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Variance

Yankel

Active member
Jan 27, 2012
398
Hello

I have a problem solving this question....

to the numbers 0 and 2, we want to add a 3rd number, such that the variance won't change. What is the 3rd number ?

(when I say variance I mean dividing by n, not by n-1)

thanks !
 

TheEmptySet

New member
Mar 15, 2012
8
Hello

I have a problem solving this question....

to the numbers 0 and 2, we want to add a 3rd number, such that the variance won't change. What is the 3rd number ?

(when I say variance I mean dividing by n, not by n-1)

thanks !
The variance is the square root of the sum of the differences from of the data points from the mean so.... $\sigma^2=\frac{1}{2}\left[(0-1)^2+(2-1)^2\right]=1$. If we add the new point $y$, we will need to solve the system of equations. $\mu=\frac{1}{3}\left[ 0+2+y\right]$ and $\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]=1$ This will give a quadratic equation in $y$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
The variance is the square root of the sum of the differences from of the data points from the mean so.... $\sigma^2=\frac{1}{2}\left[(0-1)^2+(2-1)^2\right]=1$. If we add the new point $y$, we will need to solve the system of equations. $\mu=\frac{1}{3}\left[ 0+2+y\right]$ and $\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]=1$ This will give a quadratic equation in $y$
If we divide the variance by n instead of by (n-1), we can only do so if the expected mean $\mu$ is given.
Otherwise we would lose a degree of freedom.

So the set of equations should be:

$\sigma^2=\frac{1}{2}\left[(0-\mu)^2+(2-\mu)^2\right]$
$\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]$

This does introduce the problem that we have more unknowns than equations.
 

TheEmptySet

New member
Mar 15, 2012
8
If we divide the variance by n instead of by (n-1), we can only do so if the expected mean $\mu$ is given.
Otherwise we would lose a degree of freedom.

So the set of equations should be:

$\sigma^2=\frac{1}{2}\left[(0-\mu)^2+(2-\mu)^2\right]$
$\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]$

This does introduce the problem that we have more unknowns than equations.
Maybe I am misunderstanding something but the way I read the problem was if we have the data set $\{0,2 \}$ we can calculate the mean and variance directly. Now if a new data set is created by adding one other point $\{0,2,y \}$. We can now calculate the new population mean and the new variance of this three point data set. Now we can just solve for what $y$ needs to be. I will wait for clarification from the OP on this.