1. ## Distribution Probability Question

An airline has determined that 4% of people do not show up for their flights. To avoid having empty seats, the flight is overbooked. A large jet holds 300 people. What is the probability that some travelers will not get a seat if 310 tickets are sold?

I know that the expected number would be 310 times 0.04 and i calculated the standard deviation, but it gave me the the wrong z-score. Could someone help please?

2. Originally Posted by dalbir4444
An airline has determined that 4% of people do not show up for their flights. To avoid having empty seats, the flight is overbooked. A large jet holds 300 people. What is the probability that some travelers will not get a seat if 310 tickets are sold?

use the binomial distribution here. (do you see why?)

the probability that $k$ people show up is given by:

$P(k) = {310 \choose k}(.96)^k(0.04)^{310 - k}$

you want the chances of more people showing up than there are seats for. since there are actually 300 seats, but 310 tickets sold, it means we will not have enough seats if 301 people, or 302 people, or ..., 310 people show up. that is, the probability of that happening is:

$P(k \ge 301) = P(301) + P(302) + \cdots + P(310)$

3. Assuming that "showing or not showing" for each ticket is independent of the other tickets, this is actually a binomial distribution with p= 0.96 and q= 0.04 and number of events n= 300. Its mean is np= (310)(.96)= 297.6. The standard deviation is given by $\sqrt{npq}= \sqrt{310(.96)(.04)}= 3.45022$. Are those what you got?