Because not all airline passengers show up for their reserved seat, an airline s
ID: 2871594 • Letter: B
Question
Because not all airline passengers show up for their reserved seat, an airline sells 127 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.36, and the passengers behave independently. Round your answers to two decimal places (e.g. 98.76).
(a) What is the probability that every passenger who shows up gets a seat?
(b) What is the probability that the flight departs with empty seats?
(c) What are the mean and (d) standard deviation of the number of passengers who show up?
Explanation / Answer
a) Let’s assume X is the number of passenger who shows up
If every passenger who shows up get a seat, then X<=120
Pr(X<=120) = 1-Pr(X>120)
Pr(X=127) = 127c0 0.360 0.64127 = .0000
Pr(X=126) = 126c1 0.361 0.64126 =1.7x10 -23
Pr( X= 125) = 125c2 0.362 0.64 125=5.9x10-22
Pr(X= 124) = 124C3 0.36 3 0.64 124= 4.3X10-26
Pr( X= 123) = 123C4 0.36 4 0.64 123 =2.4x10-26
Pr( X= 122) = 122 C5 0.36 5 0.64 122= 0.00
Pr ( X= 121) = 121C6 0.36 6 0.64 121=0.00
Pr (X =120 ) = 120C6 0.36 7 0.64 120 = 0.00
These Probabilities add up to = 2.3x10-26 approx
So 1- 0000 = 1
B) The flight will depart with empty seats if (X< 120)
Pr(X <120) = 1- Pr( X>=120)
Pr(X=120) =127C7 (0.36)7(0.64)120= 3.8 x 10-16
1- Pr (X>=120) = 1- (3.8x10-16)= 1
C) Available seat =120
Pr of showing up =0.64
Mean = X×0.64 =120×0.64 =76.8
Variance = 120×(0.64)(0.34)
Standard deviation = sqrt 27.648
Standard Deviation =5.258