probability - vehicle arriving and gap in between them
$begingroup$
Vehicle arriving at an intersection from one of the approach roads 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 vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.
My approach for solving above question:
If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.
The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.
probability poisson-distribution
asked Dec 20 '18 at 10:37
swapnilswapnil
335
$endgroup$
add a comment |
$begingroup$
Vehicle arriving at an intersection from one of the approach roads 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 vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.
My approach for solving above question:
If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.
The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.
probability poisson-distribution
asked Dec 20 '18 at 10:37
swapnilswapnil
335
$endgroup$
$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53
add a comment |
$begingroup$
Vehicle arriving at an intersection from one of the approach roads 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 vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.
My approach for solving above question:
If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.
The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.
probability poisson-distribution
asked Dec 20 '18 at 10:37
swapnilswapnil
335
$endgroup$
Vehicle arriving at an intersection from one of the approach roads 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 vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.
My approach for solving above question:
If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.
The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.
probability poisson-distribution
probability poisson-distribution
asked Dec 20 '18 at 10:37
swapnilswapnil
335
asked Dec 20 '18 at 10:37
swapnilswapnil
335
asked Dec 20 '18 at 10:37
swapnilswapnil
335
asked Dec 20 '18 at 10:37
swapnilswapnil
335
asked Dec 20 '18 at 10:37
swapnilswapnil
335
335
$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53
add a comment |
$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53
$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53
$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53
add a comment |
1 Answer
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$begingroup$
As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
$$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
$$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
$$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
$$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
$$ F_n(t) = lambda e^{-lambda t}.$$
Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
The probability that the gap is greater than $8$ seconds is
$$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
which you can easily evaluate.
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
$$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
$$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
$$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
$$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
$$ F_n(t) = lambda e^{-lambda t}.$$
Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
The probability that the gap is greater than $8$ seconds is
$$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
which you can easily evaluate.
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
$endgroup$
add a comment |
$begingroup$
As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
$$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
$$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
$$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
$$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
$$ F_n(t) = lambda e^{-lambda t}.$$
Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
The probability that the gap is greater than $8$ seconds is
$$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
which you can easily evaluate.
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
$endgroup$
add a comment |
$begingroup$
As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
$$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
$$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
$$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
$$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
$$ F_n(t) = lambda e^{-lambda t}.$$
Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
The probability that the gap is greater than $8$ seconds is
$$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
which you can easily evaluate.
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
$endgroup$
As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
$$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
$$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
$$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
$$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
$$ F_n(t) = lambda e^{-lambda t}.$$
Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
The probability that the gap is greater than $8$ seconds is
$$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
which you can easily evaluate.
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
edited Feb 21 at 21:17
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
answered Dec 22 '18 at 3:15
kevinkayakskevinkayaks
1558
1558
add a comment |
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$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53