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ryan90
03-23-2010, 09:15 AM
An airline has 500 places for a flight. However, due to only 95% of reservations being taken the airline usually overbooks by 20 places. When the airline has taken these 520 bookings, what is the probability that customers arrive for more than 500 bookings?

Im not sure at all how to begin with this problem?
can anyone help?

many thanks for your time :tup:

BGM
03-23-2010, 09:47 AM
Presume that the customers are independent and their probability of arrival are all 95&#37;
Then the total number of arrival N ~ Binomial(520, 0.95)
And the question require you to compute Pr{N > 500}

ryan90
03-23-2010, 09:51 AM
Presume that the customers are independent and their probability of arrival are all 95%
Then the total number of arrival N ~ Binomial(520, 0.95)
And the question require you to compute Pr{N > 500}

Can you explain this in simpler language? I really appreciate your time :)

Mean Joe
03-23-2010, 01:20 PM
Well here's a start...
Airline has booked 520 passengers, where there is a 95&#37; chance that a booked passenger will actually show up on the day of the flight. What is the probability that more than 500 people will show up?

Hint: You can also use the normal approximation for the binomial distribution.

In easy language, we expect 520 * .95 = 494 people to show up. But of course, we know that there is some variance to this number. Can you calculate the variance?

BGM
03-23-2010, 01:25 PM
So as the set up above,
let Ij be the Bernoulli random variable of the jth customer, j = 1, 2, ..., 520
i.e. Pr{Ij = 1} = 0.95 and Pr{Ij = 0} = 0.05, Ij are mutually independent

Ij is serve as the counting variable for the customer. If the jth customer really arrive,
Ij = 1; else Ij = 0. So the total number of customers arrive out of a total 520 customers
= I1 + I2 + ... + I520
And the distribution of a sum of independent and identically distributed bernoulli random
variables are just Binomial distribution with probability mass function
Pr{N = n} = (520Cn)(0.95)^n(0.05)^(520-n), n = 0, 1, ..., 520