Question: An airline finds that 7% of the people who make reservations on a certain flight do not show up for the flight. If the airline sells 185 tickets for a flight with only 180 seats, use the normal approximation to the binomial distribution to find the probability that a seat will be available for every person holding a reservation and planning to fly. (Round your answer to four decimal places.)
I am frustrated at this problem. I need to use a normal approximation to a binomial distribution.
My attempt starts by letting $n=185$ for the total number of tickets sold. Then, I let $p=0.93$ which is the probability of people showing up which leaves $q=0.07$ being no shows.
So, this is what I got…
Since $np = 185 \cdot 0.93$ and $nq = 185 \cdot 0.07$
$P(Y \leq 185) \approx P(W \leq 185.5) = P(Z \leq \frac{185.5-172.05}{\sqrt{185 \cdot 0.07 \cdot 0.93}})$
$P(Z \leq \frac{13.45}{\sqrt{12.0435}})$
$P(Z \leq \frac{13.45}{3.470374})$
$P(Z \leq 3.875)$
This is $0.9999$ but this is wrong which I don't understand because I even watched a tutorial video and followed each step.
I even use R on this as pbinom(180,185,0.93) which is $0.9969$ and that is still wrong.
What am I missing?
Best Answer
From $185$ persons only $180$ of them are supposed to show up. If $X$ is the random variable for the number of persons who show up, then $X\sim Bin(185,0.93)$. Using normal distribution with the continuity correction factor we get
$$P(X\leq 180)\color{green}{\approx} \Phi\left(\frac{180+0.5-185\cdot 0.93}{\sqrt{185\cdot 0.93\cdot 0.07}} \right)=\Phi(2.435)=0.9926=99.26\%$$
This is an approximation only. And I agree to your exact result of $0.9969=99.69\%$.