Mixing
Expected number of a Poisson-distributed variable
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 Nov 21 '18 at 13:14
V. Spitsyn
113
add a comment |
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 Nov 21 '18 at 13:14
V. Spitsyn
113
add a comment |
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 Nov 21 '18 at 13:14
V. Spitsyn
113
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 Nov 21 '18 at 13:14
V. Spitsyn
113
asked Nov 21 '18 at 13:14
V. Spitsyn
113
asked Nov 21 '18 at 13:14
V. Spitsyn
113
asked Nov 21 '18 at 13:14
V. Spitsyn
113
asked Nov 21 '18 at 13:14
V. Spitsyn
113
113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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 Nov 21 '18 at 13:31
3684
1277
I thought so too, but the right answer is $28$.
– V. Spitsyn
Nov 21 '18 at 13:41
1
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
Nov 21 '18 at 13:54
add a comment |
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 Nov 21 '18 at 17:14
DavidPM
26618
In R statistical softwareround(50*pexp(75,1.1/100))returns 28.
– BruceET
Nov 24 '18 at 23:31
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Nov 21 '18 at 13:31
3684
1277
I thought so too, but the right answer is $28$.
– V. Spitsyn
Nov 21 '18 at 13:41
1
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
Nov 21 '18 at 13:54
add a comment |
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 Nov 21 '18 at 13:31
3684
1277
I thought so too, but the right answer is $28$.
– V. Spitsyn
Nov 21 '18 at 13:41
1
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
Nov 21 '18 at 13:54
add a comment |
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 Nov 21 '18 at 13:31
3684
1277
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 Nov 21 '18 at 13:31
3684
1277
answered Nov 21 '18 at 13:31
3684
1277
answered Nov 21 '18 at 13:31
3684
1277
answered Nov 21 '18 at 13:31
3684
1277
1277
I thought so too, but the right answer is $28$.
– V. Spitsyn
Nov 21 '18 at 13:41
1
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
Nov 21 '18 at 13:54
add a comment |
I thought so too, but the right answer is $28$.
– V. Spitsyn
Nov 21 '18 at 13:41
1
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
Nov 21 '18 at 13:54
I thought so too, but the right answer is $28$.
– V. Spitsyn
Nov 21 '18 at 13:41
I thought so too, but the right answer is $28$.
– V. Spitsyn
Nov 21 '18 at 13:41
1
1
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
Nov 21 '18 at 13:54
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
Nov 21 '18 at 13:54
add a comment |
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 Nov 21 '18 at 17:14
DavidPM
26618
In R statistical softwareround(50*pexp(75,1.1/100))returns 28.
– BruceET
Nov 24 '18 at 23:31
add a comment |
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 Nov 21 '18 at 17:14
DavidPM
26618
In R statistical softwareround(50*pexp(75,1.1/100))returns 28.
– BruceET
Nov 24 '18 at 23:31
add a comment |
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 Nov 21 '18 at 17:14
DavidPM
26618
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 Nov 21 '18 at 17:14
DavidPM
26618
edited Nov 21 '18 at 17:19
answered Nov 21 '18 at 17:14
DavidPM
26618
answered Nov 21 '18 at 17:14
DavidPM
26618
answered Nov 21 '18 at 17:14
DavidPM
26618
26618
In R statistical softwareround(50*pexp(75,1.1/100))returns 28.
– BruceET
Nov 24 '18 at 23:31
add a comment |
In R statistical softwareround(50*pexp(75,1.1/100))returns 28.
– BruceET
Nov 24 '18 at 23:31
In R statistical software
round(50*pexp(75,1.1/100)) returns 28.– BruceET
Nov 24 '18 at 23:31
In R statistical software
round(50*pexp(75,1.1/100)) returns 28.– BruceET
Nov 24 '18 at 23:31
add a comment |
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