The question is as follows:
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
Basket: Mean=13.52 oz, sigma=1.88 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
It appears this is a linear combination of $W=4apples + 6pears + 1basket$
The formula in our textbook (and online) for the standard deviation of linear combinations is:
$\sigma_W=\sqrt{a^{2}\sigma_{1}^{2}+b^{2}\sigma_{1}^{2}…}$
so naturally I tried $\sigma_W=\sqrt{4^{2}\left(0.2\right)^2+6^{2}\left(0.18\right)^{2} + 1^{2}\left(1.88\right)^{2}}=2.311$
However, the correct answer is (B) 1.97 ounces.
I can get that value by using $\sigma_W=\sqrt{4\left(0.2\right)^2+6\left(0.18\right)^{2} + 1\left(1.88\right)^{2}}=1.972$
My question is why don't I need to square the coefficients?
Best Answer
The key point is that you have 11 independent random variables. Let us focus on the 4 apples and denote the corresponding random variables for the weights as $X_1,X_2,X_3$ and $X_4$. The assumption is formally
$P(X_i|X_j)=P(X_i) \ \ \forall \ i,j=\{1,2,3,4 \}, i\neq j$
If you have $n$ independent distributed random variables then the variance of the sum of the variable is equal to the sum of the variances:
$$Var\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n Var\left( X_i \right)$$
In case of the 4 apples we have
$Var\left( X_1+X_2+X_3+X_4\right)=Var(X_1)+Var(X_2)+Var(X_3)+Var(X_4)$
$=0.2^2+0.2^2+0.2^2+0.2^2=4\cdot 0.2^2=4\cdot 0.04=0.16$
It would be another case if you would calulate the variance of the four times weight of a single apple.
$Var\left(4X_1\right)=16\cdot Var\left(X_1\right)=0.64$