Question 983113
# 1


X = number of tickets sold


"Overbooked" means that more than 120 tickets have been sold. 
So when X > 120, the flight has been overbooked.
We want to find the probability P(X > 120)


Let x = 120. Find the z-score
z = (x-mu)/sigma
z = (120-90)/15
z = 30/15
z = 2


Use this z-score to find the value of P(Z < 2). I'm going to use a table, but you can use a TI calculator. The normalcdf function gets the job done.


I used a <a href="https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf">table</a> to find that P(Z < 2) = 0.97725


So,


P(Z > 2) = 1 - P(Z < 2)
P(Z > 2) = 1 - 0.97725
P(Z > 2) = 0.02275


Therefore, 


P(X > 120) = P(Z > 2)
P(X > 120) = 0.02275


The probability of overbooking is 0.02275
There's a 2.275% chance that the flight is overbooked.

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# 2


Multiply the probability with the number of flights per day.


0.02275*5 = 0.11375


So on average, we expect 0.11375 flights per day to be overbooked. On any given day, this is a good thing since the number is small. If we extend this over 30 days, then 30*0.11375 = 3.4125


So over 30 days, we expect on average roughly 3.4125 flights to be overbooked.