The problem of matching aircra
The problem of matching aircraft to passenger demand on eachflight leg is called the flight assignment problem in the airlineindustry. Suppose the demand for the 6 p.m. flight from ToledoExpress Airport to Chicago’s O’Hare Airport on Cheapfare Airlinesis normally distributed with a mean of 144 passengers and astandard deviation of 42. Round probabilities in parts (a) through(c) to four decimal places. a) Suppose a Boeing 757 with a capacityof 186 passengers is assigned to this flight. What is theprobability that the demand will exceed the capacity of thisairplane? b) What is the probability that the demand for thisflight will be at least 94 passengers but no more than 200passengers? c) What is the probability that the demand for thisflight will be less than 100 passengers? Round answers in parts (d)and (e) to the nearest whole number. d) If Cheapfare Airlines wantsto limit the probability that this flight is overbooked to 1%, howmuch capacity should the airplane that is used for this flighthave? passengers e) What is the 65th percentile of thisdistribution?
Answer:
a)
| for normal distribution z score =(X-μ)/σx | |
| here mean= μ= | 144 |
| std deviation =σ= | 42.0000 |
probability that the demand will exceed the capacityof this airplane:
| probability = | P(X>186) | = | P(Z>1)= | 1-P(Z<1)= | 1-0.8413= | 0.1587 |
b)
probability that the demand for this flight will beat least 94 passengers but no more than 200 passengers:
| probability = | P(94<X<200) | = | P(-1.19<Z<1.33)= | 0.9082-0.1170= | 0.7912 |
c)
probability that the demand for this flight will beless than 100 passengers:
| probability = | P(X<100) | = | P(Z<-1.05)= | 0.1469 |
d)
| for 99th percentile critical value of z= | 2.33 | ||
| thereforecorresponding value=mean+z*std deviation= | 241.9~ 242 | ||
e)
| for 65th percentile critical value of z= | 0.39 | ||
| thereforecorresponding value=mean+z*std deviation= | 160.4 ~160 | ||
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