1.有一批灯泡寿命(单位：h)的抽取样本为1458 1395 1562 1614 13511490 1478 1382 1536 1496求这批灯泡的平均寿命μ及寿命方差σ²的矩估计量.

1. **计算样本均值** \[ \overline{X} = \frac{1}{10} \sum_{i=1}^{10} X_i = \frac{14762}{10} = 1476.2 \] 2. **计算样本方差** 使用公式 $S^2 = \frac{1}{n} \left( \sum_{i=1}^n X_i^2 - n \overline{X}^2 \right)$ \[ \sum_{i=1}^{10} X_i^2 = 21853650, \quad \overline{X}^2 = 2179166.44 \] \[ S^2 = \frac{1}{10} \left( 21853650 - 10 \times 2179166.44 \right) = 6198.56 \] **答案：** \[ \boxed{ \begin{array}{ll} \text{平均寿命 } \mu: & 1476.2 \\ \text{寿命方差 } \sigma^2: & 6198.6 \\ \end{array} } \]