Finding the 3rd Number to Keep Variance Unchanged

[Yankel]

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]

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 $y$

 

[I like Serena]

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]

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.

 

[CellCoree]

Hello,

I understand your question and would like to provide a response. To find the 3rd number that would keep the variance unchanged, we first need to understand what variance is and how it is calculated.

Variance is a measure of how spread out a set of numbers is from the mean. It is calculated by taking the sum of the squared differences between each number and the mean, and dividing it by the total number of values in the set. In this case, the total number of values is 3 (0, 2, and the 3rd number we are trying to find).

To keep the variance unchanged, the sum of squared differences between the numbers and the mean must remain the same. This means that the 3rd number we add must have a squared difference of 0 from the mean. In other words, the 3rd number must be equal to the mean of the set (which in this case is (0+2)/2 = 1).

Therefore, the 3rd number that would keep the variance unchanged is 1. This can be verified by calculating the variance before and after adding the 3rd number. Before adding 1, the variance is (0-1)^2 + (2-1)^2 / 3 = 1. After adding 1, the variance is (0-1)^2 + (2-1)^2 + (1-1)^2 / 3 = 1. This shows that the variance remains unchanged.

I hope this explanation helps to solve your problem. If you have any further questions, please don't hesitate to ask. Happy problem solving!