# 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?

• Where did you ever use the fact that there are 180 seats on the airplane? – Ted Shifrin Feb 2 '20 at 17:14
• Blame the tutorial video that I saw last night. R saved me but I couldn't use it as an answer. – usukidoll Feb 2 '20 at 20:17

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 $$0.9969=99.69\%$$.

• The LaTeX seems to be broken (edit: nevermind I see it). I am at a loss at this. Like... I did exactly what the sample video said to do which is to add a 0.5 to the....UGHHH! >:/ The sample video took $n = 185$ and added the $0.5$ to it which produced a big z score number when I applied it to my question. – usukidoll Feb 2 '20 at 9:35
• Is all OK now or you have any further specific question? – callculus Feb 2 '20 at 9:37
• I agree with the R version, but when I put it in webassign, it just gave me the finger so I was like wow okkkk... but R is a way better way of solving for this scenario. I even practiced with similar types of questions with different numbers and I got those right, but when it came to this. I was like what is going on here??? Using R was like my worst case scenario option... just use pbinom(180,185,0.93) because it was getting ridiculous . – usukidoll Feb 2 '20 at 9:39
• No. There are sample videos sort of like watching how the problem is solved. So, I was following the tutorial. My main gripe is that I didn't know that I was supposed to use the lower number which is 180 and then add 0.5 to it. The tutor who did a similar example problem used the sample size $n=185$ and added 0.5 to it. That's where the problem was. – usukidoll Feb 2 '20 at 9:47
• @usukidoll Ok. But I have to remark that it is true that $n=185$, but $x=180$. The general formula is $$P(X\leq x)\color{green}{\approx} \Phi\left(\frac{x+0.5-n\cdot p}{\sqrt{n\cdot p\cdot(1-p)}} \right)$$ – callculus Feb 2 '20 at 9:50

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.

• I recalculated according to what you latexed and the other comment and I see that whoever did the sample video made a huge error. I do get that $P(Z \leq 2.43)$ which is $.9925$. I should've looked more at the textbook example instead because it did give the right directions and that 0.5 was to be added to 180. The video I watched added 0.5 to the sample size and everything went downhill from there. – usukidoll Feb 2 '20 at 9:46