There are two trains competing for passenger traffic of 1000 people going from Chicago to L.A. Assume that passengers are equally likely to choose each train. How many seats must a train have to assure a probability of .99 or better of having a seat for each passenger?
Since you've dealt with the central limit theorem and that trying to find the above x will be a real pain, I suggest you approximate it by a normal distribution.
Let Xi be a rv associated to each of the passengers (i is going from 1 to 1000).
Let's say Xi=1 if the passenger goes to train 1. Xi=0 otherwise (and let's consider that almost surely, none of them will fall into the water )
Xi follows a Bernoulli distribution, with parameter 1/2.
E(Xi)=1/2 and Var(Xi)=1/4
The conditions are sufficient to approximate the binomial distribution by a normal distribution.
So finally, we have, by the CLT :
Where Z follows a normal standard distribution.
And (in other words, the number of passengers that will want a seat on train 1)
This gives :
now use a z-table to find x such that is the nearest to 0.01
And will be the value you're looking for.