What is the probability that more than 100 people will crash? Due to COVID19, we have not had classes at the university. The teacher has uploaded some slides to the virtual classroom and has sent us problems that are not related to the examples on the slides.
An airplane has a one in a million chance of crashing in flight. On each flight, on average, a number of people with a normal distribution of N (60; 20) travel. If 8,000 planes fly an average of 100 trips each, what is the probability that more than 100 people will crash?
Solution:
I tried this but... (is a disaster)
Be X the ramdom variable that count the number of people in one flight
Be Y the ramdom variable that count the number of people who have flown throughout the year, then 
Y be distributed normal N(2400000; 16000000)
Idon't know is this is correct. I on´t have any idea that how to continue.
 A: Let $$X_i \sim \operatorname{Normal}(\mu = 60, \sigma = 20)$$ be the random number of people on flight $i$, for $i \in \{1, \ldots, n\}$ where $n = 8 \times 10^5$.  For this scenario, we treat each flight as if it was a unique plane, rather than the same plane.  Let $p = 10^{-6}$ be the probability of flight $i$ crashing.  Then the number of crashed flights is approximately binomial with parameters $n$ and $p$.  Since $p$ is so small and $n$ so large, we can approximate this as a Poisson distribution with intensity $\lambda = np = 4/5$, with probability mass function $$\Pr[Y = y] = e^{-\lambda}\frac{\lambda^y}{y!}.$$  Given $Y$ crashes, the total number of involved passengers $T$ is normally distributed with mean $\mu = 60Y$ and variance $\sigma^2 = 400Y$.  Thus the probability of at least $100$ passengers experiencing a crash is 
$$\begin{align*}
\Pr[T \ge 100] &= \sum_{y=1}^\infty \Pr\left[\frac{T - 60y}{20\sqrt{y}} \ge \frac{100 - 60y}{20\sqrt{y}}\right]e^{-0.8} \frac{(0.8)^y}{y!} \\
&= e^{-0.8} \sum_{y=1}^\infty \left(1 - \Phi\left(\textstyle{\frac{5-3y}{\sqrt{y}}}\right)\right)\frac{(0.8)^y}{y!} \\
&\approx e^{-0.8} \sum_{y=1}^4 \left(1 - \Phi\left(\textstyle{\frac{5-3y}{\sqrt{y}}}\right)\right)\frac{(0.8)^y}{y!} + \Pr[Y \ge 5],
\end{align*}$$ where $\Phi$ is the CDF of the standard normal distribution.  For $y \ge 5$, the CDF becomes effectively $0$ and its complement $1$, so we approximate the tail using just the Poisson terms.  This gives us a estimated numeric probability of $$\Pr[T \ge 100] \approx 0.163099 + 0.00141131 \approx 0.16451.$$

I should point out that my reason for furnishing a complete solution in this case is because I feel this question is rather unfair.  It is much more sophisticated than it seems at first glance, and to get a reasonable estimate, one needs to understand more than a basic level of knowledge of the normal, binomial, and Poisson distributions--in particular, where to approximate, and where to compute things exactly.
A: First, the number of people cannot have this normal distribution because it has a significant chance of producing a negative number :-) And of course the number of people should always be an integer. 
The interesting thing is that one crash is very unlikely to have more than 100 people crashing. Two crashes have a high probability of 101 people crashing (rough guess 0.7), 3 crashes make it very, very likely. If you go to the number of people first, you’ll get very wrong results. The number of crashed people is absolutely not normal distributed. 
Calculate the probabilities for 0,  1, 2 etc crashes and calculate the probability of 101 people being on these flights. 
