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Sagot :
Answer:
A) 0.019
B) 0.563
Step-by-step 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
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