Probability/statistics question help

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Discussion Overview

The discussion revolves around a probability question related to airline reservations, specifically focusing on the likelihood of accommodating all passengers who show up for a flight given certain parameters. The scope includes theoretical approaches using binomial, Poisson, and normal distributions, as well as practical considerations for calculating probabilities and making reservations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about how to approach the probability question regarding airline reservations.
  • Another participant shares their calculations for parts a) and b) using binomial and Poisson distributions, noting that the results are close and suggesting that this is expected due to the nature of the distributions.
  • A question is raised about the specific probability computed by the binomial distribution and the meaning of the terms used in the calculations.
  • One participant explains their logic for calculating the probability of accommodating passengers, emphasizing the need to sum probabilities for various numbers of passengers showing up.
  • Another participant discusses the use of Poisson and normal approximations, mentioning the need for a continuity correction and expressing uncertainty about how to apply it.
  • Suggestions for determining the number of reservations to sell to ensure a high probability of accommodation are offered, including trial and error methods.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the best approach to the problem, with various methods and interpretations being discussed. Some participants agree on the use of certain distributions, while others question the logic and calculations presented.

Contextual Notes

There are unresolved aspects regarding the application of continuity correction in normal approximations and the specific probabilities being calculated in the binomial distribution. Participants express varying levels of confidence in their approaches and calculations.

Who May Find This Useful

This discussion may be useful for students or individuals interested in probability theory, particularly in the context of real-world applications such as airline reservations and statistical approximations.

redruM
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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... :confused:
any help will be appreciated.

btw, if this post is in the wrong section, feel free to move it :)
 
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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.
 
Last edited:
What probability is computed by the binomial distribution? In particular, what does 5C0 * .05^0 * .95^5 mean? Is that what the problem asks?
 
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!
 
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).
 

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