Question

On a particular day, 112 of 280 passengers on a particular DTW-LAX flight used the e-ticket check-in kiosk to obtain boarding passes. In a random sample of eight passengers, use the binomial model to find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

Answer 1

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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Population of 280, which means that $N = 280$

112 use e-ticket check-in kiosk, which means that $k = 112$

Sample of eight passengers means that $n = 8$

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

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,280,8,112) = \frac{C_{112,4}*C_{168,4}}{C_{280,4}} = 0.2348$

0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.