How to calculate new standard deviation based on formula given only the mean and set standard deviations
0
$begingroup$
I currently have the mean and standard deviation, I need to calculate the new standard deviation given a formula:
µX = 9.5 and µY = 6.8
σX = 0.4 and σY = 0.1
with the equation X-Y
What steps do I go though to calculate the new mean and standard deviation?
Thanks in advance!
statistics standard-deviation means
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
$endgroup$
|
show 1 more comment
0
$begingroup$
I currently have the mean and standard deviation, I need to calculate the new standard deviation given a formula:
µX = 9.5 and µY = 6.8
σX = 0.4 and σY = 0.1
with the equation X-Y
What steps do I go though to calculate the new mean and standard deviation?
Thanks in advance!
statistics standard-deviation means
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
$endgroup$
$begingroup$
If $X$ and $Y$ are independent, then the variance of $aX+bY$ is $a^2$ times the variance of $X$ plus $b^2$ times the variance of $Y$. Here $a=1$ and $b=-1$. But the post does not say $X$ and $Y$ are independent. With just the information given, we cannot compute the variance of $X-Y$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:31
$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Jesse P Francis
Oct 31 '15 at 17:35
$begingroup$
That makes sense, however, I know that the answer to the mean is -2.7 and the standard deviation should be 0.412, however, with these variables, I cannot seem to figure out the standard deviation after the equation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:35
1
$begingroup$
@JessePFrancis, I will look over these, thanks!
$endgroup$
– Jake Orben
Oct 31 '15 at 17:36
$begingroup$
For independent $X$ and $Y$, the formula I mentioned gives that the standard deviation of $X-Y$ is $sqrt{(0.4)^2+(0.1)^2}$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:41
|
show 1 more comment
0
0
0
$begingroup$
I currently have the mean and standard deviation, I need to calculate the new standard deviation given a formula:
µX = 9.5 and µY = 6.8
σX = 0.4 and σY = 0.1
with the equation X-Y
What steps do I go though to calculate the new mean and standard deviation?
Thanks in advance!
statistics standard-deviation means
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
$endgroup$
I currently have the mean and standard deviation, I need to calculate the new standard deviation given a formula:
µX = 9.5 and µY = 6.8
σX = 0.4 and σY = 0.1
with the equation X-Y
What steps do I go though to calculate the new mean and standard deviation?
Thanks in advance!
statistics standard-deviation means
statistics standard-deviation means
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
asked Oct 31 '15 at 17:26
Jake OrbenJake Orben
101
101
$begingroup$
If $X$ and $Y$ are independent, then the variance of $aX+bY$ is $a^2$ times the variance of $X$ plus $b^2$ times the variance of $Y$. Here $a=1$ and $b=-1$. But the post does not say $X$ and $Y$ are independent. With just the information given, we cannot compute the variance of $X-Y$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:31
$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Jesse P Francis
Oct 31 '15 at 17:35
$begingroup$
That makes sense, however, I know that the answer to the mean is -2.7 and the standard deviation should be 0.412, however, with these variables, I cannot seem to figure out the standard deviation after the equation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:35
1
$begingroup$
@JessePFrancis, I will look over these, thanks!
$endgroup$
– Jake Orben
Oct 31 '15 at 17:36
$begingroup$
For independent $X$ and $Y$, the formula I mentioned gives that the standard deviation of $X-Y$ is $sqrt{(0.4)^2+(0.1)^2}$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:41
|
show 1 more comment
$begingroup$
If $X$ and $Y$ are independent, then the variance of $aX+bY$ is $a^2$ times the variance of $X$ plus $b^2$ times the variance of $Y$. Here $a=1$ and $b=-1$. But the post does not say $X$ and $Y$ are independent. With just the information given, we cannot compute the variance of $X-Y$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:31
$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Jesse P Francis
Oct 31 '15 at 17:35
$begingroup$
That makes sense, however, I know that the answer to the mean is -2.7 and the standard deviation should be 0.412, however, with these variables, I cannot seem to figure out the standard deviation after the equation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:35
1
$begingroup$
@JessePFrancis, I will look over these, thanks!
