Question 983113: A one-plane airline has a passenger capacity of 120. The number of passengers on a flight is normally distributed with a mean of 90 and a standard deviation of 15.
1. What is the probability that a flight will be “overbooked” (ie not able to get a seat on the flight)?
2. If there are 5 flights a day, every day, how many flights will be overbooked in a 30 day month?
Please help! and explain! Thanks so much!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! # 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 table 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.
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