Expected number of a Poisson-distributed variable
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Fifty spotlights have just been installed in an outdoor security system. According to the manufacturer’s specifications, these particular lights are expected to burn
out at the rate of 1.1 per one hundred hours. What is the expected number of bulbs that will fail to last for at least seventy-five hours?
Here $lambda=1.1/100$ hours. My first idea was to find the average burnout rate for $75$ hour interval, which equals $3/4 cdot lambda=0.825/75$ hours. However, I do not understand how it can help to find the expected number of bulbs that will burn out within $75$ hours.
statistics expected-value
asked 2 days ago
V. Spitsyn
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up vote
1
down vote
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Fifty spotlights have just been installed in an outdoor security system. According to the manufacturer’s specifications, these particular lights are expected to burn
out at the rate of 1.1 per one hundred hours. What is the expected number of bulbs that will fail to last for at least seventy-five hours?
Here $lambda=1.1/100$ hours. My first idea was to find the average burnout rate for $75$ hour interval, which equals $3/4 cdot lambda=0.825/75$ hours. However, I do not understand how it can help to find the expected number of bulbs that will burn out within $75$ hours.
statistics expected-value
asked 2 days ago
V. Spitsyn
63
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Fifty spotlights have just been installed in an outdoor security system. According to the manufacturer’s specifications, these particular lights are expected to burn
out at the rate of 1.1 per one hundred hours. What is the expected number of bulbs that will fail to last for at least seventy-five hours?
Here $lambda=1.1/100$ hours. My first idea was to find the average burnout rate for $75$ hour interval, which equals $3/4 cdot lambda=0.825/75$ hours. However, I do not understand how it can help to find the expected number of bulbs that will burn out within $75$ hours.
statistics expected-value
asked 2 days ago
V. Spitsyn
63
Fifty spotlights have just been installed in an outdoor security system. According to the manufacturer’s specifications, these particular lights are expected to burn
out at the rate of 1.1 per one hundred hours. What is the expected number of bulbs that will fail to last for at least seventy-five hours?
Here $lambda=1.1/100$ hours. My first idea was to find the average burnout rate for $75$ hour interval, which equals $3/4 cdot lambda=0.825/75$ hours. However, I do not understand how it can help to find the expected number of bulbs that will burn out within $75$ hours.
statistics expected-value
statistics expected-value
asked 2 days ago
V. Spitsyn
63
asked 2 days ago
V. Spitsyn
63
asked 2 days ago
V. Spitsyn
63
asked 2 days ago
V. Spitsyn
63
asked 2 days ago
V. Spitsyn
63
63
add a comment |
add a comment |
2 Answers
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0
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The expected value of a Poisson distribution is equal to $lambda$ by definition. For your particular example, $E(x)=lambda=frac{1.1*75}{100}=0.825$, so you have already found the expected number of bulbs that will burn out. A common misconception is the the expected value must be an integer but that is not actually the case.
answered 2 days ago
3684
306
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I thought so too, but the right answer is $28$.
– V. Spitsyn
2 days ago
Any rough attempts to estimate the expected number of bulbs burning out will tell you that the expected number of bulbs burning out is less that $1.1$. This is because on average only $1.1$ bulbs burn out per $100$ hours and $75$ hours is less that 100 hours.
– 3684
2 days ago
add a comment |
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Think better of the exponential distribution, which defines the probability of the time that elapses until the occurrence of a particular event (or between events) in Poisson processes.
The cumulative density function for such distribution is, for positive time $t$, as follows:
$$F(t,lambda)=1-e^{-lambda,t}$$
Our unit measure of time is 100 hours. $lambda=1.1$ and the failure threshold we are interested in is $t=0.75$. The probability that a spotlight will fail before $t=0.75$ is
$$1-e^{-1.1*0.75}$$
which is, approximately, $0.561765$. Now, since we have 50 spotlights, $50*0.561765$ is the desired output, which is approximately 28.
answered 2 days ago
DavidPM
365
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The expected value of a Poisson distribution is equal to $lambda$ by definition. For your particular example, $E(x)=lambda=frac{1.1*75}{100}=0.825$, so you have already found the expected number of bulbs that will burn out. A common misconception is the the expected value must be an integer but that is not actually the case.
answered 2 days ago
3684
306
New contributor
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I thought so too, but the right answer is $28$.
– V. Spitsyn
2 days ago
Any rough attempts to estimate the expected number of bulbs burning out will tell you that the expected number of bulbs burning out is less that $1.1$. This is because on average only $1.1$ bulbs burn out per $100$ hours and $75$ hours is less that 100 hours.
– 3684
2 days ago
add a comment |
up vote
0
down vote
The expected value of a Poisson distribution is equal to $lambda$ by definition. For your particular example, $E(x)=lambda=frac{1.1*75}{100}=0.825$, so you have already found the expected number of bulbs that will burn out. A common misconception is the the expected value must be an integer but that is not actually the case.
answered 2 days ago
3684
306
New contributor
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I thought so too, but the right answer is $28$.
