Poisson Distribution

[Josephine]

An aircraft has 116 seats. The airline found, from past experience, that on average, 2.5% of ticketed passengers will fail to check in for a flight. IF the airline sells 120 tickets for a flight, find, by using a suitable approximation, the probability that more than 116 passengers will check in for a flight. Find also the probability that there will be empty seats on the flight.

My working:
n=120, p= 2.5?
How do i get started? Thanks.

 

[galactus]

Use the normal approximation to binomial distribution.

 

[tkhunny]

Did you label this "Poisson" on purpose?

120*0.025 = 3 <== This is your parameter for your Poisson Distribution.

Pr(Overbooked), in this case, means 117, 118, 119, or 120, corresponding to failure to check in values of 3, 2, 1, and 0. For Pr(No Empty Seats), just add 116, or 4 fail to check in. For Pr(an empty seat), calculate 1 - Pr(No Empty Seats)

If \(\displaystyle \L\,Pr(n) = \frac{e^{-3}*3^{n}}{n!}\), find 1-(Pr(0)+Pr(1)+Pr(2)+Pr(3)+Pr(4)).

I get 18.47%. How close was that to your Normal or Binomial approximations?