I'm working on the following exercise:
Suppose that 18% of people who reserved a seat on a plane don't actually take the flight. Suppose that the passengers take the flight independently and that the plane has 220 available seats. There are more reservations than seats, so that less seats remain unused.
How many reservations can be made at most, so that the probability of every passenger getting a seat is $ \geq 99 $%?
We have to solve this with the De Moivre-Laplace theorem. I defined $ X_n $ as the number of boarding passengers and $ n $ as the wanted number of reservations. I wanted to approximate $ P(0\leq X_n\leq 220)\geq 0.99 $ with the standard normal distribution and solve for $ n $. I got $ P(0\leq X_n\leq 220)\approx\frac{1}{\sqrt{2\pi}}\int_{x_2}^{x_1}e^{-\frac{t^2}{2}}dt $, where $ x_1=\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}} $ and $ x_2=\frac{-0.82n}{\sqrt{n\cdot0.82\cdot0.18}} $. I'm not quite sure about that, though.
But my question would be: How to solve for $ n $? I got the hint that I can invert the distribution function by searching the standard distribution table for a value near 0.99. But I'm lost on how exactly to use this.
Thanks in advance for any help!
I´ve seen that you have made some progress. Nevertheless I post my thoughts. You can use the function for the cumulative distribution of the standard normal, which is $\Phi(z)$. See here the table. But first of all you have $P(X\leq 220)=\Phi\left(\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}} \right)\geq 0.99$
Now you have to use the inverse function.
$\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}} \geq \Phi^{-1}\left(0.99\right)$
You can now use the table at look at which value of z you have a probability of 0.99 at the table. This is approximately at $z=2.33$. I don´t know where you have read off $z=2.3$.
Or you make an equation with the integral: $\int\limits_{-\infty}^z \frac1{\sqrt{2\cdot \pi}}\cdot e^{\frac12x^2} \, dx=0.99$. With wolfram alpha I get $z=2.32634$.
So the inequality is $\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}} \geq 2.32634$
The result is