– V. Spitsyn
2 days ago
Any rough attempts to estimate the expected number of bulbs burning out will tell you that the expected number of bulbs burning out is less that $1.1$. This is because on average only $1.1$ bulbs burn out per $100$ hours and $75$ hours is less that 100 hours.
– 3684
2 days ago
add a comment |
up vote
0
down vote
up vote
0
down vote
The expected value of a Poisson distribution is equal to $lambda$ by definition. For your particular example, $E(x)=lambda=frac{1.1*75}{100}=0.825$, so you have already found the expected number of bulbs that will burn out. A common misconception is the the expected value must be an integer but that is not actually the case.
answered 2 days ago
3684
306
New contributor
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The expected value of a Poisson distribution is equal to $lambda$ by definition. For your particular example, $E(x)=lambda=frac{1.1*75}{100}=0.825$, so you have already found the expected number of bulbs that will burn out. A common misconception is the the expected value must be an integer but that is not actually the case.
answered 2 days ago
3684
306
New contributor
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
3684
306
New contributor
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
3684
306
answered 2 days ago
3684
306
306
New contributor
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I thought so too, but the right answer is $28$.
– V. Spitsyn
2 days ago
Any rough attempts to estimate the expected number of bulbs burning out will tell you that the expected number of bulbs burning out is less that $1.1$. This is because on average only $1.1$ bulbs burn out per $100$ hours and $75$ hours is less that 100 hours.
– 3684
2 days ago
add a comment |
I thought so too, but the right answer is $28$.
– V. Spitsyn
2 days ago
Any rough attempts to estimate the expected number of bulbs burning out will tell you that the expected number of bulbs burning out is less that $1.1$. This is because on average only $1.1$ bulbs burn out per $100$ hours and $75$ hours is less that 100 hours.
– 3684
2 days ago
I thought so too, but the right answer is $28$.
– V. Spitsyn
2 days ago
I thought so too, but the right answer is $28$.
– V. Spitsyn
2 days ago
Any rough attempts to estimate the expected number of bulbs burning out will tell you that the expected number of bulbs burning out is less that $1.1$. This is because on average only $1.1$ bulbs burn out per $100$ hours and $75$ hours is less that 100 hours.
– 3684
2 days ago
Any rough attempts to estimate the expected number of bulbs burning out will tell you that the expected number of bulbs burning out is less that $1.1$. This is because on average only $1.1$ bulbs burn out per $100$ hours and $75$ hours is less that 100 hours.
– 3684
2 days ago
add a comment |
up vote
0
down vote
Think better of the exponential distribution, which defines the probability of the time that elapses until the occurrence of a particular event (or between events) in Poisson processes.
The cumulative density function for such distribution is, for positive time $t$, as follows:
$$F(t,lambda)=1-e^{-lambda,t}$$
Our unit measure of time is 100 hours. $lambda=1.1$ and the failure threshold we are interested in is $t=0.75$. The probability that a spotlight will fail before $t=0.75$ is
$$1-e^{-1.1*0.75}$$
which is, approximately, $0.561765$. Now, since we have 50 spotlights, $50*0.561765$ is the desired output, which is approximately 28.
answered 2 days ago
DavidPM
365
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
Think better of the exponential distribution, which defines the probability of the time that elapses until the occurrence of a particular event (or between events) in Poisson processes.
The cumulative density function for such distribution is, for positive time $t$, as follows:
$$F(t,lambda)=1-e^{-lambda,t}$$
Our unit measure of time is 100 hours. $lambda=1.1$ and the failure threshold we are interested in is $t=0.75$. The probability that a spotlight will fail before $t=0.75$ is
$$1-e^{-1.1*0.75}$$
which is, approximately, $0.561765$. Now, since we have 50 spotlights, $50*0.561765$ is the desired output, which is approximately 28.
answered 2 days ago
DavidPM
365
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
Think better of the exponential distribution, which defines the probability of the time that elapses until the occurrence of a particular event (or between events) in Poisson processes.
The cumulative density function for such distribution is, for positive time $t$, as follows:
$$F(t,lambda)=1-e^{-lambda,t}$$
Our unit measure of time is 100 hours. $lambda=1.1$ and the failure threshold we are interested in is $t=0.75$. The probability that a spotlight will fail before $t=0.75$ is
$$1-e^{-1.1*0.75}$$
which is, approximately, $0.561765$. Now, since we have 50 spotlights, $50*0.561765$ is the desired output, which is approximately 28.
answered 2 days ago
DavidPM
365
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Think better of the exponential distribution, which defines the probability of the time that elapses until the occurrence of a particular event (or between events) in Poisson processes.
The cumulative density function for such distribution is, for positive time $t$, as follows:
$$F(t,lambda)=1-e^{-lambda,t}$$
Our unit measure of time is 100 hours. $lambda=1.1$ and the failure threshold we are interested in is $t=0.75$. The probability that a spotlight will fail before $t=0.75$ is
$$1-e^{-1.1*0.75}$$
which is, approximately, $0.561765$. Now, since we have 50 spotlights, $50*0.561765$ is the desired output, which is approximately 28.
answered 2 days ago
DavidPM
365
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered 2 days ago
DavidPM
365
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
DavidPM
365
answered 2 days ago
DavidPM
365
365
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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