Respuesta :
Answer:
0.1373 = 13.73% probability that less than 65 of the ticket holders will show up for the flight
Step-by-step explanation:
For each ticket holder, there are only two possible outcomes. Either they will show up for the flight, or they will not. Ticket holders are independent. This means that we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
An airline estimates that 95% of ticket holders actually show up for the flight.
This means that [tex]p = 0.95[/tex]
70 tickets:
This means that [tex]n = 70[/tex]
What is the probability that less than 65 of the ticket holders will show up for the flight?
This is:
[tex]P(X < 65) = 1 - P(X \geq 65)[/tex]
In which
[tex]P(X \geq 65) = P(X = 65) + P(X = 66) + P(X = 67) + P(X = 68) + P(X = 69) + P(X = 70)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 65) = C_{70,65}.(0.95)^{65}.(0.05)^{5} = 0.1348[/tex]
[tex]P(X = 66) = C_{70,66}.(0.95)^{66}.(0.05)^{4} = 0.1941[/tex]
[tex]P(X = 67) = C_{70,67}.(0.95)^{67}.(0.05)^{3} = 0.2201[/tex]
[tex]P(X = 68) = C_{70,68}.(0.95)^{68}.(0.05)^{2} = 0.1845[/tex]
[tex]P(X = 69) = C_{70,69}.(0.95)^{69}.(0.05)^{1} = 0.1016[/tex]
[tex]P(X = 70) = C_{70,70}.(0.95)^{70}.(0.05)^{0} = 0.0276[/tex]
[tex]P(X \geq 65) = P(X = 65) + P(X = 66) + P(X = 67) + P(X = 68) + P(X = 69) + P(X = 70) = 0.1348 + 0.1941 + 0.2201 + 0.1845 + 0.1016 + 0.0276 = 0.8627[/tex]
[tex]P(X < 65) = 1 - P(X \geq 65) = 1 - 0.8627 = 0.1373[/tex]
0.1373 = 13.73% probability that less than 65 of the ticket holders will show up for the flight
