Suppose you are out fishing and you want to predict the amount of time (in minutes) it takes to catch 3 fish. Let T_F~N(15; 10) be the time it takes to reel in a fish. Let T_T~ N(10; 8) be the amount of time you spend waiting for the fish to take the bait. Assume TF and TT are independent. Also, assume you start timing as soon as you start reeling in the first fish, and the time ends when you have finished catching the third fish. What is the probability that you finish catching the three fish in 60 minutes or less?
I was able to find E(T)=25 and V(T)=18 for the total time. But I am having trouble solving for the Probability(T<=60 mins).
Let's $T_F^1$, $T_F^2$ and $T_F^3$ the time it takes to reel in the first, second and third fish respectively. Lets also $T_T^2$, $T_T^3$ the amount of time you spend waiting for the second and the third fish to take the bait respectively. The total amount of time will be $$T = T_F^1 + T_T^2 + T_F^2 + T_T^3 + T_F^3$$ As those five variables are independents, $T\sim \mathcal{N}(65,46)$.
Now we compute the probability by $$P(T\leq 60) = P\left(\frac{T-65}{\sqrt{46}}\leq \frac{60-65}{\sqrt{46}}\right) = P(Z\leq-0.7372098) = \Phi(-0.7372098) = 0.2304974\,,$$ where $Z\sim\mathcal{N}(0,1)$ and $\Phi$ is the cumulative distribution function of a standard normal.