How many tickets can you sell for a plane?

Question

I'm trying to learn about limit theorems, but I have no idea how to calculate the following question:

Suppose that 15% of people don’t show up for a flight, and suppose that
their decisions are independent. How many tickets can you sell for a plane with
144 seats and be 99% sure that not too many people will show up?

This is what I got so far:

First I took 15 % of 144, and add that to the 144. $$ 144 + 21.6 = 165.6 $$

The problem is that I have no idea how to know calculate the 99%.

The book says the answer of this question is: 157

Answer 1

I think you're best off using the normal approximation to the binomial distribution. Otherwise, the inequality that you have to address is, as avid19 drily puts it, non-trivial.

Suppose that you sell $n$ tickets. You can then expect $0.85n$ of the ticketed passengers to show up. However, that is merely the mean of a distribution that is approximately normal. The standard deviation is given by

$$ \sigma = \sqrt{(0.85)(0.15)n} \doteq 0.357 \sqrt{n} $$

To guarantee that you are $99$ percent likely not to have too many passengers show up, you are looking for the one-sided $z$-test (because you don't care if too few people show up) at $99$ percent, which yields $z = 2.33$. That is,

$$ 0.85n + 2.33(0.357\sqrt{n}) \leq 144 $$

I trust you can take it from here on in.

Answer 2

If you sell $n$ tickets, what's the probability that too many people show up? Well, if $n \leq 144$ it's $0$. If $n=145$, the probability that too many people show up is $.85^{145}$ which is needless to say, tiny.

In general this is a binomial distribution. Meaning that if you sell $n$ tickets, the probability that $k$ people will show up is ${n \choose k} .85^k.15^{n-k}$.

So if you sell $n$ tickets, what's the probability that at most $144$ people arrive?

It is:

$$\sum\limits_{k=0}^{144}{n \choose k} .85^k.15^{n-k}$$

So you just need to solve the (nontrivial) inequality:

$$\sum\limits_{k=0}^{144}{n \choose k} .85^k.15^{n-k}\leq .99$$

for $n$.

You may write this in the perhaps more simplified way:

$$\sum\limits_{k=145}^{n}{n \choose k} .85^k.15^{n-k}\gt .01$$

Plug in some numbers to see what the probabilities are.

Answer 3

For each $n\in\mathbb{N}$ let $X_n \sim \mathrm{Binom}(n,0.85)$. Then each realization of $X_n$ will be the number of successes of $n$ trials, where each trial have a probability of 0.85 of succeeding. Your question reduces to finding the largest $n\in\mathbb{N}$ such that $P(X_n\leq 144)\geq 0.99$. We know the probability density function of the binomial distribution, it is: $$P(X_n=k)=p(k,n,0.85)= \frac{n!}{k!(n-k)!}0.85^k(1-0.85)^{n-k} \quad \quad \text{ for } k=0,...,n$$ Thus the probability of $(X_n\leq 144)$ is given by $$ P(X_n\leq 144) = \sum_{k=0}^{144} \frac{n!}{k!(n-k)!}0.85^k(1-0.85)^{n-k}. $$ Using R i've found out that \begin{align*} P(X_{156} \leq 144) &=0.9980396, \\ P(X_{157} \leq 144) &=0.9959136, \\ P(X_{158} \leq 144) &=0.9920622, \\ P(X_{159} \leq 144) &=0.9855424. \end{align*} Thus the largest integer $n\in\mathbb{N}$ such that $P(X_n\leq 144) \geq 0.99$ is $158$.