基础知识
方差的概念
(1)一般地,若离散型随机变量\(x\)的概率分布列为
\(X\) | \(x_1\) | \(x_2\) | ⋯ | \(x_i\) | ⋯ | \(x_n\) |
---|---|---|---|---|---|---|
\(P\) | \(p_1\) | \(p_2\) | ⋯ | \(p_i\) | ⋯ | \(p_n\) |
则称
\[D(X)=(x_1-E(X))^2p_1+(x_2-E(X))^2 p_2+\cdots+(x_n-E(X))^2 p_n=\sum_{i=1}^n\left(x_i-E(X)\right)^2 p_i \]为随机变量\(X\)的方差,有时候也记为\(V(x)\),并称\(\sqrt{D(X)}\)为随机变量\(X\)的标准差,记为\(σ(X)\).
随机变量的方差和标准差都可以度量随机变量取值与其均值的偏离程度,反映了随机变量取值的离散程度.
方差越小,随机变量的取值越集中;方差越大,随机变量的取值越分散.
方差的性质
(1)一般地,\(D(a X+b)=a^2 D(X)\).(可用方差的概念证明)
(2) \(D X=E\left(X^2\right)-E^2(X)\)
证明 \(D(X)=\left(x_1-E(X)\right)^2 p_1+\left(x_2-E(X)\right)^2 p_2+\cdots+\left(x_n-E(X)\right)^2 p_n\)
\(=\sum_{i=1}^n\left(x_i-E(X)\right)^2 p_i\)
\(=\sum_{i=1}^n\left[x_i^2-2 x_i E(X)+E^2(X)\right] p_i\)
\(=\sum_{i=1}^n\left[x_i^2 p_i-2 x_i p_i E(X)+E^2(X) p_i\right]\)
\(=\sum_{i=1}^n x_i^2 p_i-\sum_{i=1}^n 2 x_i p_i E(X)+\sum_{i=1}^n E^2(X) p_i\)
\(=E\left(X^2\right)-2 E(X) \sum_{i=1}^n x_i p_i+E^2(X) \sum_{i=1}^n p_i\)
\(=E\left(X^2\right)-2 E(X) \cdot E(X)+E^2(X) \cdot 1\)
\(=E(X^2 )-E^2 (X)\)
基本方法
【题型1】 方差的概念
【典题1】 甲、乙两名工人加工同一种零件,两人每天加工的零件数相同,所得次品数分别为\(X\),\(Y\),
\(X\)和\(Y\)的分布列如下表.
\(X\) | \(0\) | \(1\) | \(2\) |
---|---|---|---|
\(P\) | \(\dfrac{6}{10}\) | \(\dfrac{1}{10}\) | \(\dfrac{3}{10}\) |
\(Y\) | \(0\) | \(1\) | \(2\) |
---|---|---|---|
\(P\) | \(\dfrac{5}{10}\) | \(\dfrac{3}{10}\) | \(\dfrac{2}{10}\) |
试对这两名工人的技术水平进行比较.
解析 工人甲生产出次品数\(X\)的期望和方差分别为:
\(E(X)=0 \times \dfrac{6}{10}+1 \times \dfrac{1}{10}+2 \times \dfrac{3}{10}=0.7\),
\(D(X)=(0-0.7)^2 \times \dfrac{6}{10}+(1-0.7)^2 \times \dfrac{1}{10}+(2-0.7)^2 \times \dfrac{3}{10}=0.81\).
工人 乙生产出次品数\(Y\)的期望和方差分别为:
\(E(Y)=0 \times \dfrac{5}{10}+1 \times \dfrac{3}{10}+2 \times \dfrac{2}{10}=0.7\),
\(D(Y)=(0-0.7)^2 \times \dfrac{5}{10}+(1-0.7)^2 \times \dfrac{3}{10}+(2-0.7)^2 \times \dfrac{2}{10}=0.61\).
由\(E(X)=E(Y)\)知,两人出次品的平均数相同,技术水平相当,
但\(D(X)>D(Y)\),可见乙的技术比较稳定.
点拨 方差越大,数据越不稳定;方差越小,数据越稳定.
【典题2】 设\(0<a<1\),离散型随机变量\(X\)的分布列是
\(X\) | \(0\) | \(1\) | \(2\) |
---|---|---|---|
\(P\) | \(\dfrac{1-a}{2}\) | \(\dfrac{1}{2}\) | \(\dfrac{a}{2}\) |
则当\(a\)在\(\left(0, \dfrac{2}{3}\right)\)内增大时( )
A.\(D(X)\)增大 \(\qquad\) B.\(D(X)\)减小 \(\qquad\) C.\(D(X)\)先减小后增大 \(\qquad\) D.\(D(X)\)先增大后减小
解析 由离散型随机变量\(X\)的分布列得: \(E(X)=\dfrac{1}{2}+a\),
\(D(X)=\left(-\dfrac{1}{2}-a\right)^2 \times \dfrac{1-a}{2}+\left(\dfrac{1}{2}-a\right)^2 \times \dfrac{1}{2}+\left(\dfrac{3}{2}-a\right)^2 \times \dfrac{a}{2}\)\(=-a^2+a+\dfrac{1}{4}=-\left(a-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\),
\(\because 0<a<1\),\(D(X)\)先增大后减小.
故选:\(D\).
点拨 利用方差的概念求出方差,相当确定一函数关系式,再利用函数的性质确定其变化.
