1. A contestant's nine scores, after removing the highest and the lowest, have an average of 91 for the remaining seven scores. In the stem-and-leaf plot of the nine scores made on-site, one data point is unclear and is represented by $x$ in Figure 1.
Then the variance of the seven remaining scores is ( ).
(A) $\frac{116}{9}$
(B) $\frac{36}{7}$
$\left.\begin{array}{lllllllll}8 & 7 & 7 & & & & \\ 9 & 4 & 0 & 1 & 0 & x & 9 & 1\end{array} \right\rvert\,$
(C) 36
(D) $\frac{6 \sqrt{7}}{7}$
Figure 1
1. B.

From the conditions, we know that the data set is
$$
87,87,94,90,91,90,9 x, 99,91,
$$

The highest score is 99, and the lowest score is 87. The remaining data are $87, 94, 90, 91, 90, 90+x, 91$. Therefore,
$$
\begin{array}{l}
\frac{87+94+90+91+90+(90+x)+91}{7}=91 \\
\Rightarrow x=4 .
\end{array}
$$

Thus, the variance we are looking for is
$$
\begin{array}{l}
\frac{1}{7}\left[(87-91)^{2}+(94-91)^{2}+(90-91)^{2}+\right. \\
\left.(91-91)^{2}+(90-91)^{2}+(94-91)^{2}+(91-91)^{2}\right] \\
=\frac{36}{7} .
\end{array}
$$