# Variance

#### Yankel

##### Active member
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
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
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
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.