I have the joint distribution. This is how many packages of toilet paper, $\mathbf{T}$, and packages of wipes, $\mathbf{W}$, a shopper can obtain per visit.
I'm being tasked to find the standard deviation of the packages of wipes purchased, given that a shopper has bought exactly one package of toilet paper.
This is my current work. I found the conditional expectation to act as my $E[X]$, as well as applied the same idea for $E[X^2]$. I used these values to find $Var(X)$, and then square rooted for the standard deviation.
\begin{equation} \begin{split} E[W]=E[W|T=1] &= \sum_{i=1}^m a_i\cdot P(W|T=1)\\ &= 0\cdot .22 + 1\cdot .05 + 2\cdot .03\\ E[W] &= .11 \end{split} \end{equation}
\begin{equation} \begin{split} E[W^2]=E[W^2|T=1] &= \sum_{i=1}^m a_i\cdot P(W^2|T=1)\\ &= 0^2\cdot .22 + 1^2\cdot .05 + 2^2\cdot .03\\ E[W^2] &= .17 \end{split} \end{equation}
Now,
\begin{equation} \begin{split} Var(W) & = E[W^2] - E[W]^2\\ &= .17 - .11^2\\ Var(W) &= .1597 \end{split} \end{equation}
By this logic, my standard deviation should be $\sigma = \sqrt{Var(W)} = \boxed{.3973}$. My answer key however says $\sigma = .6574$