Problem 5 - Arbeitsbuch Mathematik für Ingenieure

Question

In einem Flugzeug haben 224 Passagiere Platz. Da der Fluggesellschaft aus Erfahrung bekannt ist, dass ein Passagier auf einer bestimmten Flugstrecke mit der Wahrscheinlichkeit von \(2 \%\) nicht zum Abflug erscheint, werden fïr diesen Flug 227 Buchungen vorgenommen. Es wird angenommen, dass die Entscheidungen der Passagiere, ob der Flug angetreten wird, unabhängig voneinander zustande kommen. Berechnen sie die Wahrscheinlichkeit dafür, dass alle zum Abflug erscheinenden Passagiere einen Platz erhalten, a) exakt, b) näherungsweise unter Verwendung der Poisson-Approximation.

Step-by-Step Solution

Answer

The exact probability is found using the binomial sum formula; the Poisson approximation uses a mean of 222.46 with a cumulative sum for the same limit.

Step 1 Title - Define variables and given probabilities

Given that the probability of a passenger not showing up is 2%, which means the probability of showing up is 98%. Let the total number of bookings be n = 227 and the number of seats be k = 224.

Step 2 Title - Calculate the exact probability

We need to find the probability that at most 224 passengers show up for the flight. Represent this as a binomial distribution where the number of trials is 227 and the probability of success (showing up) is 0.98. The required probability is: \[ P(X \leq 224) = \sum_{x=0}^{224} \binom{227}{x} (0.98)^x (0.02)^{227-x} \]

Step 3 Title - Calculate using Poisson approximation

For the Poisson approximation, we use the mean \( \lambda \) which is \( n \cdot p \), here \( \lambda = 227 \cdot 0.98 = 222.46 \). The Poisson probability mass function is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] We need to calculate \[ P(X \leq 224) = \sum_{x=0}^{224} \frac{222.46^x e^{-222.46}}{x!} \]

Step 4 Title - Calculate practical approximations

Using a statistical tool/software or a table, find the cumulative probabilities for both the exact binomial distribution and the Poisson approximation. They can be plugged in or computed using the distribution functions for both steps.

Key Concepts

binomial distribution

The binomial distribution is crucial for modeling situations where there are a fixed number of trials (n), each with the same probability of success (p). In this exercise, trials are the number of bookings (227), and success is a passenger showing up for the flight (98%).

We express the probability of at most 224 passengers showing up as a sum of probabilities from 0 to 224 successes:
. This way calculates the exact cumulative probability using the binomial formula:

} ,

Poisson approximation

The Poisson approximation is handy when dealing with the binomial distribution, especially for large n and small p. It's easier to compute and quite accurate. In this problem, where n=227 and p=0.98, we use the Poisson distribution with parameter \( \lambda = np = 227 \cdot 0.98 = 222.46 \)

The Poisson formula is:


\text\text\text \text\text\text \text\text\text \ P(X = k) = \ \frac{\lambda^k e^{-\lambda}}{k!} \
\text\text\text\text . <\br> . <\br>\text Applying this model, we get:

< \br>< \br>< \br>\textThe required probability is:
\ P(X = 0 \ to \ 224) \ Which can be calculated as cumulative probability.
<\br> <\br> <\br> <\br> <\br> As it\text simpler to calculate.<\br><\br>.

cumulative probability

The cumulative probability is the probability that a random variable takes on a value less than or equal to a certain number. In this case, we need to find the probability that the number of passengers showing up \( \ (X) \) is less than or equal to 224. \
Calculating cumulative probability involves summing up individual probabilities.
\For binomial distribution, \ P(X \ leq 224)= \ \sum_{x=0}^{224} \binoo{227}{x}+ (0.98)^x +(0.02)^{227-x}
<\br> In Poisson approximation: \bu\ P(X \ leq 224) =\ \sum_{k=0}^{224}\frac{\lambda^k e^{-\lambda}})\x! \ \ess
<\br> These calculations, when summed up, give us a complete view of the entire distribution\value or range.<\br>