Question

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.10, and the passengers behave independently. (a) What is the probability that every passenger who shows up can take the flight

Answer 1

Answer:

98.75% probability that every passenger who shows up can take the flight

Step-by-step explanation:

For each passenger who show up, there are only two possible outcomes. Either they can take the flight, or they do not. The probability of a passenger taking the flight is independent from other passenger. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

The probability that a passenger does not show up is 0.10:

This means that the probability of showing up is 1-0.1 = 0.9. So $p = 0.9$

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets

This means that $n = 125$

Using the approximation:

$\mu = E(X) = np = 125*0.9 = 112.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.9*0.1} = 3.354$

(a) What is the probability that every passenger who shows up can take the flight

This is $P(X \leq 120)$, so this is the pvalue of Z when X = 120.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{120 - 112.5}{3.354}$

$Z = 2.24$

$Z = 2.24$ has a pvalue of 0.9875

98.75% probability that every passenger who shows up can take the flight