Quadratic and geometric average

Question

I'd like to find the find the quadratic average and the geometric average. To do this I have these informations :

The standart deviation, the arithmetic average and the number of values.

I know the formula to get the quadratic and geometric average with values, but in this case I don't have them.

Thanks a lot for your help.

Answer 1

You can't get the geometric average when $n=3$. Knowing the arithmetic and quadratic average of three numbers $(x,y,z)$ is equivalent to knowing $x+y+z$ and $x^2+y^2+z^2$, while knowing the geometric average means knowing $xyz$. If $xyz$ were a function of the other two, its gradient at any point would be a linear combination of the other two gradients. But $$ \begin{align*} \nabla(x+y+z)&=(1,1,1)\\ \nabla(x^2+y^2+z^2)&=2(x,y,z)\\ \nabla(xyz)&=(yz,xz,xy) \end{align*} $$ At $(x,y,z)=(1,2,3)$, say, the three gradients are linearly independent.

Answer 2

It's not possible for the geometric average, there are infinitely many solutions depending on the number of observations.