How to solve this problem with Poisson distribution
$begingroup$
Problem:
A store owner observes that there are $3$ (in average) customers visiting the store per hour. He wants to find the probability that there are at least $1$ customer visiting his store in $10$ minute, using Poisson distribution.
As I read in a probability book, the solution is to use the Poisson distribution with rate $lambda = 3/6 = 1/2$ per $10$ minutes (since there are $3$ customers visiting the store per hour). But why don't we use the Poisson distribution with rate $lambda = 3$ per hour and re-state the question as "What is the probability that there are at least $6$ customers visiting the store within an hour".
I've tried both ways and they gave different answers. Some may argue that having at least $1$ customer visiting the store within $10$ minutes isn't equivalent to having at least $6$ customers visiting the store within an hour. But I think having $3$ customers visiting the store, in average, per hour isn't equivalent to having $1/2$ customer visiting the store, in average, per $10$ minutes.
Can anyone explain the solution of the book for me ?
probability-theory poisson-distribution poisson-process
edited Jan 2 at 9:33
Did
247k23223459
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
$endgroup$
add a comment |
$begingroup$
Problem:
A store owner observes that there are $3$ (in average) customers visiting the store per hour. He wants to find the probability that there are at least $1$ customer visiting his store in $10$ minute, using Poisson distribution.
As I read in a probability book, the solution is to use the Poisson distribution with rate $lambda = 3/6 = 1/2$ per $10$ minutes (since there are $3$ customers visiting the store per hour). But why don't we use the Poisson distribution with rate $lambda = 3$ per hour and re-state the question as "What is the probability that there are at least $6$ customers visiting the store within an hour".
I've tried both ways and they gave different answers. Some may argue that having at least $1$ customer visiting the store within $10$ minutes isn't equivalent to having at least $6$ customers visiting the store within an hour. But I think having $3$ customers visiting the store, in average, per hour isn't equivalent to having $1/2$ customer visiting the store, in average, per $10$ minutes.
Can anyone explain the solution of the book for me ?
probability-theory poisson-distribution poisson-process
edited Jan 2 at 9:33
Did
247k23223459
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
$endgroup$
add a comment |
$begingroup$
Problem:
A store owner observes that there are $3$ (in average) customers visiting the store per hour. He wants to find the probability that there are at least $1$ customer visiting his store in $10$ minute, using Poisson distribution.
As I read in a probability book, the solution is to use the Poisson distribution with rate $lambda = 3/6 = 1/2$ per $10$ minutes (since there are $3$ customers visiting the store per hour). But why don't we use the Poisson distribution with rate $lambda = 3$ per hour and re-state the question as "What is the probability that there are at least $6$ customers visiting the store within an hour".
I've tried both ways and they gave different answers. Some may argue that having at least $1$ customer visiting the store within $10$ minutes isn't equivalent to having at least $6$ customers visiting the store within an hour. But I think having $3$ customers visiting the store, in average, per hour isn't equivalent to having $1/2$ customer visiting the store, in average, per $10$ minutes.
Can anyone explain the solution of the book for me ?
probability-theory poisson-distribution poisson-process
edited Jan 2 at 9:33
Did
247k23223459
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
$endgroup$
Problem:
A store owner observes that there are $3$ (in average) customers visiting the store per hour. He wants to find the probability that there are at least $1$ customer visiting his store in $10$ minute, using Poisson distribution.
As I read in a probability book, the solution is to use the Poisson distribution with rate $lambda = 3/6 = 1/2$ per $10$ minutes (since there are $3$ customers visiting the store per hour). But why don't we use the Poisson distribution with rate $lambda = 3$ per hour and re-state the question as "What is the probability that there are at least $6$ customers visiting the store within an hour".
I've tried both ways and they gave different answers. Some may argue that having at least $1$ customer visiting the store within $10$ minutes isn't equivalent to having at least $6$ customers visiting the store within an hour. But I think having $3$ customers visiting the store, in average, per hour isn't equivalent to having $1/2$ customer visiting the store, in average, per $10$ minutes.
