Bus waiting time
Agneta takes the bus to school every morning. The buses goes 5 times every hour, but since Agneta has neither time table or a watch, she consider her waiting time as random. What is the probability that her shortest waiting time during a 5 day period is no more than 2.4 minutes.
I understand that Agneta's waiting time is a random variable, but I don't understand what distribution the random variable should have. If I knew the probability density function, then I would calculate the probability as
$$
P(zeta < 2.4) = int_0^{2.4} f_{zeta}(x)dx
$$
where
$$
zeta in xi_1 + xi_2 + xi_3 + xi_4 + xi_5
$$
where $xi_1,xi_2,...,xi_5$ are the probabilities for each day. Could it be done like this, or is their an easier way of doing this that I am totally missing?
probability
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
add a comment |
Agneta takes the bus to school every morning. The buses goes 5 times every hour, but since Agneta has neither time table or a watch, she consider her waiting time as random. What is the probability that her shortest waiting time during a 5 day period is no more than 2.4 minutes.
I understand that Agneta's waiting time is a random variable, but I don't understand what distribution the random variable should have. If I knew the probability density function, then I would calculate the probability as
$$
P(zeta < 2.4) = int_0^{2.4} f_{zeta}(x)dx
$$
where
$$
zeta in xi_1 + xi_2 + xi_3 + xi_4 + xi_5
$$
where $xi_1,xi_2,...,xi_5$ are the probabilities for each day. Could it be done like this, or is their an easier way of doing this that I am totally missing?
probability
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
Its a scaled beta distribution. See en.wikipedia.org/wiki/…
– Coolwater
Nov 28 '18 at 10:03
I would think it natural to assume that the busses leave at uniform intervals five times every hour, which means that the waiting time for any specific day is uniform on $[0, 12]$.
– Arthur
Nov 28 '18 at 10:13
@Arthur So, you mean that $$ begin{cases} f(t) = frac{1}{5}, 0 < t < 12 \ 0, annars end{cases} $$ or what?
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:30
That is not a probaility distribution function, it doesn't integrate to 1. The $frac15$ must be replaced with $frac1{12}$. But the remainder of your approach is a mystery to me , because summing up the probabilities makes no sense. Better think about what the probability is for any random day to have a waiting time of at least 2.4 min and try to go from there.
– Ingix
Nov 28 '18 at 10:47
My bad, of course it should be $$ f(t) = begin{cases} frac{1}{12}, 0 < t < 12 \ 0, otherwise end{cases} $$
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:50
add a comment |
Agneta takes the bus to school every morning. The buses goes 5 times every hour, but since Agneta has neither time table or a watch, she consider her waiting time as random. What is the probability that her shortest waiting time during a 5 day period is no more than 2.4 minutes.
I understand that Agneta's waiting time is a random variable, but I don't understand what distribution the random variable should have. If I knew the probability density function, then I would calculate the probability as
$$
P(zeta < 2.4) = int_0^{2.4} f_{zeta}(x)dx
$$
where
$$
zeta in xi_1 + xi_2 + xi_3 + xi_4 + xi_5
$$
where $xi_1,xi_2,...,xi_5$ are the probabilities for each day. Could it be done like this, or is their an easier way of doing this that I am totally missing?
probability
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
Agneta takes the bus to school every morning. The buses goes 5 times every hour, but since Agneta has neither time table or a watch, she consider her waiting time as random. What is the probability that her shortest waiting time during a 5 day period is no more than 2.4 minutes.
I understand that Agneta's waiting time is a random variable, but I don't understand what distribution the random variable should have. If I knew the probability density function, then I would calculate the probability as
$$
P(zeta < 2.4) = int_0^{2.4} f_{zeta}(x)dx
$$
where
$$
zeta in xi_1 + xi_2 + xi_3 + xi_4 + xi_5
$$
where $xi_1,xi_2,...,xi_5$ are the probabilities for each day. Could it be done like this, or is their an easier way of doing this that I am totally missing?
probability
probability
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
asked Nov 28 '18 at 10:01
Kristoffer Jerzy Linder
316
316
Its a scaled beta distribution. See en.wikipedia.org/wiki/…
– Coolwater
Nov 28 '18 at 10:03
I would think it natural to assume that the busses leave at uniform intervals five times every hour, which means that the waiting time for any specific day is uniform on $[0, 12]$.
