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Nat2105 [25]
4 years ago
6

The problem of matching aircraft to passenger demand on each flight leg is called the flight assignment problem in the airline i

ndustry. Suppose the demand for the 6 p.m. flight from Toledo Express Airport to Chicago's O'Hare Airport on Cheapfare Airlines is normally distributed with a mean of 132 passengers and a standard deviation of 42. Round probabilities in parts (a) through (c) to four decimal places.
a) Suppose a Boeing 757 with a capacity of 183 passengers is assigned to this flight. What is the probability that the demand will exceed the capacity of this airplane?

b) What is the probability that the demand for this flight will be at least 80 passengers but no more than 200 passengers?

c) What is the probability that the demand for this flight will be less than 100 passengers?

Round answers in parts (d) and (e) to the nearest whole number.

d) If Cheapfare Airlines wants to limit the probability that this flight is overbooked to 3%, how much capacity should the airplane that is used for this flight have? passengers

e) What is the 79th percentile of this distribution?
Mathematics
1 answer:
Darina [25.2K]4 years ago
7 0

Answer:

Step-by-step explanation:

Given that the demand for the 6 p.m. flight from Toledo Express Airport to Chicago's O'Hare Airport on Cheapfare Airlines is normally distributed with a mean of 132 passengers and a standard deviation of 42

Let X be the no of passengers who report

X is N(132, 42)

Or Z is \frac{x-132}{42}

a) Suppose a Boeing 757 with a capacity of 183 passengers is assigned to this flight.

the probability that the demand will exceed the capacity of this airplane

=P(X>183) = P(Z>1.21) =

=0.5-0.3869\\=0.1131

b)  the probability that the demand for this flight will be at least 80 passengers but no more than 200 passengers

=P(80\leq x\leq 200)\\= P(-1.23\leq z\leq 1.62)\\

=0.4474+0.3907

=0.8381

c) the probability that the demand for this flight will be less than 100 passengers

=P(x

d) If Cheapfare Airlines wants to limit the probability that this flight is overbooked to 3%, how much capacity should the airplane that is used for this flight have? passengers

=P(Z>c)=0.03\\c=1.88\\X=132+1.88(42)\\=210.96

e) 79th percentile of this distribution

=132+0.81(42)\\= 166.02

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   P(X)           42%            50%         8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c)   P(win first game)  = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

P(X = 0)  =  P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game)  × P(win second game)] + [ P(win first game)  × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value  \mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66

f) Variance \sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844

Standard deviation \sigma=\sqrt{variance} = \sqrt{0.3844}=0.62

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