I am having a problem understanding why the below is calculating the variance when the exercise is asking about the standard deviation.
The standard deviation of the time you take for each statistics homework problem is 1.5 minutes, and it is 2 minutes for each chemistry problem. What is the standard deviation of the time you expect to spend on statistics and physics homework for the week if you have 5 statistics and 4 chemistry homework problems assigned?
Variability of a linear combination of two independent random variables:
V(aX + bY) = $a^2$ ×V(X) + $b^2$ ×V(Y)
The standard deviation of the linear combination is the square root of the variance.
So we will use the above formula to calculate the exercise:
V(5S + 4C) = $5^2$ × V(S) + $4^2$ × V(C)
= 25 × $1.5^2$ + 16 × $2^2$
= 56.25 + 64
= 120.25
In my mind we are clearly finding the variance here, not the standard deviation, as that would be the sqrt of the 120.25, still the answer is 120.25, why?
Edit: Also, the formula says that $a^2$ and $b^2$ but still they do power of two on both 5 and 4 and V(S) and V(C).
As you have correctly determined, the answer to the question as written is the square root of the variance you have calculated. Your reasoning is correct.
It would seem that you are right and the answer book is wrong.
$$ \sigma_S = 1.5, \sigma_C = 2 $$ Therefore $$ V(S) = (1.5)^2, V(C) = 2^2 $$
Scaling a random variable by $a$ scales the variance by $a^2$, so $V(aS) = a^2V(S)$