Answer:
a) 0.019
b) 0.563
c) x = 1.966 hours
Detailed explanation:
E(X) = 1
The probability function for an exponential random variable can be expressed as
P(X=x) = λ e^(-λ.x)
The cumulative distribution function is defined as
P(X ≤ x) = 1 - e^(-λ.x)
a) This refers to the time gaps between the arrivals of smaller aircraft at a county airport, which follows an exponential distribution.
To find the number of planes landing every hour, the Poisson distribution formula is used as it is more suited for discrete systems.
The Poisson formula is defined as
P(X = x) = (e^-λ)(λˣ)/x!
For this case, λ = 1 aircraft per hour.
The likelihood of having more than three aircraft arrivals in one hour is expressed as P(X > 3)
P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
P(X > 3) = 1 - 0.98101 = 0.01899 = 0.019 rounded to three decimal places
b) For 30 separate one-hour intervals selected, what is the probability that none of the intervals contain more than three arrivals?
The possibility that a single 1-hour interval has three or fewer arrivals equals 1 - P(X > 3)
= 1 - 0.01899 = 0.98101
The likelihood that none of the thirty 1-hour intervals exceed three arrivals is (0.98101)³⁰ = 0.5626 = 0.563 rounded to three decimal places
c) Calculate the duration of an interval (in hours) where the chance of no arrivals is 0.14
To solve this, we will apply the cumulative distribution function for an exponential random variable
P(X ≤ x) = 1 - e^(-λ.x)
P(X > x) = 1 - P(X ≤ x)
P(X > x) = e^(-λ.x)
Here, λ = 1, x =?,
0.14 = e⁻ˣ
e⁻ˣ = 0.14
Taking the natural logarithm: In e⁻ˣ = In 0.14 = -1.966
-x = -1.966
x = 1.966 hours
Hope this Helps!!!
