Respuesta :
Answer:
(a) The probability that more than three arrive in one hour is approximately 0.018988
(b) The probability that no interval contains more than 3 arrivals in 30 separate one-hour intervals is approximately 0.56264
(c) The length of time such that the probability of no arrival is 0.1 is approximately 2.30259 hours
Step-by-step explanation:
The distribution of the arrival time of a small aircraft = Exponential distribution
The mean arrival time, μ = 1 hour
The pdf of a exponential distribution, f(x) = λ·e^(-λ·x)
The mean, μ = 1/λ
∴ 1 = 1/λ
λ = 1 hour
∴ For the exponential distribution, f(x) = 1·e^(-x)
(a) The Poisson probability is given as follows;
[tex]P(X = k) = \dfrac{\lambda^k \times e^{-k}}{k!}[/tex]
The probability for more than 3 is given as follows;
[tex]P(X > 3) = 1 - \dfrac{1^0 \times e^{-1}}{0!} - \dfrac{1^1 \times e^{-1}}{1!} - \dfrac{1^2 \times e^{-1}}{2!} - \dfrac{1^3 \times e^{-1}}{3!} \approx 0.018988[/tex]
(b) P(X ≤ 3) for 30 1 hour intervals = P(X ≤ 3)³⁰ = (1 - 0.018988)³⁰ ≈ 0.56264
(c) When the probability that no arrival occurs is 0.1, we have;
0.1 P(X > x) = f(x) = 1·e^(-x) = e^(-x)
-x·ln(e) = ln(0.1)
∴ x = -㏑(0.1) ≈ 2.30259
The length of time such that the probability that no arrival occur during the interval is 0.1, x ≈ 2.30259 hours.
