Airlines often sell more tickets on a flight than there are available seats. The

Question

Airlines often sell more tickets on a flight than there are available seats. The reason is that a certain percentage of ticketed passengers don’t show up . A new airline has 180 seats with no show at 5%. Binomial distribution can describe the situation of no show versus succeeded to show passengers . The binomial distribution can be approximated with a normal distribution with mean equals to n multiplied by P and standard deviation ( square root of ( n.p .(1-p)
1- If the airline sells 185 tickets what is the probability of having an overbooking situation ? 2- if the airline sells 200 tickets what is the probability of having empty seats ? 3- how many tickets shoudl the airline sells if the airline wants a 2% probability of overbooking situation ?


Airlines often sell more tickets on a flight than there are available seats. The reason is that a certain percentage of ticketed passengers don’t show up . A new airline has 180 seats with no show at 5%. Binomial distribution can describe the situation of no show versus succeeded to show passengers . The binomial distribution can be approximated with a normal distribution with mean equals to n multiplied by P and standard deviation ( square root of ( n.p .(1-p)
1- If the airline sells 185 tickets what is the probability of having an overbooking situation ? 2- if the airline sells 200 tickets what is the probability of having empty seats ? 3- how many tickets shoudl the airline sells if the airline wants a 2% probability of overbooking situation ?



1- If the airline sells 185 tickets what is the probability of having an overbooking situation ? 2- if the airline sells 200 tickets what is the probability of having empty seats ? 3- how many tickets shoudl the airline sells if the airline wants a 2% probability of overbooking situation ?

Explanation / Answer

1)

X = binom(185 ,0.95)

P(X > 180)

1-pnorm(180.5,185*0.95,sqrt(185*0.05*0.95))
[1] 0.05453767

2)

P(X<180)

X - binom(200,0.95)

pnorm(180,200*0.95,sqrt(200*0.05*0.95))
[1] 0.000588433

3)

let the number of ticket the airline sells = n

P(X > 180) = 0.02

P( Z > (180 - n * 0.95) / sqrt( n * 0.05 * 0.95)) = 0.02

P(Z > z) = 0.02

z = 2.054

hence

n = 183.097

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.