irlines routinely overbook flights based on the expectation that some fraction of booked passengerswill not show up for each flight. For a p

Question

irlines routinely overbook flights based on the expectation that some fraction of booked passengerswill not show up for each flight. For a particular flight, there are only 50 seats, but the airline has sold52 tickets. Assume that a booked passenger will not show for the flight with probability 5%.

[2 points] Let X be the number of passengers that arrive for the flight. Under what assump- tion(s) is X a Binomial random variable?

[2 points] For the remaining questions, we will assume X is Binomial. What are the values of the relevant parameters?

[3 points]Using the Binomial frequency function, find the exact probability that 51 passengers arrive. Also find the exact probability that 52 passengers arrive.

[1 points] What is the probability that there will not be enough seats on the plane?

[5 points] If there are not enough seats, then the airline has to pay a penalty of 500$ to each passenger who cannot ride. In other words, the total penalty is 500$ if 51 passengers arrive and 1000$ if 52 passengers arrive. Find the expected value and the variance of the amount the airline will have to pay.

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Autumn 2 weeks 2021-11-23T18:02:29+00:00 1 Answer 0 views 0

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    2021-11-23T18:03:59+00:00

    Answer:

    a) For this problem to be truly a binomial distribution problem,

    – There has to be only two outcomes, that is, a booked passenger shows up or not.

    – The probability given for a booked passenger not showing up has to be the same for every booked passenger

    – The numbers involved, that is, total number of seats, total number of booked passengers and total number of booked passengers that show up have to be finite and independent.

    b) n = total number of sample spaces = total number of booked passengers = 52

    x = Number of successes required = Total number of booked passengers that show up

    p = probability of success = probability that a passenger shows up = 0.95 (that is, 1 – 0.05)

    q = probability of failure = probability that a passenger doesn’t show up = 0.05

    c) Probability that 51 booked passengers show up = 0.190

    Probability that 52 booked passenger show up = 0.0694

    d) Probability that there will not be enough seats = 0.2594

    e) The expected amount and variance for each flight are $164.4 and 89872.64 respectively.

    Step-by-step explanation:

    a) For this problem to be truly a binomial distribution problem,

    – There has to be only two outcomes, that is, a booked passenger shows up or not.

    – The probability given for a booked passenger not showing up has to be the same for every booked passenger

    – The numbers involved, that is, total number of seats, total number of booked passengers and total number of booked passengers that show up have to be finite and independent.

    b) Binomial distribution function is represented by

    P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

    n = total number of sample spaces = total number of booked passengers = 52

    x = Number of successes required = Total number of booked passengers that show up

    p = probability of success = probability that a passenger shows up = 0.95 (that is, 1 – 0.05)

    q = probability of failure = probability that a passenger doesn’t show up = 0.05

    c) Probability that 51 booked passengers show up

    n = 52, x = 51, p = 0.95 q = 0.05

    P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

    P(X = 51) = ⁵²C₅₁ (0.95)⁵¹ (0.05)⁵²⁻⁵¹ = 0.190

    Probability that 52 booked passengers show up

    n = 52, x = 52, p = 0.95 q = 0.05

    P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

    P(X = 52) = ⁵²C₅₂ (0.95)⁵² (0.05)⁵²⁻⁵² = 0.0694

    d) Probability that there will not be enough seats = (Probability that 51 booked passengers show up) + (Probability that 52 booked passenger show up) = 0.19 + 0.0694 = 0.2594

    e) Expected value paid to overbooked passengers

    P(X=xᵢ) = pᵢ

    For 1 overbooked passenger,

    P(X=$500) = 0.190

    For a B

    P(X=$1000) = 0.0694

    But E(X) = Σ xᵢpᵢ

    E(X) = (0.19 × 500) + (0.0694 × 1000) = $164.4

    Expected value to be paid per flight = $164.4

    Variance = Var(X) = Σx²p − (xbar)² = [(500²×0.19) + (1000²×0.0694)] – 164.4² = $89872.64

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