Question

he time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the answers to 3 decimal places. (a) What is the probability that more than three aircraft arrive within an hour? Enter your answer in accordance to the item a) of the question statement (b) If 30 separate one-hour intervals are chosen, what is the probability that no interval contains more than three arrivals? Enter your answer in accordance to the item b) of the question statement

Answer 1

A) 0.019

B) 0.563

Explanation:

a) We will use Poisson distribution formula to solve this;

The formula is given as;

P(X = x) = ((e^-λ) × (λˣ))/x!

Mean is 1. Thus;

λ = 1 aircraft/hour.

Thus, the probability that more than three aircrafts will arrive within an hour is written as; P(X > 3)

Thus;

P(X > 3) = 1 - P(X ≤ 3)

Thus;

1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

Solving through online calculator, we have;

P(X > 3) = 1 - 0.98101

P(X > 3) = 0.01899

To 3 decimal places, we have; P(X > 3)= 0.019

b) Probability of one 1-hour interval not containing more than 3 arrivals is, let's first find;

P(X ≤ 3) = 1 - P(X > 3)

P(X ≤ 3) = 1 - 0.01899

P(X ≤ 3) = 0.98101

Since there are 30 one-hour intervals, then we have;

Probability that none of the thirty 1-hour intervals will contain more than 3 arrivals;

(P ≤ 3) = (0.98101)³⁰

(P ≤ 3) = 0.5626

Approximating to 3 decimal places, we have;

(P ≤ 3) = 0.563