MHB Finding the 3rd Number to Keep Variance Unchanged

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To maintain the variance when adding a third number to the set {0, 2}, the new number must be calculated based on the existing mean and variance. The variance of the original set is 1, calculated using the formula for variance dividing by n. Adding a new number y requires solving a system of equations that includes the mean and variance of the new set {0, 2, y}. The discussion highlights the challenge of having more unknowns than equations, emphasizing the need for the expected mean to avoid losing a degree of freedom. Clarification from the original poster is awaited to ensure accurate understanding of the problem.
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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 !
 
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Yankel said:
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 $y$
 
TheEmptySet said:
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.
 
I like Serena said:
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.
 
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