# Is it possible for the median to equal the stddev?

1. Dec 29, 2004

### ktpr2

Is it possible to have a set where the average is some number and the standard deviation from the average is equal to the median value of that same set? If so, could you show me such a set?

2. Dec 29, 2004

### Tom Mattson

Staff Emeritus
Yes.

I'll show you a trivial example and then give you a hint. But you have to try it for yourself.

The trivial example is the set {0}. Both the standard deviation and the median are zero.

Now if you want a more substantial example, just use the definitions of standard deviation and median:

&sigma;2=&Sigma;(xi-&mu;)/N
m=middle value (if N is odd) or average of 2 middle values (if N is even)

Let N be odd for simplicity. In fact, let N=3. That means that the median is x2. So all you have to do is set &sigma; equal to x2 and choose some values for x1 and x2. Then use your equation to solve for x3.

3. Dec 29, 2004

### ktpr2

well i'm not sure where the squared part of the summation bit went when you removed the sqaure root by squaring both sides, but...

with {64, 127, x3}, we have

127^2 = ( (64-μ)+(127-μ)+(x3-μ) ) / 3

μ = (64+127+x3)/3

Now as soon as I pick a number for μ I've inadvertedly picked my stddev. I can solve for μ but I have a feeling you didn't intend for me to do all that algebraic manipulation.

And of course if i pick whatever value I want for μ the std. dev and median do not equal another. So what am I doing wrong?

EDIT - as an aside, is it possible for the mean+stddev to equal the median?

Last edited: Dec 29, 2004
4. Dec 29, 2004

### ktpr2

okay

For $$\mu+\sigma = median$$

Well I did some algebraic exploration of my own and I found sets like $$\{x, x+1,x+1\}$$ to work "good enough." I failed to mention that my values for x are $$0\geq x \geq255$$ and $$x \in N$$ so I just round any reals i get in calculating