TITLE: The probability that more than 10500 passengers travel with buses in 72 hours
QUESTION [5 upvotes]: Question:
Buses go from a terminal to a destination city with a rate of $10$ bus per hour. The number of passengers on each bus is independent of the other buses and assumed to follow this distribution: $10$ passengers with a probability of $0.6$, $20$ passengers with a probability of $0.2$, and $30$ passengers with a probability of $0.2$.
What is the probability that in $72$ hours, more than $10500$ passengers reach the destination using the buses in this terminal?
(Hint: you should use normal approximation)

Note: I've seen a lot of similar questions. However, this one asks for the probability of something related to number of passengers, and not a probability which is related to the waiting time of the buses. My problem is that I cannot even understand what random variable we are looking for. For instance, can I conclude that in $72$ hours, $720$ buses will pass? If yes, then what? How should I proceed? I mean, first of all I need somebody to rewrite the question in a  mathematical way.

REPLY [2 votes]: Here is how you would work through this using normal distribution.
Step1: create Expected Value Table for numbers of passengers in a bus (number of passengers in buses are independent).
$\begin {array}{|c|c|c|c|}
\hline
X = x & p(x) & xp(x)&(x-\mu)^2 p(x) \\ \hline
10&0.6&6& \\ \hline
20&0.2&4& \\ \hline
30&0.2&6& \\ \hline
&&16& \\ \hline
\end{array}$
$\mathbb{E}(X) = 16$.
The mean, $\mu$, of a discrete probability function is the expected value.
So, $\mu = 16$.
Step2: now using the value of $\mu$, fill the last column in above table and that gives you the variance. I will leave this as an exercise for you but it comes to,
$\sigma^2 = 64 \implies \sigma = 8, \ $ where $\sigma$ is the standard deviation.
Step3: find standard or z-score
We are asked the probability of $720$ bus carrying more than $10500$ passengers which on an average $ \gt \frac{10500}{720} = \frac{175}{12}$
$z = \frac{x-\mu}{\sigma} = \frac{175/12 - 16}{8} = - \frac{17}{96}$
Step4: Read the z-table
Given that $z \approx -0.18$, read the row which says $0.1$ and the column which says $0.08$ and you get a value of $0.0714$.
Now note that $z$ is left of $z=0$ and as we have to find probability of $x \gt \frac{175}{12}$, the desired probability is
$ \approx 0.5 + 0.0714 = 0.5714$
Alternatively you can use an online tool like WolframAlpha for evaluating the below integral:
$ \displaystyle \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-17/96} e^{-z^2/2} dz \approx 0.4297$
This is the probability of buses carrying less than $10500$ passengers. So subtracting from $1$ gives you the desired probability.