Respuesta :
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
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 [tex]p = 0.9[/tex]
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets
This means that [tex]n = 125[/tex]
Using the approximation:
[tex]\mu = E(X) = np = 125*0.9 = 112.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.9*0.1} = 3.354[/tex]
(a) What is the probability that every passenger who shows up can take the flight
This is [tex]P(X \leq 120)[/tex], so this is the pvalue of Z when X = 120.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 112.5}{3.354}[/tex]
[tex]Z = 2.24[/tex]
[tex]Z = 2.24[/tex] has a pvalue of 0.9875
98.75% probability that every passenger who shows up can take the flight
