# The arithmetic mean of $X$ when arithmetic mean of $X^2 = 29$.

Sorry if my question is a beginner because of my mathematical knowledge is low.

arithmetic mean is :

$$\overline x=\dfrac{x_1+x_2+\cdots+x_n}n$$

What method can solve it?

$$\overline{X^2}=29\quad\Rightarrow\,\text{Not }\left(\overline X\right)^2 \\\,\\ \overline X=\boxed{?}\qquad\qquad\qquad\quad\,\,\,$$

I got the square root but answer is not right. For example :

$$\overline X=\dfrac{1+2}2=\dfrac32=1.5 \\\,\\ \overline{X^2}=\dfrac{1+4}2=\dfrac52=2.5 \\\,\\ \sqrt{2.5}\approx1.581\ne1.5$$

• You can't solve it, you need more information. All that you know is that $\lvert \overline{x}\rvert \leqslant \sqrt{\overline{x^2}}$. – Daniel Fischer Nov 29 '13 at 12:35
• If you knew the variance of these numbers then you would almost be able to get the mean (upto a sign). – fretty Dec 30 '13 at 11:16

An example of why it's impossible with the given information. Suppose $x_1 = \sqrt{29}$, $x_2 = \sqrt{29}$. Then $\overline{x^2} = 29$, and $\overline{x} = \sqrt{29}$.

Now suppose $x_1 = \sqrt{29}$, $x_2 = -\sqrt{29}$. Again $\overline{x^2} = 29$, but now $\overline{x} = 0$. So we have two sets with the same $\overline{x^2}$ but different $\overline{x}$. Therefore, without more data, it's impossible to get $\overline{x}$ from $\overline{x^2}$.