设总体X服从 (0-1) 分布, X=1 =1/3-|||-X1,···,xn为来自X的样本,-|||-则 overline (x)= [填空1], overline (x)= 填空2]-|||-则 (S)^2= [填空3]

步骤 1：计算总体X的期望值E(X)
由于X服从(0-1)分布，且$P\{ X=1\} =1/3$，则$P\{ X=0\} =2/3$。根据期望的定义，$E(X) = 0 \times P\{ X=0\} + 1 \times P\{ X=1\} = 0 \times \frac{2}{3} + 1 \times \frac{1}{3} = \frac{1}{3}$。

步骤 2：计算总体X的方差D(X)
根据方差的定义，$D(X) = E(X^2) - [E(X)]^2$。由于X服从(0-1)分布，$E(X^2) = E(X) = \frac{1}{3}$，因此$D(X) = \frac{1}{3} - (\frac{1}{3})^2 = \frac{1}{3} - \frac{1}{9} = \frac{2}{9}$。

步骤 3：计算样本均值$\overline{X}$的期望值$E(\overline{X})$
样本均值$\overline{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$，其中$X_i$是来自总体X的样本。根据期望的线性性质，$E(\overline{X}) = E(\frac{1}{n} \sum_{i=1}^{n} X_i) = \frac{1}{n} \sum_{i=1}^{n} E(X_i) = \frac{1}{n} \times n \times E(X) = E(X) = \frac{1}{3}$。

步骤 4：计算样本均值$\overline{X}$的方差$D(\overline{X})$
根据方差的性质，$D(\overline{X}) = D(\frac{1}{n} \sum_{i=1}^{n} X_i) = \frac{1}{n^2} \sum_{i=1}^{n} D(X_i) = \frac{1}{n^2} \times n \times D(X) = \frac{1}{n} D(X) = \frac{1}{n} \times \frac{2}{9} = \frac{2}{9n}$。

步骤 5：计算样本方差$S^2$的期望值$E(S^2)$
样本方差$S^2 = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \overline{X})^2$。根据方差的性质，$E(S^2) = E(\frac{1}{n-1} \sum_{i=1}^{n} (X_i - \overline{X})^2) = \frac{1}{n-1} \sum_{i=1}^{n} E((X_i - \overline{X})^2) = \frac{1}{n-1} \times n \times D(X) = \frac{n}{n-1} \times \frac{2}{9} = \frac{2}{9}$。