Finding the probability using a normal distrubtion.

Question

I have a stats question that says, "An airline flies airplanes that hold 100 passengers. Typically, some 10% of the passengers with reservations do not show up for the flight. The airline generally overbooks fights in an attempt to fill them."

And the question is asking 4 parts, the first being "Find the probability that a flight booked for 100 people flies full." The second being, "Find the probability that for a flight booked for 100 passengers, between 90 and 100 passengers inclusively will show up for the flight" Third being, "Find the probability that for a flight with 105 reservations, everyone will get a seat." and fourth "If the flight is booked for 106 passengers, can the airline be at least 95% sure that everyone will get a seat?"

The only hints that I have about this question are that it is a normal distribution because that is what the homework assignment is about, but I cannot even think of how to start this problem because the mean and standard deviation do not seem to be given.

Answer 1

I'm pretty sure you are supposed to use the normal approximation to the normal, as suggested by @Karl. I hope you used continuity corrections to give that the best chance to work. Here are exact binomial answers from R statistical software to compare with your approximate answers:

 dbinom(100, 100, .9)             # (a)
 2.65614e-05                      # essentially 0
 diff(pbinom(c(89,100), 100, .9)) # (b)
 0.5831555
 pbinom(100, 105, .9)             # (c)
 0.9832837
 pbinom(100, 106, .9)             # (d)
 0.9602006                        # Yes

Just to show that the normal approximation is pretty good, but not exact, here is the normal approximation to the answer for (d). Your answer using normal tables may be a little different, because that requires a bit of rounding that software does not.

 # normal approx for (d)
 mu = 106*.9;  sg = sqrt(mu*.1)
 pnorm(100.5, mu, sg)
 0.9506497    

Using a normal approximation for such an extreme value of $p$ (extreme meaning far from 1/2) does not assure accuracy in the second decimal place when $n = 100.$

Another approach would be to model the number of no-shows as approximately $Y \sim Pois(\lambda).$ In the last part we would have $\lambda = 106(0.1) = 10.6$ and we would need at least 6 no shows or $P(Y \ge 6) = 1 - P(Y \le 5) = 0.952.$

 1-ppois(5, 10.6)
 0.9524726