题目
设X_(1),X_(2),X_(3)是相互独立的随机变量,且E(X_(i))=1,D(X_(i))=8(i=1,2,3),令overline(X)=(1)/(3)(X_(1)+X_(2)+X_(3)),则估计P(|overline(X)-1|A. (5)/(8)B. (5)/(6)C. 以上都不对D. (5)/(7)
设$X_{1},X_{2},X_{3}$是相互独立的随机变量,且$E(X_{i})=1,D(X_{i})=8(i=1,2,3)$,令$\overline{X}=\frac{1}{3}(X_{1}+X_{2}+X_{3})$,则估计$P(|\overline{X}-1|< 4)\geq()$。
A. $\frac{5}{8}$
B. $\frac{5}{6}$
C. 以上都不对
D. $\frac{5}{7}$
题目解答
答案
B. $\frac{5}{6}$
解析
步骤 1:计算$\overline{X}$的期望和方差
由于$X_{1},X_{2},X_{3}$是相互独立的随机变量,且$E(X_{i})=1,D(X_{i})=8(i=1,2,3)$,则$\overline{X}=\frac{1}{3}(X_{1}+X_{2}+X_{3})$的期望和方差分别为:
\[E(\overline{X}) = E\left(\frac{1}{3}(X_{1}+X_{2}+X_{3})\right) = \frac{1}{3}(E(X_{1})+E(X_{2})+E(X_{3})) = \frac{1}{3}(1+1+1) = 1\]
\[D(\overline{X}) = D\left(\frac{1}{3}(X_{1}+X_{2}+X_{3})\right) = \frac{1}{9}(D(X_{1})+D(X_{2})+D(X_{3})) = \frac{1}{9}(8+8+8) = \frac{8}{3}\]
步骤 2:应用切比雪夫不等式
根据切比雪夫不等式,对于任意随机变量$X$,有$P(|X-E(X)| \geq k\sigma) \leq \frac{1}{k^2}$,其中$\sigma$是$X$的标准差。对于$\overline{X}$,我们有$E(\overline{X}) = 1$,$D(\overline{X}) = \frac{8}{3}$,则$\sigma = \sqrt{\frac{8}{3}}$。设$k\sigma = 4$,则$k = \frac{4}{\sigma} = \frac{4}{\sqrt{\frac{8}{3}}} = \sqrt{6}$。因此,根据切比雪夫不等式,我们有:
\[P(|\overline{X}-1| \geq 4) \leq \frac{1}{k^2} = \frac{1}{6}\]
步骤 3:计算$P(|\overline{X}-1|< 4)$
根据步骤 2 的结果,我们有:
\[P(|\overline{X}-1| < 4) = 1 - P(|\overline{X}-1| \geq 4) \geq 1 - \frac{1}{6} = \frac{5}{6}\]
由于$X_{1},X_{2},X_{3}$是相互独立的随机变量,且$E(X_{i})=1,D(X_{i})=8(i=1,2,3)$,则$\overline{X}=\frac{1}{3}(X_{1}+X_{2}+X_{3})$的期望和方差分别为:
\[E(\overline{X}) = E\left(\frac{1}{3}(X_{1}+X_{2}+X_{3})\right) = \frac{1}{3}(E(X_{1})+E(X_{2})+E(X_{3})) = \frac{1}{3}(1+1+1) = 1\]
\[D(\overline{X}) = D\left(\frac{1}{3}(X_{1}+X_{2}+X_{3})\right) = \frac{1}{9}(D(X_{1})+D(X_{2})+D(X_{3})) = \frac{1}{9}(8+8+8) = \frac{8}{3}\]
步骤 2:应用切比雪夫不等式
根据切比雪夫不等式,对于任意随机变量$X$,有$P(|X-E(X)| \geq k\sigma) \leq \frac{1}{k^2}$,其中$\sigma$是$X$的标准差。对于$\overline{X}$,我们有$E(\overline{X}) = 1$,$D(\overline{X}) = \frac{8}{3}$,则$\sigma = \sqrt{\frac{8}{3}}$。设$k\sigma = 4$,则$k = \frac{4}{\sigma} = \frac{4}{\sqrt{\frac{8}{3}}} = \sqrt{6}$。因此,根据切比雪夫不等式,我们有:
\[P(|\overline{X}-1| \geq 4) \leq \frac{1}{k^2} = \frac{1}{6}\]
步骤 3:计算$P(|\overline{X}-1|< 4)$
根据步骤 2 的结果,我们有:
\[P(|\overline{X}-1| < 4) = 1 - P(|\overline{X}-1| \geq 4) \geq 1 - \frac{1}{6} = \frac{5}{6}\]