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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















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:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















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:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














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:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$






share|cite|improve this answer











$endgroup$



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:
$$Cov(X,Y)=sum_{i=1}^{n}frac{(x_i-bar x)(y_i-bar y)}{sigma_xsigma_y}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 31 '15 at 17:45

























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


















  • $begingroup$
    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


















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