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 $X$ ?










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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















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 $X$ ?










share|cite|improve this question






















  • 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













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 $X$ ?










share|cite|improve this question













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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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


















  • 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










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 $$
Var(X) = 2000^2 * Var(C)
$$






share|cite|improve this answer























  • 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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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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 $$
Var(X) = 2000^2 * Var(C)
$$






share|cite|improve this answer























  • 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















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 $$
Var(X) = 2000^2 * Var(C)
$$






share|cite|improve this answer























  • 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













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 $$
Var(X) = 2000^2 * Var(C)
$$






share|cite|improve this answer














The method you used is correct. Notice that adding a constant does not change the variance of a random variable. Therefore $$
Var(X) = 2000^2 * Var(C)
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • 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


















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