10 000 people are invited to an event, with probability $\frac{4}{5}$ of showing up. What is the minimum amount of chairs needed to promise that for a probability of at least 0.95 that all people that come will have a place to sit?
Ok so we can define:
X=number of people that show up
$X$~$Bin(10 000, 0.8)$
And we can use Laplace-De Moivre theorem and approximate it to a normal distribution
~N(8000, 40)
From here, I have difficulties understanding the solution manual.
They state that: $Pr(x \lt C)\ge 0.95$
Correct me if I'm wrong, this means that the probability that less than C people show up, has to be greater than 0.95. I'm having troubles wrapping my head around this and would appreciate it if someone could mayb elaborate.
If the question is confusing let me know and I'll do my best to reword it better.
In order to provide enough chairs for everyone with probability $0.95$, you need to provide at least as many chairs than the number of people that you would expect to arrive in $95\%$ of cases. That is, you need precisely $$P(x\leq C) \geq 0.95$$ since otherwise there is more than a $5\%$ chance that more people arrive than there are chairs!
Note that there is one caveat: if we replace $0.95$ with something much closer to $100%$, say $1-10^{-5}$, you might end up with more than $10,000$ chairs being needed, which is incorrect. This is an example of how the Laplace-De Moivre theorem can fail for extreme tail probabilities, but with these numbers the power of the theorem is pretty clear.
If you were really planning an event like this, though, I would advise you to book a hall with fixed capacity instead. Handling over $8000$ chairs sounds scary!