Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our 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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
