Probability and combining random variables Adrian cycles to school with a mean time of 20 minutes and a standard deviation of 5 minutes. Pamela walks to school with a mean time of 30 minutes and a standard deviation of 2 minutes. They each calculate the total time it takes them to get to school over a five-day week. What is the expected mean and standard deviation of the difference in the total weekly journey times, assuming journey times are independent? 
Can someone help with this probability question please? 
 A: Let $X_i$ be the time that it takes Adrian to go to school on the $i$-th day of the week, for $i=1,2,3,4,5$. Then, the total weekly journey time $X$ of Adrian can be expressed as $$X=X_1+X_2+X_3+X_4+X_5$$ Due to linearity of expectation $$E[X]=E[X_1]+E[X_2]+E[X_3]+E[X_4]+E[X_5]=20+\ldots+20=5\cdot20=100$$ Due to independence of journey times $$\begin{align*}Var(X)&=Var(X_1)+Var(X_2)+Var(X_3)+Var(X_4)+Var(X_5)=\\&=5^2+\ldots+5^2=5\cdot5^2=125\end{align*}$$
where $Var(X)$ denotes the variance of $X$. The standard deviation is the (positive) root of the variance $$σ_X=\sqrt{Var(X)}=\sqrt{125}$$ (Note that independence is necessary to use this formula for the variance but is not needed to use the formula for the expectation).
Similarly let $Y_i$ be the time that it takes Pamela to go to school on the $i$-th day of the week, for $i=1,2,3,4,5$. Then, the total weekly journey time $Y$ of Pamela can be expressed as $$Y=Y_1+Y_2+Y_3+Y_4+Y_5$$ and as above  $$E[X]=E[Y_1]+E[Y_2]+E[Y_3]+E[Y_4]+E[Y_5]=5\cdot30=150$$ and  $$Var(Y)=Var(Y_1)+Var(Y_2)+Var(Y_3)+Var(Y_4)+Var(Y_5)=5\cdot2^2=20$$
and therefore $$σ_Y=\sqrt{Var(Y)}=\sqrt{20}$$ 
Now that you have the expected times and standard deviations of the individual total journey times $X$ of Adrian and $Y$ of Pamela, you can proceed to answer the actual question of the exercise which is to calculate the expected value and the standard deviation of their difference $X-Y$, i.e. $$E[X-Y] \quad \text{ and } \quad σ_{X-Y}=\sqrt{Var(X-Y)}$$ (note that the choice $X-Y$ instead of $Y-X$ is not specified in the exercise. So if you take $Y-X$ the sign of the expectation will be exactly the opposite but the variance will be the same.) Now, using the properties of the expectation and the variance for independent random variables you obtain as above that $$E[X-Y]=E[X]-E[Y]=100-150=-50$$ and $$Var(X-Y)=Var(X)\color{blue}{+}Var(Y)=125+20=145$$ and therefore $$σ_{X-Y}=\sqrt{145}$$ (note that the "-" becomes a "+" when calculating the variance!) 
