AP Statistics 2.11: Normal Mean and SD

Let W represent the weight of a randomly selected full carton of eggs, P the weight of the packaging, and $X_{i}$ the weight of the i th egg, for $i=1,2 \ldots, 12$.
Note that $W=P+X_{1}+X_{2}+\ldots+X_{12}$.
Properties of expected values establish that $\mathrm{E}(W)=\mathrm{E}(P)+\mathrm{E}\left(X_{1}\right)+\ldots+\mathrm{E}\left(X_{12}\right)$.
Because all 12 eggs have the same mean weight, this becomes $\mathrm{E}(W)=\mathrm{E}(P)+12 \times \mathrm{E}\left(X_{i}\right)$.
We were told that E(W)=840 and E(P)=20, so we can solve

Because of independence, properties of variance establish that
Var(W)=Var(P)+Var(X_1)+Var(X_2)++Var(X_12).
Because all 12 eggs have the same variance of their weights, this becomes
Var(W)=Var(P)+12 x Var(X_i).
We were told that $\operatorname{SD}(W)=7.9$ and $\operatorname{SD}(P)=1.7$. Therefore, $\operatorname{Var}(W)=(7.9)^{2}=62.41$ and $\operatorname{Var}(P)=(1.7)^{2}=2.89$.
We can solve $62.41=2.89+12 \times \operatorname{Var}\left(X_{i}\right)$ to find $\operatorname{Var}\left(X_{i}\right)=\frac{62.41-2.89}{12}=4.96$. Thus, $\operatorname{SD}\left(X_{i}\right)=\sqrt{(4.96)} \approx 2.23$ grams.