4. Given that the average of the data $x_{1}, x_{2}, \ldots, x_{10}$ is 6, and the standard deviation is $\sqrt{2}$, then the range of the average of the data $x_{1}, x_{2}, \ldots, x_{5}$ is $\qquad$
4. $[6-\sqrt{2}, 6+\sqrt{2}]$

Analysis: Let $t_{i}=x_{i}-6(i=1,2, \ldots, 10)$, since the average of $x_{i}$ is 6, we have $x_{1}+\ldots+x_{10}=60$, thus $t_{1}+\ldots+t_{10}=0$. The standard deviation is $\sqrt{2}$, so $t_{1}^{2}+t_{2}^{2}+\ldots+t_{10}^{2}=20$. By the $A M-G M$ inequality, $t_{1}^{2}+\ldots+t_{5}^{2} \geq \frac{1}{5}\left(t_{1}+\ldots+t_{5}\right)^{2}, t_{6}^{2}+\ldots+t_{10}^{2} \geq \frac{1}{5}\left(t_{6}+\ldots+t_{10}\right)^{2}=\frac{1}{5}\left(t_{1}+\ldots+t_{5}\right)^{2}$, hence $\left(t_{1}+\ldots+t_{5}\right)^{2} \leq 50$, which gives $t_{1}+\ldots+t_{5} \in[-5 \sqrt{2}, 5 \sqrt{2}]$. And $\frac{x_{1}+\ldots+x_{5}}{5}=\frac{t_{1}+\ldots+t_{5}}{5}+6$, therefore $6-\sqrt{2} \leq \frac{x_{1}+\ldots+x_{5}}{5} \leq 6+\sqrt{2}$. The left equality holds when $x_{1}=\ldots=x_{5}=6-\sqrt{2}, x_{6}=\ldots=x_{10}=6+\sqrt{2}$, and the right equality holds when $x_{1}=\ldots=x_{5}=6+\sqrt{2}, x_{6}=\ldots=x_{10}=6-\sqrt{2}$.