随机变量X的概率分布律为 X=k =dfrac (1)(n) ,k=1,2,···, n,则 D(X)=（）.A. X=k =dfrac (1)(n) ,k=1,2,···, n,则 D(X)= B. X=k =dfrac (1)(n) ,k=1,2,···, n,则 D(X)= C. X=k =dfrac (1)(n) ,k=1,2,···, n,则 D(X)= D. X=k =dfrac (1)(n) ,k=1,2,···, n,则 D(X)=

步骤 1：计算期望值 E(X)
随机变量X的概率分布律为$P\{ X=k\} =\dfrac {1}{n}$，$k=1,2,\ldots,n$。因此，期望值E(X)为：
$$E(X) = \sum_{k=1}^{n} k \cdot P\{ X=k\} = \sum_{k=1}^{n} k \cdot \dfrac {1}{n} = \dfrac {1}{n} \sum_{k=1}^{n} k = \dfrac {1}{n} \cdot \dfrac {n(n+1)}{2} = \dfrac {n+1}{2}$$
步骤 2：计算方差 D(X)
方差D(X)为：
$$D(X) = E(X^2) - [E(X)]^2$$
其中，$E(X^2)$为：
$$E(X^2) = \sum_{k=1}^{n} k^2 \cdot P\{ X=k\} = \sum_{k=1}^{n} k^2 \cdot \dfrac {1}{n} = \dfrac {1}{n} \sum_{k=1}^{n} k^2 = \dfrac {1}{n} \cdot \dfrac {n(n+1)(2n+1)}{6} = \dfrac {(n+1)(2n+1)}{6}$$
因此，方差D(X)为：
$$D(X) = E(X^2) - [E(X)]^2 = \dfrac {(n+1)(2n+1)}{6} - \left(\dfrac {n+1}{2}\right)^2 = \dfrac {(n+1)(2n+1)}{6} - \dfrac {(n+1)^2}{4}$$
$$D(X) = \dfrac {2(n+1)(2n+1) - 3(n+1)^2}{12} = \dfrac {(n+1)(4n+2 - 3n - 3)}{12} = \dfrac {(n+1)(n-1)}{12} = \dfrac {n^2-1}{12}$$