Find the expected value and the variance of the time at which the plumber completes the project
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This is a problem 8 from Chapter 8 "Introduction to probability" by Anderson:
Our faucet is broken, and a plumber has been called. The arrival time
of the plumber is uniformly distributed between 1pm and 7pm. The time
required to fix the broken faucet is then exponentially distributed
with mean 30 minutes. Supposing that the two times are independent,
find the expected value and the variance of the time at which the
plumber completes the project.
I've solved it, but I am not sure if I interpreted the statement of the problem (translated into the language of probability distribution) correctly. So let X be the arrival time of the plumber, then $Xsim Unif[1,7]$. Let Y be the time it takes to fix the broken faucet, then $Ysim Exp(2)$. Let Z be the time at which the plumber completes the project, then $Z=X+Y$. Now $E[Z]=E[X+Y]=E[X]+E[Y]=frac{1+7}{2}+frac{1}{2}=4.5$
$Var(Z) = text{(since X and Y are independent)} Var(X)+Var(Y) = frac{{(7-1)}^2}{12}+frac{1}{2^2}=3.25$
Are there any mistakes in my solution? Thanks.
probability proof-verification
asked Dec 5 at 2:51
dxdydz
1929
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This is a problem 8 from Chapter 8 "Introduction to probability" by Anderson:
Our faucet is broken, and a plumber has been called. The arrival time
of the plumber is uniformly distributed between 1pm and 7pm. The time
required to fix the broken faucet is then exponentially distributed
with mean 30 minutes. Supposing that the two times are independent,
find the expected value and the variance of the time at which the
plumber completes the project.
I've solved it, but I am not sure if I interpreted the statement of the problem (translated into the language of probability distribution) correctly. So let X be the arrival time of the plumber, then $Xsim Unif[1,7]$. Let Y be the time it takes to fix the broken faucet, then $Ysim Exp(2)$. Let Z be the time at which the plumber completes the project, then $Z=X+Y$. Now $E[Z]=E[X+Y]=E[X]+E[Y]=frac{1+7}{2}+frac{1}{2}=4.5$
$Var(Z) = text{(since X and Y are independent)} Var(X)+Var(Y) = frac{{(7-1)}^2}{12}+frac{1}{2^2}=3.25$
Are there any mistakes in my solution? Thanks.
probability proof-verification
asked Dec 5 at 2:51
dxdydz
1929
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is a problem 8 from Chapter 8 "Introduction to probability" by Anderson:
Our faucet is broken, and a plumber has been called. The arrival time
of the plumber is uniformly distributed between 1pm and 7pm. The time
required to fix the broken faucet is then exponentially distributed
with mean 30 minutes. Supposing that the two times are independent,
find the expected value and the variance of the time at which the
plumber completes the project.
I've solved it, but I am not sure if I interpreted the statement of the problem (translated into the language of probability distribution) correctly. So let X be the arrival time of the plumber, then $Xsim Unif[1,7]$. Let Y be the time it takes to fix the broken faucet, then $Ysim Exp(2)$. Let Z be the time at which the plumber completes the project, then $Z=X+Y$. Now $E[Z]=E[X+Y]=E[X]+E[Y]=frac{1+7}{2}+frac{1}{2}=4.5$
$Var(Z) = text{(since X and Y are independent)} Var(X)+Var(Y) = frac{{(7-1)}^2}{12}+frac{1}{2^2}=3.25$
Are there any mistakes in my solution? Thanks.
probability proof-verification
asked Dec 5 at 2:51
dxdydz
1929
This is a problem 8 from Chapter 8 "Introduction to probability" by Anderson:
Our faucet is broken, and a plumber has been called. The arrival time
of the plumber is uniformly distributed between 1pm and 7pm. The time
required to fix the broken faucet is then exponentially distributed
with mean 30 minutes. Supposing that the two times are independent,
find the expected value and the variance of the time at which the
plumber completes the project.
I've solved it, but I am not sure if I interpreted the statement of the problem (translated into the language of probability distribution) correctly. So let X be the arrival time of the plumber, then $Xsim Unif[1,7]$. Let Y be the time it takes to fix the broken faucet, then $Ysim Exp(2)$. Let Z be the time at which the plumber completes the project, then $Z=X+Y$. Now $E[Z]=E[X+Y]=E[X]+E[Y]=frac{1+7}{2}+frac{1}{2}=4.5$
$Var(Z) = text{(since X and Y are independent)} Var(X)+Var(Y) = frac{{(7-1)}^2}{12}+frac{1}{2^2}=3.25$
Are there any mistakes in my solution? Thanks.
probability proof-verification
probability proof-verification
asked Dec 5 at 2:51
dxdydz
1929
asked Dec 5 at 2:51
dxdydz
1929
asked Dec 5 at 2:51
dxdydz
1929
asked Dec 5 at 2:51
dxdydz
1929
asked Dec 5 at 2:51
dxdydz
1929
1929
add a comment |
add a comment |
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Looks good. You have identified the distributions -and their mean and variance- correctly, and used the right equations -including the effect of independance-. If the calculations are okay, then you have it.
answered Dec 5 at 2:59
Graham Kemp
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1 Answer
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1 Answer
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up vote
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Looks good. You have identified the distributions -and their mean and variance- correctly, and used the right equations -including the effect of independance-. If the calculations are okay, then you have it.
answered Dec 5 at 2:59
Graham Kemp
84.7k43378
add a comment |
up vote
0
down vote
Looks good. You have identified the distributions -and their mean and variance- correctly, and used the right equations -including the effect of independance-. If the calculations are okay, then you have it.
answered Dec 5 at 2:59
Graham Kemp
84.7k43378
add a comment |
up vote
0
down vote
up vote
0
down vote
Looks good. You have identified the distributions -and their mean and variance- correctly, and used the right equations -including the effect of independance-. If the calculations are okay, then you have it.
answered Dec 5 at 2:59
Graham Kemp
84.7k43378
Looks good. You have identified the distributions -and their mean and variance- correctly, and used the right equations -including the effect of independance-. If the calculations are okay, then you have it.
answered Dec 5 at 2:59
Graham Kemp
84.7k43378
answered Dec 5 at 2:59
Graham Kemp
84.7k43378
answered Dec 5 at 2:59
Graham Kemp
84.7k43378
answered Dec 5 at 2:59
Graham Kemp
84.7k43378
84.7k43378
add a comment |
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