal N(11,5.45)$$

Then you have

$$P(Z>11)=1-P(Z\leq 11)=1-\Phi\left( \frac{11-3\mu_x-6\mu_y}{\sqrt{9Var(X)+36Var(Y)}} \right)$$

$$=1-\Phi\left( \frac{11-10.8}{\sqrt{9\cdot 0.5^2+36\cdot 0.3^2}} \right)$$

$$=1-\Phi\left( \frac{0.2}{2.34307} \right)=1-\Phi(0.0854)$$