So i have the following question:
The times that patients spend in a doctor’s surgery have mean 5 minutes, and standard deviation 2 minutes. On one particular day, the doctor sees 30 patients during his surgery which starts at 4.30pm. Find an approximate probability that she finishes with her last patient before 6.50pm. State clearly any assumptions you need to make.
I have so far got to the following: Let X be the time (in minutes) spent at the surgery. From the question we have $\text{E}(X) = 5$ and $\text{Var}(X) = 4$. Using a random samply 30 from $X_1, X_2, \ldots , X_{30}$.
Let $$Y = \sum_{i=1}^{30} X_i$$ By the Central Limit Theorem $Y ∼ N(\mu = 150, \sigma^2 = 120)$ approximately. As surgery starts at 4 : 30, and the doctor desires to finish surgery at 6 : 50, we desire $P(Y < 140)$
However I am not sure where to go from here? In the answer it says $P(Y < 140)=0.1807$ could someone explain how this was calculated? Thanks
$P(Y<140)=P(Z<\frac{140-150}{\sqrt{120}}$)