# 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.

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.
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Lance Long
You can't get the geometric average when $n=3$. Knowing the arithmetic and quadratic average of three numbers $\left(x,y,z\right)$ 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
$\begin{array}{rl}\mathrm{\nabla }\left(x+y+z\right)& =\left(1,1,1\right)\\ \mathrm{\nabla }\left({x}^{2}+{y}^{2}+{z}^{2}\right)& =2\left(x,y,z\right)\\ \mathrm{\nabla }\left(xyz\right)& =\left(yz,xz,xy\right)\end{array}$
At $\left(x,y,z\right)=\left(1,2,3\right)$, say, the three gradients are linearly independent.
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prkosnognm
It's not possible for the geometric average, there are infinitely many solutions depending on the number of observations.