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An observer counts 300 vehicles/hour at a specific highway location. Assume that the vehicle arrival at the Location is Poisson’s distribution, the probability of having one vehicle arriving over a 30 second time interval is.......
Correct answer is between '0.20,0.21'. Can you explain this answer?
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An observer counts 300 vehicles/hour at a specific highway location. A...
Average number of vehicles in 30 seconds interval:
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An observer counts 300 vehicles/hour at a specific highway location. A...
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To find the probability of a certain number of vehicles arriving in an hour, we can use the Poisson distribution formula:
P(X = k) = (e^(-λ) * λ^k) / k!
where λ is the mean number of arrivals in an hour, and k is the number of arrivals we want to find the probability for.
Since the observer counts 300 vehicles in an hour, we can assume that λ = 300.
Now, let's find the probability of exactly 200 vehicles arriving in an hour:
P(X = 200) = (e^(-300) * 300^200) / 200!
Using a calculator, we get:
P(X = 200) = 0.028
So the probability of exactly 200 vehicles arriving in an hour is 0.028 or 2.8%.
We can also find the probability of at least 250 vehicles arriving in an hour by adding up the probabilities of 250, 251, 252, and so on, up to infinity:
P(X ≥ 250) = ∑(e^(-300) * 300^k) / k!
where the sum is taken from k = 250 to infinity.
This is a bit tedious to calculate by hand, so using a calculator or statistical software is recommended.
Using a calculator, we get:
P(X ≥ 250) = 0.039
So the probability of at least 250 vehicles arriving in an hour is 0.039 or 3.9%.
To find the probability of a certain number of vehicles arriving in an hour, we can use the Poisson distribution formula:
P(X = k) = (e^(-λ) * λ^k) / k!
where λ is the mean number of arrivals in an hour, and k is the number of arrivals we want to find the probability for.
Since the observer counts 300 vehicles in an hour, we can assume that λ = 300.
Now, let's find the probability of exactly 200 vehicles arriving in an hour:
P(X = 200) = (e^(-300) * 300^200) / 200!
Using a calculator, we get:
P(X = 200) = 0.028
So the probability of exactly 200 vehicles arriving in an hour is 0.028 or 2.8%.
We can also find the probability of at least 250 vehicles arriving in an hour by adding up the probabilities of 250, 251, 252, and so on, up to infinity:
P(X ≥ 250) = ∑(e^(-300) * 300^k) / k!
where the sum is taken from k = 250 to infinity.
This is a bit tedious to calculate by hand, so using a calculator or statistical software is recommended.
Using a calculator, we get:
P(X ≥ 250) = 0.039
So the probability of at least 250 vehicles arriving in an hour is 0.039 or 3.9%.
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An observer counts 300 vehicles/hour at a specific highway location. Assume that the vehicle arrival at the Location is Poisson’s distribution, the probability of having one vehicle arriving over a 30 second time interval is.......Correct answer is between '0.20,0.21'. Can you explain this answer?
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