$endgroup$
– Jake Orben
Oct 31 '15 at 17:36
$begingroup$
For independent $X$ and $Y$, the formula I mentioned gives that the standard deviation of $X-Y$ is $sqrt{(0.4)^2+(0.1)^2}$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:41
$begingroup$
If $X$ and $Y$ are independent, then the variance of $aX+bY$ is $a^2$ times the variance of $X$ plus $b^2$ times the variance of $Y$. Here $a=1$ and $b=-1$. But the post does not say $X$ and $Y$ are independent. With just the information given, we cannot compute the variance of $X-Y$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:31
$begingroup$
If $X$ and $Y$ are independent, then the variance of $aX+bY$ is $a^2$ times the variance of $X$ plus $b^2$ times the variance of $Y$. Here $a=1$ and $b=-1$. But the post does not say $X$ and $Y$ are independent. With just the information given, we cannot compute the variance of $X-Y$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:31
$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Jesse P Francis
Oct 31 '15 at 17:35
$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Jesse P Francis
Oct 31 '15 at 17:35
$begingroup$
That makes sense, however, I know that the answer to the mean is -2.7 and the standard deviation should be 0.412, however, with these variables, I cannot seem to figure out the standard deviation after the equation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:35
$begingroup$
That makes sense, however, I know that the answer to the mean is -2.7 and the standard deviation should be 0.412, however, with these variables, I cannot seem to figure out the standard deviation after the equation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:35
1
1
$begingroup$
@JessePFrancis, I will look over these, thanks!
$endgroup$
– Jake Orben
Oct 31 '15 at 17:36
$begingroup$
@JessePFrancis, I will look over these, thanks!
$endgroup$
– Jake Orben
Oct 31 '15 at 17:36
$begingroup$
For independent $X$ and $Y$, the formula I mentioned gives that the standard deviation of $X-Y$ is $sqrt{(0.4)^2+(0.1)^2}$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:41
$begingroup$
For independent $X$ and $Y$, the formula I mentioned gives that the standard deviation of $X-Y$ is $sqrt{(0.4)^2+(0.1)^2}$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:41
|
show 1 more comment
1 Answer
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oldest
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$begingroup$
Variance($V$) of $(X-Y)=V(X)+V(Y)-2Cov(X,Y)$ where $Cov(X,Y)$ is the covariance between $X$ and $Y$.
Mean($mu$) of $(X-Y)=mu(X)-mu(Y)$
So you need information regarding the covariance to calculate the answers properly if $X$ and $Y$ are not independent.
EDIT:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
$endgroup$
$begingroup$
Unfortunately, that is all of the data that I was given by my professor and from what I have seen it is possible to calculate this number, however, I still cannot seem to calculate the standard deviation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:39
$begingroup$
What is the standard deviation according to answer?
$endgroup$
– SchrodingersCat
Oct 31 '15 at 17:45
add a comment |
Your Answer
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1 Answer
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1 Answer
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0
$begingroup$
Variance($V$) of $(X-Y)=V(X)+V(Y)-2Cov(X,Y)$ where $Cov(X,Y)$ is the covariance between $X$ and $Y$.
Mean($mu$) of $(X-Y)=mu(X)-mu(Y)$
So you need information regarding the covariance to calculate the answers properly if $X$ and $Y$ are not independent.
EDIT:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
$endgroup$
$begingroup$
Unfortunately, that is all of the data that I was given by my professor and from what I have seen it is possible to calculate this number, however, I still cannot seem to calculate the standard deviation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:39
$begingroup$
What is the standard deviation according to answer?
$endgroup$
– SchrodingersCat
Oct 31 '15 at 17:45
add a comment |
0
$begingroup$
Variance($V$) of $(X-Y)=V(X)+V(Y)-2Cov(X,Y)$ where $Cov(X,Y)$ is the covariance between $X$ and $Y$.