– Arthur
Nov 28 '18 at 10:13
@Arthur So, you mean that $$ begin{cases} f(t) = frac{1}{5}, 0 < t < 12 \ 0, annars end{cases} $$ or what?
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:30
That is not a probaility distribution function, it doesn't integrate to 1. The $frac15$ must be replaced with $frac1{12}$. But the remainder of your approach is a mystery to me , because summing up the probabilities makes no sense. Better think about what the probability is for any random day to have a waiting time of at least 2.4 min and try to go from there.
– Ingix
Nov 28 '18 at 10:47
My bad, of course it should be $$ f(t) = begin{cases} frac{1}{12}, 0 < t < 12 \ 0, otherwise end{cases} $$
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:50
add a comment |
Its a scaled beta distribution. See en.wikipedia.org/wiki/…
– Coolwater
Nov 28 '18 at 10:03
I would think it natural to assume that the busses leave at uniform intervals five times every hour, which means that the waiting time for any specific day is uniform on $[0, 12]$.
– Arthur
Nov 28 '18 at 10:13
@Arthur So, you mean that $$ begin{cases} f(t) = frac{1}{5}, 0 < t < 12 \ 0, annars end{cases} $$ or what?
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:30
That is not a probaility distribution function, it doesn't integrate to 1. The $frac15$ must be replaced with $frac1{12}$. But the remainder of your approach is a mystery to me , because summing up the probabilities makes no sense. Better think about what the probability is for any random day to have a waiting time of at least 2.4 min and try to go from there.
– Ingix
Nov 28 '18 at 10:47
My bad, of course it should be $$ f(t) = begin{cases} frac{1}{12}, 0 < t < 12 \ 0, otherwise end{cases} $$
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:50
Its a scaled beta distribution. See en.wikipedia.org/wiki/…
– Coolwater
Nov 28 '18 at 10:03
Its a scaled beta distribution. See en.wikipedia.org/wiki/…
– Coolwater
Nov 28 '18 at 10:03
I would think it natural to assume that the busses leave at uniform intervals five times every hour, which means that the waiting time for any specific day is uniform on $[0, 12]$.
– Arthur
Nov 28 '18 at 10:13
I would think it natural to assume that the busses leave at uniform intervals five times every hour, which means that the waiting time for any specific day is uniform on $[0, 12]$.
– Arthur
Nov 28 '18 at 10:13
@Arthur So, you mean that $$ begin{cases} f(t) = frac{1}{5}, 0 < t < 12 \ 0, annars end{cases} $$ or what?
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:30
@Arthur So, you mean that $$ begin{cases} f(t) = frac{1}{5}, 0 < t < 12 \ 0, annars end{cases} $$ or what?
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:30
That is not a probaility distribution function, it doesn't integrate to 1. The $frac15$ must be replaced with $frac1{12}$. But the remainder of your approach is a mystery to me , because summing up the probabilities makes no sense. Better think about what the probability is for any random day to have a waiting time of at least 2.4 min and try to go from there.
– Ingix
Nov 28 '18 at 10:47
That is not a probaility distribution function, it doesn't integrate to 1. The $frac15$ must be replaced with $frac1{12}$. But the remainder of your approach is a mystery to me , because summing up the probabilities makes no sense. Better think about what the probability is for any random day to have a waiting time of at least 2.4 min and try to go from there.
