Airline overbooking is a common practice. Many people make reservations on several flights due to uncertain plans and then cancel at the last minute or simply fail to show up. Eagle Air is small commuter airline. Their planes hold only 25 people. Past records indicate that 20% of the people making a reservation do not show up for the flight. Suppose that Eagle Air decides to book 18 people for each flight.
a) Determine the probability that on any given flight, at least one passenger holding a reservation will not have a seat.
b)What is the probability that there will be one or more empty seats for any one flight?
c) Determine the mean and standard deviation for this random variable.
To solve this problem, we will use the binomial probability distribution.
a) To determine the probability that on any given flight, at least one passenger holding a reservation will not have a seat, we need to find the probability that exactly 0 passengers out of 18 do not show up for the flight. The probability that any given passenger does not show up is 0.20.
The probability that a passenger does show up is 1 - 0.20 = 0.80.
Using the binomial probability formula, the probability that exactly k successes occur out of n trials is given by:
P(X = k) = (n choose k) * p^k * (1 - p)^(n-k)
In this case, n = 18, k = 0, and p = 0.80. Plugging these values into the formula, we get:
P(X = 0) = (18 choose 0) * 0.80^0 * (1 - 0.80)^(18-0)
P(X = 0) = 1 * 1 * 0.20^18
P(X = 0) = 0.012
Therefore, the probability that on any given flight, at least one passenger holding a reservation will not have a seat is approximately 0.012.
b) To find the probability that there will be one or more empty seats for any one flight, we need to find the probability of success (a passenger showing up) for fewer than 18 trials (passengers). In other words, we want to find the probability that at least one passenger does not show up.
Using the complement rule, the probability of no empty seats is equal to 1 minus the probability of at least one empty seat:
P(X ≥ 1) = 1 - P(X = 0)
P(X ≥ 1) = 1 - 0.012
P(X ≥ 1) ≈ 0.988
Therefore, the probability that there will be one or more empty seats for any one flight is approximately 0.988.
c) The mean and standard deviation for a binomial distribution are given by the formulas:
Mean (μ) = n * p
Standard deviation (σ) = sqrt(n * p * (1 - p))
In this case, n = 18 and p = 0.80. Plugging these values into the formulas, we get:
Mean (μ) = 18 * 0.80 = 14.4
Standard deviation (σ) = sqrt(18 * 0.80 * (1 - 0.80)) ≈ 1.87
Therefore, the mean number of passengers who show up for each flight is 14.4, and the standard deviation is approximately 1.87.
To determine the probability, mean, and standard deviation for the given scenario, we will use the concept of binomial distribution.
a) To find the probability that at least one passenger holding a reservation will not have a seat on any given flight, we need to find the probability of 1, 2, 3 ..., 18 passengers not showing up and then subtract it from 1 (to get the probability of at least one not showing up).
The probability that a passenger does not show up on any given flight is 20% or 0.2. So, the probability that a passenger does show up is 1 - 0.2 = 0.8.
P(at least one passenger not showing up) = 1 - P(all passengers showing up)
P(at least one passenger not showing up) = 1 - (0.8)^18
Calculating this, we find:
P(at least one passenger not showing up) ≈ 0.9423
Therefore, the probability that on any given flight, at least one passenger holding a reservation will not have a seat is approximately 0.9423 or 94.23%.
b) To find the probability of there being one or more empty seats on any one flight, we need to find the probability of all passengers showing up for the flight and subtract it from 1.
P(one or more empty seats) = 1 - P(all passengers showing up)
P(one or more empty seats) = 1 - (0.2)^18
Calculating this, we find:
P(one or more empty seats) ≈ 0.9998
Therefore, the probability that there will be one or more empty seats for any one flight is approximately 0.9998 or 99.98%.
c) The mean (expected value) and standard deviation of the binomial distribution can be calculated as follows:
Mean (μ) = n * p
μ = 18 * 0.2
μ = 3.6
Standard deviation (σ) = sqrt(n * p * (1 - p))
σ = sqrt(18 * 0.2 * (1 - 0.2))
σ ≈ sqrt(2.88)
σ ≈ 1.7
Therefore, the mean for this random variable is 3.6 and the standard deviation is approximately 1.7.