Probability/statistics question help

[redruM]

i am having some difficulties for a question on probability. it is:

from past experience an airline know that, on average, 5% of passengers holding a reservation for a flight will not turn up. their planes hold 90 people, and for each flight, they allow 95 reservations to be made.
a) if the plane is fully booked, what is the exact probability that everyone who shows for a flight can be accommodated?
b) calculate the same probability using a poisson approximation.
c) repeat the process using a normal approximation with and without the correction for continuity
d) compare all the above results
e) how many reservations should be sold , so that the probability that the airline can accommodate everyone who appears is at least 99%

i have no idea what so ever of how to start off. how do i approach this question...
any help will be appreciated.

btw, if this post is in the wrong section, feel free to move it :)

 

[redruM]

can someone check my solutions to the first 2 parts?

i used binomial distribution for a) and poisson distribution for b) and they came out pretty close. so since the poisson distribution is an approximation of the binomial distribution, we would expect them to be very similar?

a) 5C0 x (0.05)^0 x (0.95)^5
= 0.773780737

b) lambda = 5 x 0.05 = 0.25
X = 0

P(X=0) = (e^-0.25 x 0.25^0)/0!
= 0.778800783

any suggestions will be appreciated.
thanks.

 

[Hurkyl]

What probability is computed by the binomial distribution? In particular, what does 5C0 * .05^0 * .95^5 mean? Is that what the problem asks?

 

[redruM]

The logic i used to work out this problem was that:

P(0 extra people are coming to be accommodated) :
5C0 * .05^0 * .95^5 ( 5 combination 0 * 0.05^0 * 0.95^5)

n = 5 , p = 0.05, q = 0.95

I am not very sure of the logic??

If anyone has any suggestions please help!

 

[broegger]

I don't know if this is the right approach, but anyway here goes:

a) If everybody is to be boarded there must be at most 90 people out of the 95 that shows up. On the other hand 89, 88, 87... people is also fine; so, to get the probability in question you need to sum up the probabilities that n people shows up for n = 0,\ldots,90 - like this:

{\small \left( \begin{array}{c} 95 \\ 90 \end{array} \right) \cdot 0.95^{90} \cdot 0.05^5 + \left( \begin{array}{c} 95 \\ 89 \end{array} \right) \cdot 0.95^{89} \cdot 0.05^6 + \ldots + \left( \begin{array}{c} 95 \\ 0 \end{array} \right) \cdot 0.95^0 \cdot 0.05^{90}}

(this tedious calculation can be done using the binomial distribution function available on most calculators)

b) The approximation is a Poisson-distribution with \lambda = np = 95 \cdot 0.95. Again; you should use the Poisson distribution function (not just the density function with x = 90) since the number of passengers that shows up can be anything from 0 to 90.

c) The approximation is a N(np,np(1-p))-distribution and similar to a) and b) you should use the distribution function to calculate the probability.

I don't how to "correct for continuity"...

e) I think the most effective way is just trial and error. Use the expression in a) with different values of n (instead of n = 95) or use one of the approximations in b) or c).