Mean($mu$) of $(X-Y)=mu(X)-mu(Y)$
So you need information regarding the covariance to calculate the answers properly if $X$ and $Y$ are not independent.
EDIT:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
$endgroup$
$begingroup$
Unfortunately, that is all of the data that I was given by my professor and from what I have seen it is possible to calculate this number, however, I still cannot seem to calculate the standard deviation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:39
$begingroup$
What is the standard deviation according to answer?
$endgroup$
– SchrodingersCat
Oct 31 '15 at 17:45
add a comment |
0
0
0
$begingroup$
Variance($V$) of $(X-Y)=V(X)+V(Y)-2Cov(X,Y)$ where $Cov(X,Y)$ is the covariance between $X$ and $Y$.
Mean($mu$) of $(X-Y)=mu(X)-mu(Y)$
So you need information regarding the covariance to calculate the answers properly if $X$ and $Y$ are not independent.
EDIT:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
$endgroup$
Variance($V$) of $(X-Y)=V(X)+V(Y)-2Cov(X,Y)$ where $Cov(X,Y)$ is the covariance between $X$ and $Y$.
Mean($mu$) of $(X-Y)=mu(X)-mu(Y)$
So you need information regarding the covariance to calculate the answers properly if $X$ and $Y$ are not independent.
EDIT:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
edited Oct 31 '15 at 17:45
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
answered Oct 31 '15 at 17:37
SchrodingersCatSchrodingersCat
22.3k52862
22.3k52862
$begingroup$
Unfortunately, that is all of the data that I was given by my professor and from what I have seen it is possible to calculate this number, however, I still cannot seem to calculate the standard deviation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:39
$begingroup$
What is the standard deviation according to answer?
$endgroup$
– SchrodingersCat
Oct 31 '15 at 17:45
add a comment |
$begingroup$
Unfortunately, that is all of the data that I was given by my professor and from what I have seen it is possible to calculate this number, however, I still cannot seem to calculate the standard deviation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:39
$begingroup$
What is the standard deviation according to answer?
$endgroup$
– SchrodingersCat
Oct 31 '15 at 17:45
$begingroup$
Unfortunately, that is all of the data that I was given by my professor and from what I have seen it is possible to calculate this number, however, I still cannot seem to calculate the standard deviation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:39
$begingroup$
Unfortunately, that is all of the data that I was given by my professor and from what I have seen it is possible to calculate this number, however, I still cannot seem to calculate the standard deviation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:39
$begingroup$
What is the standard deviation according to answer?
$endgroup$
– SchrodingersCat
Oct 31 '15 at 17:45
$begingroup$
What is the standard deviation according to answer?
$endgroup$
– SchrodingersCat
Oct 31 '15 at 17:45
add a comment |
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$begingroup$
If $X$ and $Y$ are independent, then the variance of $aX+bY$ is $a^2$ times the variance of $X$ plus $b^2$ times the variance of $Y$. Here $a=1$ and $b=-1$. But the post does not say $X$ and $Y$ are independent. With just the information given, we cannot compute the variance of $X-Y$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:31
$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Jesse P Francis
Oct 31 '15 at 17:35
$begingroup$
That makes sense, however, I know that the answer to the mean is -2.7 and the standard deviation should be 0.412, however, with these variables, I cannot seem to figure out the standard deviation after the equation.
$endgroup$
– Jake Orben
Oct 31 '15 at 17:35
1
$begingroup$
@JessePFrancis, I will look over these, thanks!
$endgroup$
– Jake Orben
Oct 31 '15 at 17:36
$begingroup$
For independent $X$ and $Y$, the formula I mentioned gives that the standard deviation of $X-Y$ is $sqrt{(0.4)^2+(0.1)^2}$.
$endgroup$
– André Nicolas
Oct 31 '15 at 17:41