Compute the probability of waiting more than $30$ seconds for the next call.
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Question: At ABC insurance company, suppose the patient insurance inquiries arrive at mean rate of $2.2$ calls per minute. Compute the probability of waiting more than $30$ seconds for the next call.
I am confused here.
When I first read the question, I thought of modelling the situation using Poisson distribution because of $2.2$ calls per minute.
However, the question is asking for time between two calls, which is not discrete.
If I follow my initial thought to solve the question, then define $X$ to be the number of calls per $30$ seconds, then from the question, $lambda = 2.2 /2 = 1.1.$
Then I think we need to calculate $mathbbP(X = 1).$
probability poisson-distribution
asked Sep 6 at 5:08
Idonknow
3,115644109
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up vote
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down vote
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Question: At ABC insurance company, suppose the patient insurance inquiries arrive at mean rate of $2.2$ calls per minute. Compute the probability of waiting more than $30$ seconds for the next call.
I am confused here.
When I first read the question, I thought of modelling the situation using Poisson distribution because of $2.2$ calls per minute.
However, the question is asking for time between two calls, which is not discrete.
If I follow my initial thought to solve the question, then define $X$ to be the number of calls per $30$ seconds, then from the question, $lambda = 2.2 /2 = 1.1.$
Then I think we need to calculate $mathbbP(X = 1).$
probability poisson-distribution
asked Sep 6 at 5:08
Idonknow
3,115644109
The Poisson model is appropriate. Under this model, the time until the next call follows an exponential distribution with mean $frac12.2$.
– Yuta
Sep 6 at 5:11
1
Waiting more than $30$ seconds for the next call means there is no calls in the first $30$-second interval. So the required probability should be $mathbbP(X=0)$.
– Yuta
Sep 6 at 5:17
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question: At ABC insurance company, suppose the patient insurance inquiries arrive at mean rate of $2.2$ calls per minute. Compute the probability of waiting more than $30$ seconds for the next call.
I am confused here.
When I first read the question, I thought of modelling the situation using Poisson distribution because of $2.2$ calls per minute.
However, the question is asking for time between two calls, which is not discrete.
If I follow my initial thought to solve the question, then define $X$ to be the number of calls per $30$ seconds, then from the question, $lambda = 2.2 /2 = 1.1.$
Then I think we need to calculate $mathbbP(X = 1).$
probability poisson-distribution
asked Sep 6 at 5:08
Idonknow
3,115644109
Question: At ABC insurance company, suppose the patient insurance inquiries arrive at mean rate of $2.2$ calls per minute. Compute the probability of waiting more than $30$ seconds for the next call.
I am confused here.
When I first read the question, I thought of modelling the situation using Poisson distribution because of $2.2$ calls per minute.
However, the question is asking for time between two calls, which is not discrete.
If I follow my initial thought to solve the question, then define $X$ to be the number of calls per $30$ seconds, then from the question, $lambda = 2.2 /2 = 1.1.$
Then I think we need to calculate $mathbbP(X = 1).$
probability poisson-distribution
probability poisson-distribution
asked Sep 6 at 5:08
Idonknow
3,115644109
asked Sep 6 at 5:08
Idonknow
3,115644109
edited Sep 6 at 5:11
asked Sep 6 at 5:08
Idonknow
3,115644109
asked Sep 6 at 5:08
Idonknow
3,115644109
asked Sep 6 at 5:08
Idonknow
3,115644109
3,115644109
The Poisson model is appropriate. Under this model, the time until the next call follows an exponential distribution with mean $frac12.2$.
– Yuta
Sep 6 at 5:11
1
Waiting more than $30$ seconds for the next call means there is no calls in the first $30$-second interval. So the required probability should be $mathbbP(X=0)$.
– Yuta
Sep 6 at 5:17
add a comment |Â
The Poisson model is appropriate. Under this model, the time until the next call follows an exponential distribution with mean $frac12.2$.
– Yuta
Sep 6 at 5:11
1
Waiting more than $30$ seconds for the next call means there is no calls in the first $30$-second interval. So the required probability should be $mathbbP(X=0)$.
