Finding the variance of Linear combinations Here are two questions of similar style from my past CIE A level exams, 

Now I am unsure how to find the variance in each case, 
If X and Y are independent random variables, the variance of $aX+bY$, where a and b are constants, is $a^{2}Var(X)+b^{2}Var(Y)$ right?
So for the first question, $Var(F)$ should be $Var(F)=5.6^2 +\frac{1}{2}^2\cdot 12.4^2 $,
which is consistent with the answers provided. 
But for the second question, to find the required probability we must first evaluate $E(5W+5H)$ and $Var(5W+5H)$ 
So 
$Var(5W+5H) = 5^2 Var(W) + 5^2 Var(H)$
subsituting given values,
$Var(5W+5H) = 5^2 \cdot 4.8^2 + 5^2 \cdot 5.2^2 = 1252$,
but that is inconsistent with the answers, which state $ Var = 5(4.8^2 + 5.2^2) = 250.4$
I cant seem to understand where I am going wrong in the second question, isn't that question identical to the first one, so shouldn't the variance be calculated using $a^{2}Var(X)+b^{2}Var(Y)$
Please help, I am really confused here...  
 A: For the second question you have ten independent random variables as each day's journeys could be different. 
So you are looking for $$Var(W_1+W_2+W_3+W_4+W_5+H_1+H_2+H_3+H_4+H_5)$$ $$=Var(W_1)+Var(W_2)+Var(W_3)+Var(W_4)+Var(W_5)$$ $$+Var(H_1)+Var(H_2)+Var(H_3)+Var(H_4)+Var(H_5)$$ $$=5Var(W)+5Var(H)$$
A: The difference in the two questions is that in the first question, a single candidate's total score is a linear combination of two individual scores; but in the second question, the total traveling time over a five-day period is the sum of the traveling time for each day for five days, and these values in general will be different from day to day.  In other words, $5(W+H)$ represents a random variable that is equal to five times the total time of a single day.  But this isn't what is asked; it is asking for the total time for five different days.
Here is a simpler example.  If we suppose that the heights of a particular variety of tomato plant are normally distributed with mean $\mu = 50$ and standard deviation $\sigma = 4.5$, and I have 100 such plants in my greenhouse, then the total height of those 100 plants is normally distributed with mean $100 \mu = 5000$ and standard deviation $\sqrt{100} \sigma = 45$.  But that is not the same as measuring a single tomato plant and multiplying its measurement by $100$.  In this latter case, the mean is still $100\mu$ but the standard deviation is $100\sigma = 450$.
