Question

3.5.13 Because not all airline passengers show up for their reserved seat, an airline sells 120 tickets for a flight that holds only 115 passengers. The probability that a passenger does not show up is 0.05, and the passengers behave independently. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that every passenger who shows up can take the flight? Enter your answer in accordance to the item a) of the question statement (b) What is the probability that the flight departs with at least one empty seat? Enter your answer in accordance to the item b) of the question statement

Answer 1

Answer:

a) = 0.7357

b) P = 0.5832

Step-by-step explanation:

Given Data:

Total  number of passenger is $= n\times p = 120\times 0.95$

Total  number of passenger is = 114

standard deviation $= \sqrt(n\times p\times (1-p))$

standard deviation = 2.387

probability of passenger ready to take flight is

$P = P(X< = 115) = P[Z<(\frac{115-114}{2.387})] = P(Z<0.63)$

from standard Z table

   = 0.7357

b) probability of flight with one empty sear

P = P(X<= 114)

P = P(Z<0.21)

From standard Z table

P = 0.5832