– Yuta
Sep 6 at 5:17
The Poisson model is appropriate. Under this model, the time until the next call follows an exponential distribution with mean $frac12.2$.
– Yuta
Sep 6 at 5:11
The Poisson model is appropriate. Under this model, the time until the next call follows an exponential distribution with mean $frac12.2$.
– Yuta
Sep 6 at 5:11
1
1
Waiting more than $30$ seconds for the next call means there is no calls in the first $30$-second interval. So the required probability should be $mathbbP(X=0)$.
– Yuta
Sep 6 at 5:17
Waiting more than $30$ seconds for the next call means there is no calls in the first $30$-second interval. So the required probability should be $mathbbP(X=0)$.
– Yuta
Sep 6 at 5:17
add a comment |Â
1 Answer
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I believe you can solve using either.
Using Poisson
Let $X$ be the number of calls in $30$ seconds. Then $X sim Poisson(lambda = 1.1)$ and $P(X=0) = e^-1.1$
Using Exponential
Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.
That means $E[Y] = 27.27$ which means $frac1lambda = 27.27$ so $lambda = .0366$ and then $P(Y > 30) = 1-P(Y le 30) = 1-(1-e^-30 *.0366) = e^-1.1$
I think the key thing is that $lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.
answered Sep 6 at 5:55
HJ_beginner
679215
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I believe you can solve using either.
Using Poisson
Let $X$ be the number of calls in $30$ seconds. Then $X sim Poisson(lambda = 1.1)$ and $P(X=0) = e^-1.1$
Using Exponential
Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.
That means $E[Y] = 27.27$ which means $frac1lambda = 27.27$ so $lambda = .0366$ and then $P(Y > 30) = 1-P(Y le 30) = 1-(1-e^-30 *.0366) = e^-1.1$
I think the key thing is that $lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.
answered Sep 6 at 5:55
HJ_beginner
679215
add a comment |Â
up vote
1
down vote
accepted
I believe you can solve using either.
Using Poisson
Let $X$ be the number of calls in $30$ seconds. Then $X sim Poisson(lambda = 1.1)$ and $P(X=0) = e^-1.1$
Using Exponential
Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.
That means $E[Y] = 27.27$ which means $frac1lambda = 27.27$ so $lambda = .0366$ and then $P(Y > 30) = 1-P(Y le 30) = 1-(1-e^-30 *.0366) = e^-1.1$
I think the key thing is that $lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.
answered Sep 6 at 5:55
HJ_beginner
679215
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I believe you can solve using either.
Using Poisson
Let $X$ be the number of calls in $30$ seconds. Then $X sim Poisson(lambda = 1.1)$ and $P(X=0) = e^-1.1$
Using Exponential
Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.
That means $E[Y] = 27.27$ which means $frac1lambda = 27.27$ so $lambda = .0366$ and then $P(Y > 30) = 1-P(Y le 30) = 1-(1-e^-30 *.0366) = e^-1.1$
I think the key thing is that $lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.
answered Sep 6 at 5:55
HJ_beginner
679215
I believe you can solve using either.
Using Poisson
Let $X$ be the number of calls in $30$ seconds. Then $X sim Poisson(lambda = 1.1)$ and $P(X=0) = e^-1.1$
Using Exponential
Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.
That means $E[Y] = 27.27$ which means $frac1lambda = 27.27$ so $lambda = .0366$ and then $P(Y > 30) = 1-P(Y le 30) = 1-(1-e^-30 *.0366) = e^-1.1$
I think the key thing is that $lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.
answered Sep 6 at 5:55
HJ_beginner
679215
edited Sep 6 at 6:23
answered Sep 6 at 5:55
HJ_beginner
679215
answered Sep 6 at 5:55
HJ_beginner
679215
answered Sep 6 at 5:55
HJ_beginner
679215
679215
add a comment |Â
add a comment |Â
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The Poisson model is appropriate. Under this model, the time until the next call follows an exponential distribution with mean $frac12.2$.
– Yuta
Sep 6 at 5:11
1
Waiting more than $30$ seconds for the next call means there is no calls in the first $30$-second interval. So the required probability should be $mathbbP(X=0)$.
– Yuta
Sep 6 at 5:17