– Ingix
Nov 28 '18 at 10:47
My bad, of course it should be $$ f(t) = begin{cases} frac{1}{12}, 0 < t < 12 \ 0, otherwise end{cases} $$
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:50
My bad, of course it should be $$ f(t) = begin{cases} frac{1}{12}, 0 < t < 12 \ 0, otherwise end{cases} $$
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:50
add a comment |
1 Answer
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It's safe to assume from common sense that bus arrivals distributed uniformly throughout the hour and not random. That means the interval between two busses is 12 minutes, and Agneta comes at a random point in that interval. So the distribution of her waiting time is uniform $tsim U_{[0,12]}$. The probability that minimum is less that 2.4 is
$$
P(min t_i<2.4) = 1-P(min t_i >2.4) = 1-prod P(t_i>2.4)=1-left(frac{12-2.4}{12}right)^5=0.67
$$
If the busses are random and their timetable works like Poisson proceess, then the distribution of waiting time would be exponential.
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
add a comment |
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1 Answer
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It's safe to assume from common sense that bus arrivals distributed uniformly throughout the hour and not random. That means the interval between two busses is 12 minutes, and Agneta comes at a random point in that interval. So the distribution of her waiting time is uniform $tsim U_{[0,12]}$. The probability that minimum is less that 2.4 is
$$
P(min t_i<2.4) = 1-P(min t_i >2.4) = 1-prod P(t_i>2.4)=1-left(frac{12-2.4}{12}right)^5=0.67
$$
If the busses are random and their timetable works like Poisson proceess, then the distribution of waiting time would be exponential.
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
add a comment |
It's safe to assume from common sense that bus arrivals distributed uniformly throughout the hour and not random. That means the interval between two busses is 12 minutes, and Agneta comes at a random point in that interval. So the distribution of her waiting time is uniform $tsim U_{[0,12]}$. The probability that minimum is less that 2.4 is
$$
P(min t_i<2.4) = 1-P(min t_i >2.4) = 1-prod P(t_i>2.4)=1-left(frac{12-2.4}{12}right)^5=0.67
$$
If the busses are random and their timetable works like Poisson proceess, then the distribution of waiting time would be exponential.
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
add a comment |
It's safe to assume from common sense that bus arrivals distributed uniformly throughout the hour and not random. That means the interval between two busses is 12 minutes, and Agneta comes at a random point in that interval. So the distribution of her waiting time is uniform $tsim U_{[0,12]}$. The probability that minimum is less that 2.4 is
$$
P(min t_i<2.4) = 1-P(min t_i >2.4) = 1-prod P(t_i>2.4)=1-left(frac{12-2.4}{12}right)^5=0.67
$$
If the busses are random and their timetable works like Poisson proceess, then the distribution of waiting time would be exponential.
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
It's safe to assume from common sense that bus arrivals distributed uniformly throughout the hour and not random. That means the interval between two busses is 12 minutes, and Agneta comes at a random point in that interval. So the distribution of her waiting time is uniform $tsim U_{[0,12]}$. The probability that minimum is less that 2.4 is
$$
P(min t_i<2.4) = 1-P(min t_i >2.4) = 1-prod P(t_i>2.4)=1-left(frac{12-2.4}{12}right)^5=0.67
$$
If the busses are random and their timetable works like Poisson proceess, then the distribution of waiting time would be exponential.
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
answered Nov 28 '18 at 10:49
Vasily Mitch
1,33837
1,33837
add a comment |
add a comment |
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Its a scaled beta distribution. See en.wikipedia.org/wiki/…
– Coolwater
Nov 28 '18 at 10:03
I would think it natural to assume that the busses leave at uniform intervals five times every hour, which means that the waiting time for any specific day is uniform on $[0, 12]$.
– Arthur
Nov 28 '18 at 10:13
@Arthur So, you mean that $$ begin{cases} f(t) = frac{1}{5}, 0 < t < 12 \ 0, annars end{cases} $$ or what?
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:30
That is not a probaility distribution function, it doesn't integrate to 1. The $frac15$ must be replaced with $frac1{12}$. But the remainder of your approach is a mystery to me , because summing up the probabilities makes no sense. Better think about what the probability is for any random day to have a waiting time of at least 2.4 min and try to go from there.
– Ingix
Nov 28 '18 at 10:47
My bad, of course it should be $$ f(t) = begin{cases} frac{1}{12}, 0 < t < 12 \ 0, otherwise end{cases} $$
– Kristoffer Jerzy Linder
Nov 28 '18 at 10:50