Question

6. An airliner has a capacity for 250 passengers. If the airline company over book a flight with 260 reservations, what is the probability that there will be not enough seats to accommodate all passengers. Assume that the probability that a randomly selected passenger does not shows up to the airport is 0.035.  
   1. Use the binomial distribution (5 points)

2. Use the Normal distribution (5 points)

Answer 1

Solutions Let x measures who show-up noi of passengers for a flight for each passengers ; 01035 Probability of we have nod Op. showing up is a a binomial sandam manable, h = 260 of showing P = 1-0.038 >0.965 ( probability Up.) tf x < 250, fueryone gets a real So, PCX >250 ) = 16 not get reads P(x>280) - PCX = 291) + P(X=252) +PCX = 253) + Pex = 254) +PCX = 255) + P( X = 256) + P ( X = 257) + P(X= 258) + P ( X = 259) + P(x=260) > 0.13402 + 0.13197 +0.1505 +0.08742 +0.056717 6:03054+0.0/31 +0.0042 +0.00089 +0.0ooo95 0.574005 Binomial Probability formule, P(x>286)=0.5740 P(x-2) C. px. (1-P) na [Probability that there will not be enough seats) x has a Normal distribution , 4. = np = 260 x0.965= 250.9 ( Mean) o= Inpli-p] = $260x 0.965 x (1-0,968) = 2,96 tion So, 250 - 250 g PC * 3250) - PC* P(x-> 2.96 =P(ZJ-0.304054) = 1-Plze-0.304) sl-0138056 = 0.61944 P(x > 250) 0.6194 , probability that there will be not be enough seats to accomodate s standard dara