Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.
Step-by-step explanation:
Hypergeometric distribution:
The probability of x sucesses is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
x is the number of sucesses.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
[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]
In this question:
Population of 280, which means that [tex]N = 280[/tex]
112 use e-ticket check-in kiosk, which means that [tex]k = 112[/tex]
Sample of eight passengers means that [tex]n = 8[/tex]
Find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.
This is P(X = 4). So
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 4) = h(4,280,8,112) = \frac{C_{112,4}*C_{168,4}}{C_{280,4}} = 0.2348[/tex]
0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
