Poisson Distribution Variance Problem
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Here is the question-
The number of computer servers that break down during a month is a Poisson Random Variable with parameter $lambda = 2$. The cost of repairing one server is 2000 and also there is a fixed overhead cost of 10000 given as salary to the technician. If $X$ is the total expenditure made on repairs during a month find expectation and variance of $X$.
I managed to calculate the expectation as follows
$$mathbb{E}[X] = 10000 + 2000*mathbb{E}[C]$$ where $C$ is the number of computer servers that break down.
$$mathbb{E}[X] = 10000 + 2000*2 = 14000$$
However I can’t seem to calculate the variance of $X$, I know that $Var(C) = 2$.
The following method I know is wrong
$$Var(X) = (2000)^2*Var(C) = (2000)^2*2$$
So, how to calculate the same for $X$ ?
probability probability-distributions random-variables poisson-distribution variance
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Here is the question-
The number of computer servers that break down during a month is a Poisson Random Variable with parameter $lambda = 2$. The cost of repairing one server is 2000 and also there is a fixed overhead cost of 10000 given as salary to the technician. If $X$ is the total expenditure made on repairs during a month find expectation and variance of $X$.
I managed to calculate the expectation as follows
$$mathbb{E}[X] = 10000 + 2000*mathbb{E}[C]$$ where $C$ is the number of computer servers that break down.
$$mathbb{E}[X] = 10000 + 2000*2 = 14000$$
However I can’t seem to calculate the variance of $X$, I know that $Var(C) = 2$.
The following method I know is wrong
$$Var(X) = (2000)^2*Var(C) = (2000)^2*2$$
So, how to calculate the same for $X$ ?
probability probability-distributions random-variables poisson-distribution variance
the variance of $c+N$ where $N$ is a RV and $c$ is a constant is equal to the variance of $N$. Extremely useful and basic property of variance. Can you combine this with the other property you listed for scaled RVs to finish?
– LoveTooNap29
Nov 24 at 0:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here is the question-
The number of computer servers that break down during a month is a Poisson Random Variable with parameter $lambda = 2$. The cost of repairing one server is 2000 and also there is a fixed overhead cost of 10000 given as salary to the technician. If $X$ is the total expenditure made on repairs during a month find expectation and variance of $X$.
I managed to calculate the expectation as follows
$$mathbb{E}[X] = 10000 + 2000*mathbb{E}[C]$$ where $C$ is the number of computer servers that break down.
$$mathbb{E}[X] = 10000 + 2000*2 = 14000$$
However I can’t seem to calculate the variance of $X$, I know that $Var(C) = 2$.
The following method I know is wrong
$$Var(X) = (2000)^2*Var(C) = (2000)^2*2$$
So, how to calculate the same for $X$ ?
probability probability-distributions random-variables poisson-distribution variance
Here is the question-
The number of computer servers that break down during a month is a Poisson Random Variable with parameter $lambda = 2$. The cost of repairing one server is 2000 and also there is a fixed overhead cost of 10000 given as salary to the technician. If $X$ is the total expenditure made on repairs during a month find expectation and variance of $X$.
I managed to calculate the expectation as follows
$$mathbb{E}[X] = 10000 + 2000*mathbb{E}[C]$$ where $C$ is the number of computer servers that break down.
$$mathbb{E}[X] = 10000 + 2000*2 = 14000$$
However I can’t seem to calculate the variance of $X$, I know that $Var(C) = 2$.
The following method I know is wrong
$$Var(X) = (2000)^2*Var(C) = (2000)^2*2$$
So, how to calculate the same for $X$ ?
probability probability-distributions random-variables poisson-distribution variance
probability probability-distributions random-variables poisson-distribution variance
asked Nov 24 at 0:12
user601297
895
895
the variance of $c+N$ where $N$ is a RV and $c$ is a constant is equal to the variance of $N$. Extremely useful and basic property of variance. Can you combine this with the other property you listed for scaled RVs to finish?
– LoveTooNap29
Nov 24 at 0:16
add a comment |
the variance of $c+N$ where $N$ is a RV and $c$ is a constant is equal to the variance of $N$. Extremely useful and basic property of variance. Can you combine this with the other property you listed for scaled RVs to finish?
– LoveTooNap29
Nov 24 at 0:16
the variance of $c+N$ where $N$ is a RV and $c$ is a constant is equal to the variance of $N$. Extremely useful and basic property of variance. Can you combine this with the other property you listed for scaled RVs to finish?
– LoveTooNap29
Nov 24 at 0:16
the variance of $c+N$ where $N$ is a RV and $c$ is a constant is equal to the variance of $N$. Extremely useful and basic property of variance. Can you combine this with the other property you listed for scaled RVs to finish?
– LoveTooNap29
Nov 24 at 0:16
add a comment |
1 Answer
1
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0
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The method you used is correct. Notice that adding a constant does not change the variance of a random variable. Therefore $$
Var(X) = 2000^2 * Var(C)
$$
Hey! I checked the answer given at the back of the text and it is 16000. That is why I say that it is not correct. I don’t know how they’re getting 16000 though.
– user601297
Nov 24 at 9:47
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The method you used is correct. Notice that adding a constant does not change the variance of a random variable. Therefore $$
Var(X) = 2000^2 * Var(C)
$$
Hey! I checked the answer given at the back of the text and it is 16000. That is why I say that it is not correct. I don’t know how they’re getting 16000 though.
– user601297
Nov 24 at 9:47
add a comment |
up vote
0
down vote
The method you used is correct. Notice that adding a constant does not change the variance of a random variable. Therefore $$
Var(X) = 2000^2 * Var(C)
$$
Hey! I checked the answer given at the back of the text and it is 16000. That is why I say that it is not correct. I don’t know how they’re getting 16000 though.
– user601297
Nov 24 at 9:47
add a comment |
up vote
0
down vote
up vote
0
down vote
The method you used is correct. Notice that adding a constant does not change the variance of a random variable. Therefore $$
Var(X) = 2000^2 * Var(C)
$$
The method you used is correct. Notice that adding a constant does not change the variance of a random variable. Therefore $$
Var(X) = 2000^2 * Var(C)
$$
edited Nov 24 at 3:26
answered Nov 24 at 1:59
user609189
314
314
Hey! I checked the answer given at the back of the text and it is 16000. That is why I say that it is not correct. I don’t know how they’re getting 16000 though.
– user601297
Nov 24 at 9:47
add a comment |
Hey! I checked the answer given at the back of the text and it is 16000. That is why I say that it is not correct. I don’t know how they’re getting 16000 though.
– user601297
Nov 24 at 9:47
Hey! I checked the answer given at the back of the text and it is 16000. That is why I say that it is not correct. I don’t know how they’re getting 16000 though.
– user601297
Nov 24 at 9:47
Hey! I checked the answer given at the back of the text and it is 16000. That is why I say that it is not correct. I don’t know how they’re getting 16000 though.
– user601297
Nov 24 at 9:47
add a comment |
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the variance of $c+N$ where $N$ is a RV and $c$ is a constant is equal to the variance of $N$. Extremely useful and basic property of variance. Can you combine this with the other property you listed for scaled RVs to finish?
– LoveTooNap29
Nov 24 at 0:16