SOLUTION: 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

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.
==============================================================
# 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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