某样本方差的计算公式是 $S^{2}=\frac{1}{16}\left[\left(x_{1}-3\right)^{2}+\left(x_{2}-3\right)^{2}+\cdots+\left(x_{16}-3\right)^{2}\right]$, 则它的样本容量是 \$ \qquad \$ , 样本的平均数是 \$ \qquad \$ , 样本的平方和是 176 时, 标准差是 \$ \qquad \$ .
$\because$ 公式 $S^{2}=\frac{1}{16}\left[\left(x_{1}-3\right)^{2}+\left(x_{2}-3\right)^{2}+\cdots+\left(x_{16}-3\right)^{2}\right]$,

$\therefore$ 它的样本容量是 16 , 样本的平均数是 3 ,

$\because$ 样本的平方和是 176 ,

$\therefore x_{1}^{2}+x_{2}^{2}+\cdots+x_{16}^{2}=176$,

$\because$ 样本的平均数是 3 ,

$\therefore x_{1}+x_{2}-\cdots+x_{16}=16 \times 3=48$,

$$
\begin{gathered}
\therefore S^{2}=\frac{1}{16}\left[\left(x_{1}-3\right)^{2}+\left(x_{2}-3\right)^{2}+\cdots+\left(x_{16}-3\right)^{2}\right] \\
=\frac{1}{16}\left[x_{1}^{2}-6 x_{1}+9+x_{2}^{2}-6 x_{2}+9+\cdots+x_{16}^{2}-6 x_{16}+x_{16}^{2}\right] \\
=\frac{1}{16}\left[x_{1}^{2}+x_{2}^{2}+\cdots+x_{16}^{2}-6 x_{1}-6 x_{2}-\cdots-6 x_{16}+9 \times 16\right] \\
=\frac{1}{16}(176-6 \times 48+9 \times 16)
\end{gathered}
$$

$$
=2
$$

$\therefore$ 标准差是 $\sqrt{2}$.

故答案为: $16,3, \sqrt{2}$