1. ## Train Problem

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?

2. Originally Posted by morganfor
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?
Let X be the random variable number of passengers wanting a seat in a train.

X ~ Binomial(n = 1000, p = 1/2).

Your job is to find the smallest value of x such that $\displaystyle \Pr (X > x ) \leq 0.01$.

Trial and error using technology is the most efficient approach.

3. Hello,

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 :

$\displaystyle \mathbb{P}\left(\frac{Y-n/2}{\sqrt{n}/4}>x\right) \approx \mathbb{P}(Z>x)$

Where Z follows a normal standard distribution.
And $\displaystyle Y=X_1+X_2+\dots+X_n$ (in other words, the number of passengers that will want a seat on train 1)
And n=1000

This gives :

$\displaystyle \mathbb{P}\left(Y>\frac{\sqrt{1000} x}{4}+500\right)\approx \mathbb{P}(Z>x)$

now use a z-table to find x such that $\displaystyle \mathbb{P}(Z>x)$ is the nearest to 0.01
And $\displaystyle \frac{\sqrt{1000} x}{4}+500$ will be the value you're looking for.