4. (10 points) Arrange $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ and the average of these 6 fractions in ascending order, then this average value is in the $\qquad$th position.
4. (10 points) Arrange $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ and the average of these 6 fractions in ascending order, then this average value is in the $\qquad$th position.

【Analysis】First, find the average of these 6 fractions, then arrange them to get the result.
【Solution】Solution: $\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right) \div 6$
$$
\begin{array}{l}
=\left[\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}\right)\right] \div 6 \\
=\left[1+\frac{83}{140}\right] \div 6 \\
\approx 1.593 \div 6 \\
=0.2655 ; \\
\frac{1}{7}<\frac{1}{6}<\frac{1}{5}<\frac{1}{4}<0.2655<\frac{1}{3}<\frac{1}{2} .
\end{array}
$$

So this average value is in the 5th position when arranged in ascending order.
Therefore, the answer is: 5