life expectancy of new computer with exponential distribution
$begingroup$
I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.
Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?
what I have:
f(x) = $lambda cdot e^{-lambda cdot x}$
given: $lambda = 3, x = 8$
so simply
$= 3 cdot e^{-3 cdot 8}
= 3cdot e^{-24}$?
sorry if this is too low level question. But I'm a bit confused.
probability statistics probability-distributions exponential-distribution
edited Dec 19 '18 at 18:59
twnly
742112
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
$endgroup$
add a comment |
$begingroup$
I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.
Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?
what I have:
f(x) = $lambda cdot e^{-lambda cdot x}$
given: $lambda = 3, x = 8$
so simply
$= 3 cdot e^{-3 cdot 8}
= 3cdot e^{-24}$?
sorry if this is too low level question. But I'm a bit confused.
probability statistics probability-distributions exponential-distribution
edited Dec 19 '18 at 18:59
twnly
742112
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
$endgroup$
add a comment |
$begingroup$
I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.
Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?
what I have:
f(x) = $lambda cdot e^{-lambda cdot x}$
given: $lambda = 3, x = 8$
so simply
$= 3 cdot e^{-3 cdot 8}
= 3cdot e^{-24}$?
sorry if this is too low level question. But I'm a bit confused.
probability statistics probability-distributions exponential-distribution
edited Dec 19 '18 at 18:59
twnly
742112
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
$endgroup$
I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.
Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?
what I have:
f(x) = $lambda cdot e^{-lambda cdot x}$
given: $lambda = 3, x = 8$
so simply
$= 3 cdot e^{-3 cdot 8}
= 3cdot e^{-24}$?
sorry if this is too low level question. But I'm a bit confused.
probability statistics probability-distributions exponential-distribution
probability statistics probability-distributions exponential-distribution
edited Dec 19 '18 at 18:59
twnly
742112
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
edited Dec 19 '18 at 18:59
twnly
742112
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
edited Dec 19 '18 at 18:59
twnly
742112
edited Dec 19 '18 at 18:59
twnly
742112
edited Dec 19 '18 at 18:59
twnly
742112
742112
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
asked Dec 19 '18 at 6:39
Daniel KimDaniel Kim
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.
By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.
You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).
Hope this helps.
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
$endgroup$
$begingroup$
Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
$endgroup$
– Daniel Kim
Dec 19 '18 at 7:05
$begingroup$
@DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
$endgroup$
– Aditya Dua
Dec 19 '18 at 7:13
add a comment |
$begingroup$
As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$
Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
begin{align*}
P(X > 8) &= 1 - P(X leq 8) \
& = 1 - F_X(8) \
& = 1 - (1-e^{-frac{1}{3}x}) \
& = e^{-frac{8}{3}}
end{align*}
This says that there is approximately 7% chance that the computer is still working after 8 years.
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
$endgroup$
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
$begingroup$
If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.
By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.
You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).
Hope this helps.
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
$endgroup$
$begingroup$
Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
$endgroup$
– Daniel Kim
Dec 19 '18 at 7:05
$begingroup$
@DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
$endgroup$
– Aditya Dua
Dec 19 '18 at 7:13
add a comment |
$begingroup$
If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.
By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.
You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).
Hope this helps.
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
$endgroup$
$begingroup$
Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
$endgroup$
– Daniel Kim
Dec 19 '18 at 7:05
$begingroup$
@DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
$endgroup$
– Aditya Dua
Dec 19 '18 at 7:13
add a comment |
$begingroup$
If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.
By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.
You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).
Hope this helps.
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
$endgroup$
If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.
By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.
You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).
Hope this helps.
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
answered Dec 19 '18 at 6:48
Aditya DuaAditya Dua
1,01418
1,01418
$begingroup$
Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
$endgroup$
– Daniel Kim
Dec 19 '18 at 7:05
$begingroup$
@DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
$endgroup$
– Aditya Dua
Dec 19 '18 at 7:13
add a comment |
$begingroup$
Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
$endgroup$
– Daniel Kim
Dec 19 '18 at 7:05
$begingroup$
@DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
$endgroup$
– Aditya Dua
Dec 19 '18 at 7:13
$begingroup$
Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
$endgroup$
– Daniel Kim
Dec 19 '18 at 7:05
$begingroup$
Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
$endgroup$
– Daniel Kim
Dec 19 '18 at 7:05
$begingroup$
@DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
$endgroup$
– Aditya Dua
Dec 19 '18 at 7:13
$begingroup$
@DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
$endgroup$
– Aditya Dua
Dec 19 '18 at 7:13
add a comment |
$begingroup$
As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$
Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
begin{align*}
P(X > 8) &= 1 - P(X leq 8) \
& = 1 - F_X(8) \
& = 1 - (1-e^{-frac{1}{3}x}) \
& = e^{-frac{8}{3}}
end{align*}
This says that there is approximately 7% chance that the computer is still working after 8 years.
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
$endgroup$
add a comment |
$begingroup$
As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$
Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
begin{align*}
P(X > 8) &= 1 - P(X leq 8) \
& = 1 - F_X(8) \
& = 1 - (1-e^{-frac{1}{3}x}) \
& = e^{-frac{8}{3}}
end{align*}
This says that there is approximately 7% chance that the computer is still working after 8 years.
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
$endgroup$
add a comment |
$begingroup$
As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$
Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
begin{align*}
P(X > 8) &= 1 - P(X leq 8) \
& = 1 - F_X(8) \
& = 1 - (1-e^{-frac{1}{3}x}) \
& = e^{-frac{8}{3}}
end{align*}
This says that there is approximately 7% chance that the computer is still working after 8 years.
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
$endgroup$
As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$
Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
begin{align*}
P(X > 8) &= 1 - P(X leq 8) \
& = 1 - F_X(8) \
& = 1 - (1-e^{-frac{1}{3}x}) \
& = e^{-frac{8}{3}}
end{align*}
This says that there is approximately 7% chance that the computer is still working after 8 years.
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
answered Dec 19 '18 at 17:27
Amin SammaraAmin Sammara
186
186
add a comment |
add a comment |
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