Conditional Mean/Variance Help 
Suppose you are deciding between three different transportation options, each with an average traveling time of $\mu_j$ hours and a standard deviation $\sigma_j$ hours. You randomly choose between the three options with equal probabilities. Let $T$ be the length of your trip. $\newcommand{\Var}{\operatorname{Var}}$Find $\operatorname E(T)$ and $\Var(T)$.

So, your expected travel time is simply $\operatorname E(T) = (\mu_1 + \mu_2 + \mu_3) \cdot \frac{1}{3}.$
However, I'm a bit stuck on finding the variance. I understand you use the following formula, but don't know the interior values: $\Var(T) = \operatorname E(\Var(T\mid j)) + \Var(\operatorname E(T\mid j))$.
$\Var(T\mid j)$, I believe, should be $\sigma_1^2 + \sigma_2^2 + \sigma_3^2$, but am stuck moving forward. any help?
 A: Here, let $J$ be the random choice of transportation. Then, $\text{Var}(T \mid J)$ and $E[T \mid J]$ represent the variance and expectation of $T$ given you have chosen the $J$th transportation option and thus ought to depend on the value of $J$.
Can you compute these two "interior" values?

 $E[T \mid J=j] = \mu_j$ and $\text{Var}(T \mid J=j) = \sigma_j^2$.

Then, $E[\text{Var}(T \mid J)]$ is easy to compute, since it is $\frac{1}{3} \text{Var}[T \mid J=1] + \frac{1}{3} \text{Var}[T \mid J=2] +\frac{1}{3} \text{Var}[T \mid J=3]$.
For the variance term, just think of how you would compute the variance of a random variable that takes the values $E[T \mid J=1]$, $E[T \mid J=2]$, $E[T \mid J=3]$ with equal probability.
A: Note that $Var(T)=E(T^2)-(E(T))^2$ you already have $E(T)$, now note that $E(T^2)=E(E(T^2/j))$.
Use the fact that $E(T^2/j)=Var(T/j)+(E(T/j))^2$ and that should do it.
A: Let $\displaystyle J = \begin{cases} \mu_1 \\ \mu_2 & \text{each with probability } 1/3. \\ \mu_3 \end{cases}$
Then $\begin{cases} \operatorname{var(T\mid J=\mu_1)} = \sigma_1, \\ \operatorname{var(T\mid J=\mu_2)} = \sigma_2, \\ \operatorname{var(T\mid J=\mu_3)} = \sigma_3.  \end{cases}$
These each happen with probability $1/3,$ so $\operatorname{E}(T\mid J) = \begin{cases} \mu_1, \\ \mu_2, & \text{each with probability }1/3. \\  \mu_3. \end{cases}$
I.e. $\operatorname{E}(T\mid J) = J.$
Therefore $\operatorname{var}(\operatorname E(T\mid J)) = \operatorname{var}(J).$
Then we have $\begin{cases} \operatorname{var}(T\mid J=\mu_1) = \sigma_1, \\ \operatorname{var}(T\mid J=\mu_3) = \sigma_2, \\ \operatorname{var}(T\mid J=\mu_3) = \sigma_3, \end{cases}$ and so $\operatorname{E}(\operatorname{var}( T\mid J)) = (\sigma_1^2+\sigma_2^2+\sigma_3^2)/3.$
Thus we have
\begin{align}
\operatorname{var}(T) & = \operatorname{E}(\operatorname{var}(T\mid J)) + \operatorname{var}(\operatorname{E}(T\mid J)) \\[10pt]
& = \frac{\sigma_1^2+\sigma_2^2+\sigma_3^2} 3 + \operatorname{var} (J).
\end{align}
The first term is the "unexplained" component of the variance. The second term is "explained" by the differences among $\mu_1,\mu_2,\mu_3.$
