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.

$$ \begin{array}{|c|c|c|} \hline \hskip{2em} \text{Item}\hskip{2em} & \hskip{2em}\text{Mean}\hskip{2em} & \hskip{2em}\text{Standard Deviation}\hskip{2em} \\ \hline \text{Apple} & 4.72 \text{ ounces} & 0.20 \text{ ounces} \\ \hline \text{Pear} & 5.41 \text{ ounces} & 0.18 \text{ ounces} \\ \hline \text{Basket} & 13.25 \text{ ounces} & 1.88 \text{ ounces} \\ \hline \end{array} $$

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 \(W\) ?

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To obtain the standard deviation of \(W\), we can take the approach of adding the sample variances (\(s^2\)) of each item. We can square root the combined sample variance to obtain the standard deviation of \(W\).

$$ W = 4A + 6P + B $$ $$ W = A + A + A + A + P + P + P + P + P + P + B $$ $$ s_W^2 = 4(s_A)^2+ 6(s_P)^2 + s_B^2 $$ $$ s_W^2 = 4(0.20)^2 + 6(0.18)^2+1.88^2 $$ $$ s_W^2 = 3.8888 $$ $$ s_W = \sqrt{3.8888} $$ $$ s_W \approx \boxed{1.972} $$