nstructor: Sujin K Spring 2017 Quiz 3 Probability and Statistics I Airlines some

Question

nstructor: Sujin K Spring 2017 Quiz 3 Probability and Statistics I Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable Yas the number of ticketed passengers who actually show up for the flight. The probability mass fund- tion of Yappears in the accompanying table. y 45 46 47 48 49 50 51 52 53 54 55 p() .05.10 .12.14 .25 .17 .06 .05 .03 .02 .01 a. Probability that the flight will not accommodate at least 3 passengers all ticketed passengers who show up b. Probability that the flight will not accommodate at most 3 passengers all ticketed passengers who show up c. Probability that at least 3 passengers on the stand-by will be accommodated d. Probability that at most 3 passengers on the stand-by will be accommodated

Explanation / Answer

Answer: answer to 1st question and subparts

1. From the information provided Y: random variable number of ticketed passengers who actually show-up for the flight.

The sum of all individual probabilities add to 1.

a. Probability that the flight will not accommodate atleast 3 passengers all ticketed passengers who show up.

Since maximum number of passengers that can be accommodated is 50, therefore any more number of passenger that turns up can’t be accommodated in the flight

This is equivalent to : P(3 passengers not accommodated) + P(4 passengers not accommodated) + P(5 passengers not accommodated) = P(53 passengers appear) + P(54 passengers appear) + P(55 passenger appear) = 0.03+0.02+0.01 = 0.06

b. Probability that the flight will not accommodate atmost 3 passengers all ticketed passengers who show up.

Since maximum number of passengers that can be accommodated is 50, therefore any more number of passenger that turns up can’t be accommodated in the flight

This is equivalent to : P(1 passengers not accommodated) + P(2 passengers not accommodated) + P(3 passengers not accommodated) = P(51 passengers appear) + P(52 passengers appear) + P(53 passenger appear) = 0.06+0.05+0.03 = 0.14

c. Probability that atleast 3 passengers on the stand-by list will be accommodated.

The stand-by list will have maximum of 5 passengers (51-55). A standby passenger will be accommodated if a ticketed passenger from list of 50 accommodated passenger does not turn up.

So probability that atleast 3 passengers on stand-by list will be accommodated = P(3 passengers on standby list accommodated) + P(4 passengers on standby list accommodated) + P(5 passengers on standby list accommodated) = P(y=47) + P(y=46) + P(y=45) = (0.12) + (0.10) + (0.05) = 0.27

d. Probability that atmost 3 passengers on the stand-by list will be accommodated.

The stand-by list will have maximum of 5 passengers (51-55). A standby passenger will be accommodated if a ticketed passenger from list of 50 accommodated passenger does not turn up.

So probability that atmost 3 passengers on stand-by list will be accommodated = P(1 passengers on standby list accommodated) + P(2 passengers on standby list accommodated) + P(3 passengers on standby list accommodated) = P(y=49) + P(y=48) + P(y=47) = (0.25) + (0.14) + (0.12) = 0.51

y p(y) 45 0.05 46 0.1 47 0.12 48 0.14 49 0.25 50 0.17 51 0.06 52 0.05 53 0.03 54 0.02 55 0.01 Total 1

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