Using Normal and Binomial approximations for airline tickets 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?
 A: 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 $0.9969=99.69\%$.
A: You need to define your variables and understand what they represent.
$n$ is the sample size for a binomial random variable.  Here, it represents the population of tickets sold to presumptive flyers.  In this case, $n = 185$.
$p$ is the probability that a randomly selected ticket sold will actually be used; i.e., the person who bought the ticket will board the plane and take the flight.  In this case, $p = 0.93$:  for any randomly selected ticket, the chance it will be used is $93\%$ and the question implies that this is independent of all other tickets bought.
$Y$ is a binomial random variable that counts the random number of tickets used among the $n$ tickets sold.  This value is probabilistic because whether a ticket is used depends on a random outcome of the buyer's choice whether to fly or not.
Next, you need to determine what the question is asking for, and how to express this in terms of the notation we have defined.  Here, the question clearly states that only $180$ seats are available, thus if the random number of tickets used $Y$ exceeds $180$, there are not enough seats for all of the passengers.  So, to ensure the flight can seat all who show up, we require $Y \le 180$, and the probability of this is $$\Pr[Y \le 180].$$
Since $$Y \sim \operatorname{Binomial}(n = 185, p = 0.93)$$ with $$\Pr[Y = y] = \binom{n}{y} p^y (1-p)^{n-y} = \binom{185}{y} (0.93)^y (1-0.93)^{185-y},$$ we can simply write this as $$\Pr[Y \le 180] = 1 - \Pr[Y \ge 181] = 1 - \sum_{y=181}^{185} \binom{185}{y} (0.93)^y (1 - 0.93)^{185-y}.$$  There are only five terms in this sum so it is not too difficult to compute with a hand calculator.  This gives the exact probability that is desired.
A normal approximation is possible although it is not guaranteed that it will be accurate to at least four decimal places.  We do this by approximating the distribution of $Y$ with a normal distribution whose mean and variance match the mean and variance of $Y$; i.e. let $$X \sim \operatorname{Normal}(\mu = np, \sigma^2 = np(1-p)).$$  Then with continuity correction applied, $$\Pr[Y \le 180] \approx \Pr[X \le 180.5] = \Pr\left[\frac{X - \mu}{\sigma} \le \frac{180.5 - (185)(0.93)}{\sqrt{(185)(0.93)(0.07)}}\right] = \Pr[Z \le \,???],$$ where $Z \sim \operatorname{Normal}(0,1)$ is a standard normal random variable.  The quantity $???$ is the $z$-score that you would need to look up in a statistical table, or use a calculator.
