Binomial Probabilty Distribution?

A large airplane keeps track of the number of no shows for one of its most important commuter flgihts. The aircraft used for this flight has 36 seats, but the airline routinely overbooks the flight by 4 passengers. over time the airline has found that if 40 passengers have been booked, then the number of ticketed passengers who do not show up is a random variable with the following prob. dis.


x= 0- 1 -2 - 3 -4 -5 -6 - 7 -8


P(x)= .05- .08 -0.13 -0.23 -0.18 -0.13 -0.08 -0.06 -0.06





a)probabilty at least 3 passengers do not show uo


b)probabilty that the number of no shows is at least 2 bu tno more than 5


c)the probabilty that every ticketed passenger who shows up gets a seat

Binomial Probabilty Distribution?
This is just a probability distribution. This is not Binomial distribution.


x=no of no shows.


x---- p(x)


0--- 0.05


1---0.08


2---0.13


3---0.23


4---0.18


5---0.13


6---0.08


7---0.06


8---0.06


Note that the probabilities add up to 1.


a) 7 --- 0.06 means probbility of 7 no shows is 0.06 etc.


a) p(atleat 3)=p(x=3)+p(x=4)+.....+p(x=8)


add up all p(x) for x=3 through 8.


b)p(x=2)+p(x=3)+p(x=4)+p(x=5)


c)p(x=4)=0.13