1987 · 数学三 · 第 22 题 · 解答题

已知离散型随机变量

X

的概率分布为：

P {X = 1} = 0.2, P {X = 2} = 0.3, P {X = 3} = 0.5.

答案

$F (x) = ⎩ ⎨ ⎧ 0, 0.2, 0.5, 1, x < 1; 1 \leq x < 2; 2 \leq x < 3; x \geq 3.$

$E (X) = \frac{23}{10}$

$D (X) = \frac{61}{100}$

查看解析展开

由 $F (x) = P (X \leq x)$ 并注意分布函数右连续，得到

F (x) = ⎩ ⎨ ⎧ 0, 0.2, 0.5, 1, x < 1, 1 \leq x < 2, 2 \leq x < 3, x \geq 3.

E (X) = 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.5 = 2.3 = \frac{23}{10} .

先算

E (X^{2}) = 1^{2} \times 0.2 + 2^{2} \times 0.3 + 3^{2} \times 0.5 = 5.9 = \frac{59}{10} .

因此

D (X) = E (X^{2}) - [E (X)]^{2} = \frac{59}{10} - (\frac{23}{10})^{2} = \frac{61}{100} .