The probability that more than 10500 passengers travel with buses in 72 hours 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.
 A: 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.
A: Following the hint they gave you...
During the period of  72 hours, the number of buses follows a poisson distribution with mean 720
$$X\sim Po(720)$$
The conditional random variable counting the total number of passengers is
$$Y|X=x=\begin{cases}
0.6,  & \text{if $y=10x$ } \\
0.2,  & \text{if $y=20x$ } \\
0.2,  & \text{if $y=30x$ } 
\end{cases}$$
with expectation $\mathbb{E}(Y|X=x)=16x$
thus the expectation of the total passengers in 72 hours is
$$\mathbb{E}(Y)=\mathbb{E}(\mathbb{E}(Y|X))=\mathbb{E}(16X)=16\times720=11520$$
with similar reasoning you can calulate $\mathbb{V}(Y)$ and the solve the problem using the Central Limit Theorem

To calculate the variance let's calculate $\mathbb{E}[Y^2]$ first
$$\mathbb{E}[Y^2]=\mathbb{E}[\mathbb{E}[Y^2|X]]=\mathbb{E}(320X^2)=320\cdot(720+720^2)=166,118,400$$
thus the variance is $\mathbb{V}[Y]=33,408,000$
to calculate the probability to have at least 10500 passengers you can use the normal approx
$$\mathbb{P}[Y>10500]=1-\Phi\left(\frac{10,500.5-11,520}{\sqrt{33,408,000}}  \right)=1-\Phi(-0.176)\approx 57.0\%$$
