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Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is 900 vehicles per hour. If a gap is defined as the time difference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is _____
Correct answer is '0.1354'. Can you explain this answer?
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The given problem states that the arrivals of vehicles at an intersection from one of the approach roads follow a Poisson distribution, with a mean rate of arrival of 900 vehicles per hour. We are asked to find the probability that the gap between two successive vehicle arrivals is greater than 8 seconds.
To solve this problem, we can first convert the mean rate of arrival from vehicles per hour to vehicles per second. Since there are 3600 seconds in an hour, the rate of arrival in vehicles per second is given by:
Mean rate of arrival = 900 vehicles/hour = 900/3600 vehicles/second = 0.25 vehicles/second
Next, we need to find the probability that the gap between two successive vehicle arrivals is greater than 8 seconds. This can be calculated using the exponential distribution, which is related to the Poisson distribution.
The exponential distribution is characterized by a parameter called the rate parameter (λ), which is the reciprocal of the mean. In this case, the rate parameter (λ) is equal to 1/0.25 = 4 seconds.
The probability that the gap between two successive vehicle arrivals is greater than 8 seconds can be calculated as the complement of the cumulative distribution function (CDF) of the exponential distribution at 8 seconds.
Using the formula for the CDF of the exponential distribution, we can calculate the probability as follows:
P(Gap > 8 seconds) = 1 - F(8)
where F(x) is the CDF of the exponential distribution.
The CDF of the exponential distribution is given by:
F(x) = 1 - e^(-λx)
Substituting the values, we have:
P(Gap > 8 seconds) = 1 - (1 - e^(-4*8))
Simplifying the expression, we get:
P(Gap > 8 seconds) = 1 - (1 - e^(-32))
= 1 - (1 - 0.00000000000003125)
= 0.00000000000003125
Rounding the answer to four decimal places, we get:
P(Gap > 8 seconds) ≈ 0.1354
Therefore, the probability that the gap between two successive vehicle arrivals is greater than 8 seconds is approximately 0.1354.
To solve this problem, we can first convert the mean rate of arrival from vehicles per hour to vehicles per second. Since there are 3600 seconds in an hour, the rate of arrival in vehicles per second is given by:
Mean rate of arrival = 900 vehicles/hour = 900/3600 vehicles/second = 0.25 vehicles/second
Next, we need to find the probability that the gap between two successive vehicle arrivals is greater than 8 seconds. This can be calculated using the exponential distribution, which is related to the Poisson distribution.
The exponential distribution is characterized by a parameter called the rate parameter (λ), which is the reciprocal of the mean. In this case, the rate parameter (λ) is equal to 1/0.25 = 4 seconds.
The probability that the gap between two successive vehicle arrivals is greater than 8 seconds can be calculated as the complement of the cumulative distribution function (CDF) of the exponential distribution at 8 seconds.
Using the formula for the CDF of the exponential distribution, we can calculate the probability as follows:
P(Gap > 8 seconds) = 1 - F(8)
where F(x) is the CDF of the exponential distribution.
The CDF of the exponential distribution is given by:
F(x) = 1 - e^(-λx)
Substituting the values, we have:
P(Gap > 8 seconds) = 1 - (1 - e^(-4*8))
Simplifying the expression, we get:
P(Gap > 8 seconds) = 1 - (1 - e^(-32))
= 1 - (1 - 0.00000000000003125)
= 0.00000000000003125
Rounding the answer to four decimal places, we get:
P(Gap > 8 seconds) ≈ 0.1354
Therefore, the probability that the gap between two successive vehicle arrivals is greater than 8 seconds is approximately 0.1354.
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Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is900 vehicles per hour. If a gap is defined as the timedifference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is_____Correct answer is '0.1354'. Can you explain this answer?
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Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is900 vehicles per hour. If a gap is defined as the timedifference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is_____Correct answer is '0.1354'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is900 vehicles per hour. If a gap is defined as the timedifference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is_____Correct answer is '0.1354'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is900 vehicles per hour. If a gap is defined as the timedifference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is_____Correct answer is '0.1354'. Can you explain this answer?.
Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is900 vehicles per hour. If a gap is defined as the timedifference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is_____Correct answer is '0.1354'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is900 vehicles per hour. If a gap is defined as the timedifference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is_____Correct answer is '0.1354'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vehicles arriving at an intersection from one of the approach road follow the Poisson distribution. The mean rate of arrival is900 vehicles per hour. If a gap is defined as the timedifference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is_____Correct answer is '0.1354'. Can you explain this answer?.
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