# [Math] Question on linear combination of random variables

probabilitystatistics

I am looking at this question and I am still confused after reading the accepted answer.
This is the question:

A company ships gift baskets that contain apples and pears. The distributions of weight for the apples, the pears, and the baskets are each approximately normal. The mean and standard deviation for each distribution is shown in the table below. The weights of the items are assumed to be independent.

Apples: Mean=4.72 oz, sigma=0.2 oz

Pears: Mean=5.41 oz, sigma=0.18 oz

Let the random variable W represent the total weight of 4 apples, 6 pears, and 1 basket. Which of the following is closest to the standard deviation of W?

(A) 1.90 ounces

(B) 1.97 ounces

(C) 2.26 ounces

(D) 3.76 ounces

(E) 3.83 ounces

Specifically, I was looking at these notes:

It's basically the same question but the method that the questioner used in the hyperlinked question is the same method used in these notes I found.

$$\newcommand{\Var}{\operatorname{Var}}$$Let $$X$$, $$Y$$, and $$Z$$ be the random weights of an apple, pear, and basket respectively.
If I understand correctly, you think that the weight of four apples should be $$4X$$. That is not the case. This like saying "Randomly generate the value of $$X$$ and then multiply that by four. That is the weight of the four apples". Do you see how that is not what we want? What we actually want is "Randomly generate four weights independently, $$X_1$$, $$X_2$$, $$X_3$$, and $$X_4$$. The sum of these is the weight of the four apples (combined)."
Notice that if we take the variance of the $$4X$$, we get $$\Var(4X) = 4^2\Var(X) = 16(0.2)^2 = 0.64$$ whereas $$\Var(X_1+\dotsb+X_4) = \Var(X_1)+\dotsb+\Var(X_4) = 4\Var(X_1) = 4(0.2)^2 = 0.16,$$ where the second equality is true by independence, and we notice that $$\Var(X_1) = \dotsb = \Var(X_4)$$ for the second equality because the $$X_i$$ follow the same distribution.
All together, the variance of the weight will be $$\Var(W) = \Var(X_1+\dotsb + X_4 + Y_1+\dotsb+Y_6+ Z_1),$$
which is not $$\Var(4X+6Y+Z)$$, and I'll leave you to work out the details.