Can anyone explain the solution of the book for me ?
probability-theory poisson-distribution poisson-process
probability-theory poisson-distribution poisson-process
edited Jan 2 at 9:33
Did
247k23223459
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
edited Jan 2 at 9:33
Did
247k23223459
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
edited Jan 2 at 9:33
Did
247k23223459
edited Jan 2 at 9:33
Did
247k23223459
edited Jan 2 at 9:33
Did
247k23223459
247k23223459
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
asked Jan 2 at 8:52
HOANG GIANGHOANG GIANG
43
43
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The correct way to state the problem while using the $lambda = 3 text{hr}^{-1}$ rate is to ask what is the probability that at least $1$ customer arrives within $t = frac{1}{6} text{hr}$.
Because of the form of the probability distribution for the Poisson process $left(frac{(lambda t)^n e^{-lambda t}}{n!}right)$, as long as the $lambda t$ term and $n$ terms don't change, you'll have the same probability distribution. But trying to change $n$ while keeping $n/t$ constant will clearly change the probability distribution.
As a freebie, the probability that there is at least $1$ customer in the shop within ten minutes is:
$$mathbb{P}[N(1/3)geq1] = 1 - mathbb{P}[N(1/3)=0] = 1 - e^{-1/2}$$
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
The correct way to state the problem while using the $lambda = 3 text{hr}^{-1}$ rate is to ask what is the probability that at least $1$ customer arrives within $t = frac{1}{6} text{hr}$.
Because of the form of the probability distribution for the Poisson process $left(frac{(lambda t)^n e^{-lambda t}}{n!}right)$, as long as the $lambda t$ term and $n$ terms don't change, you'll have the same probability distribution. But trying to change $n$ while keeping $n/t$ constant will clearly change the probability distribution.
As a freebie, the probability that there is at least $1$ customer in the shop within ten minutes is:
$$mathbb{P}[N(1/3)geq1] = 1 - mathbb{P}[N(1/3)=0] = 1 - e^{-1/2}$$
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
$endgroup$
add a comment |
$begingroup$
The correct way to state the problem while using the $lambda = 3 text{hr}^{-1}$ rate is to ask what is the probability that at least $1$ customer arrives within $t = frac{1}{6} text{hr}$.
Because of the form of the probability distribution for the Poisson process $left(frac{(lambda t)^n e^{-lambda t}}{n!}right)$, as long as the $lambda t$ term and $n$ terms don't change, you'll have the same probability distribution. But trying to change $n$ while keeping $n/t$ constant will clearly change the probability distribution.
As a freebie, the probability that there is at least $1$ customer in the shop within ten minutes is:
$$mathbb{P}[N(1/3)geq1] = 1 - mathbb{P}[N(1/3)=0] = 1 - e^{-1/2}$$
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
$endgroup$
add a comment |
$begingroup$
The correct way to state the problem while using the $lambda = 3 text{hr}^{-1}$ rate is to ask what is the probability that at least $1$ customer arrives within $t = frac{1}{6} text{hr}$.
Because of the form of the probability distribution for the Poisson process $left(frac{(lambda t)^n e^{-lambda t}}{n!}right)$, as long as the $lambda t$ term and $n$ terms don't change, you'll have the same probability distribution. But trying to change $n$ while keeping $n/t$ constant will clearly change the probability distribution.
As a freebie, the probability that there is at least $1$ customer in the shop within ten minutes is:
$$mathbb{P}[N(1/3)geq1] = 1 - mathbb{P}[N(1/3)=0] = 1 - e^{-1/2}$$
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
$endgroup$
The correct way to state the problem while using the $lambda = 3 text{hr}^{-1}$ rate is to ask what is the probability that at least $1$ customer arrives within $t = frac{1}{6} text{hr}$.
Because of the form of the probability distribution for the Poisson process $left(frac{(lambda t)^n e^{-lambda t}}{n!}right)$, as long as the $lambda t$ term and $n$ terms don't change, you'll have the same probability distribution. But trying to change $n$ while keeping $n/t$ constant will clearly change the probability distribution.
As a freebie, the probability that there is at least $1$ customer in the shop within ten minutes is:
$$mathbb{P}[N(1/3)geq1] = 1 - mathbb{P}[N(1/3)=0] = 1 - e^{-1/2}$$
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
answered Jan 2 at 9:24
aghostinthefiguresaghostinthefigures
1,2351216
1,2351216
add a comment |
add a comment |
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