Assume that the time used for someone driving to work has a normal distribution with expected value $E(X) = 27$ and standard deviation $\sigma = 2.5$. Driving from work, we have the same distribution, but now with $E(X) = 31.5$ and $\sigma = 2.5$.
What is the probability that a given individual will use more than a total of $61.5$ minutes going to and from work on a given day?
(Time used to and from work on a given day and time used to and from work on different days are indepedent).
I'm confused about how I'm supposed to combine these two different distributions. I tried using $E(X) = 27-31.5 = -4.5$ and $\sigma = \sqrt{2.5^2+2.5^2} = 3.5355$ but this doesn't make sense when I want to find $P(X>61.5$).
Any ideas?
Let $X$ denote the time driving to work and $Y$ the time going back home.
Then $Z:=X+Y$ is the total, and has normal distribution with expectation: $$\mathbb EZ=\mathbb E(X+Y)=\mathbb EX+\mathbb EY=27+31.5$$
If moreover $X$ and $Y$ are independent then:$$\mathsf{Var}(Z)=\mathsf{Var}(X)+\mathsf{Var}(Y)=2.5^2+2.5^2$$
The distribution of $Z$ is determined by expectation and variance, so this together enables you to find $P(Z>61.5)$.