【巩固练习】
1.已知\(X\)的分布列为
\(X\) | \(-1\) | \(0\) | \(1\) |
---|---|---|---|
\(P\) | \(0.5\) | \(3.0\) | \(2.0\) |
则\(D(X)\)等于( )
A.\(0.7\) \(\qquad \qquad \qquad \qquad\) B.\(0.61\) \(\qquad \qquad \qquad \qquad\) C.\(-0.3\) \(\qquad \qquad \qquad \qquad\) D.\(0\)
2.已知随机变量\(\xi\)的分布列为下表所示,若\(E\xi =\dfrac{1}{4}\),则\(D\xi =\)( )
\(\xi\) | \(-1\) | \(0\) | \(1\) |
---|---|---|---|
\(P\) | \(\dfrac{1}{3}\) | \(a\) | \(b\) |
A.\(\dfrac{5}{6}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{41}{48}\) \(\qquad \qquad \qquad \qquad\)C.1 \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{2}{3}\)
3.(多选)设随机变量\(X\)的分布列为,
\(X\) | \(0\) | \(1\) | \(2\) |
---|---|---|---|
\(P\) | \(a\) | \(\dfrac{b}{2}\) | \(\dfrac{b}{2}\) |
其中\(ab≠0\).则下列说法正确的是( )
A.\(a+b=1\) \(\qquad \qquad\) B. \(E(X)=26\) \(\qquad \qquad\) C. \(D(X)\)先增大后减小 \(\qquad \qquad\) D. \(D(X)\)有最小值
4.2013年4月1日至7日是江西省“爱鸟周”,主题是“秀美江西,让鸟儿自由飞翔”.为更好 地保护鄱阳湖候鸟资源,需评测保护区的管理水平.
现甲、乙两个野生动物保护区有相同的自然环境,且候鸟的种类和数量也大致相等,两个保护区内每个季度发现违反保护条例的事件次数的分布列分别为:
\(X\) | \(0\) | \(1\) | \(2\) | \(3\) |
---|---|---|---|---|
\(P\) | \(0.3\) | \(0.3\) | \(0.2\) | \(0.2\) |
\(Y\) | \(0\) | \(1\) | \(2\) |
---|---|---|---|
\(P\) | \(0.1\) | \(0.5\) | \(0.4\) |
试评定这两个保护区的管理水平.
参考答案
- 答案
解析 \(E(X)=-1×0.5+0×0.3+1×0.2=-0.3\),
\(D(X)=0.5 \times(-1+0.3)^2+0.3 \times(0+0.3)^2+0.2 \times(1+0.3)^2=0.61\). - 答案 \(B\)
解析 由\(E(\xi )=-1×\dfrac{1}{3}+0×a+1×b=\dfrac{1}{4}\),整理得 \(b=\dfrac{7}{12}\),
由\(\dfrac{1}{3}+a+b=1\),\(a=1-\dfrac{1}{3}-\dfrac{7}{12}=\dfrac{1}{12}\),
\(\therefore D(\xi)=\left(-1-\dfrac{1}{4}\right)^2 \times \dfrac{1}{3}+\left(0-\dfrac{1}{4}\right)^2 \times \dfrac{1}{12}+\left(1-\dfrac{1}{4}\right)^2 \times \dfrac{7}{12}=\dfrac{41}{48}\).
故选:\(B\). - 答案 \(AC\)
解析 由题意可知\(a+\dfrac{b}{2}+\dfrac{b}{2}=1\),即\(a+b=1\),所以\(A\)正确;
\(E(X)=0×a+1×\dfrac{b}{2}+2×\dfrac{b}{2}=\dfrac{3b}{2}\),所以\(B\)不正确;
\(D(X)=a\left(0-\dfrac{3 b}{2}\right)^2+\dfrac{b}{2}\left(1-\dfrac{3 b}{2}\right)^2+\dfrac{b}{2}\left(2-\dfrac{3 b}{2}\right)^2=-\dfrac{9}{4} b^2+\dfrac{29}{8} b\),\(b\in (0,1)\),
\(D(X)\)是二次函数,对称轴为:\(b=\dfrac{29}{36}\in (0,1)\),
所以\(D(X)\)先增大后减小,所以\(C\)正确;\(D\)不正确;
故选:\(AC\). - 答案 乙保护区的管理较好一些
解析 甲保护区内的违规次数\(Y\)的数学期望和方差为:
\(E(X)=0×0.3+1×0.3+2×0.2+3×0.2=1.3\),
\(D(X)=(0-1.3)^2 \times 0.3+(1-1.3)^2 \times 0.3+(2-1.3)^2 \times 0.2\)
\(+(3-1.3)^2 \times 0.2=1.21\).
乙保护区内的违规次数\(Y\)的数学期望和方差为:
\(E(Y)=0×0.1+1×0.5+2×0.4=1.3\),
\(D(Y)=(0-1.3)^2 \times 0.1+(1-1.3)^2 \times 0.5+(2-1.3)^2 \times 0.4=0.41\).
因为\(E(X)=E(Y)\),\(D(X)>D(Y)\),所以两个保护区内每个季度发生的违规事件的平均次数相同,但甲保护区内的违规事件次数相对分散和波动,乙保护区内的违规事件次数更加集中和稳定.相对而言,乙保护区的管理较好一些.
【题型2】 方差的性质
【典题1】 (多选)设离散型随机变量\(X\)的分布列为
\(X\) | \(1\) | \(2\) | \(3\) | \(4\) |
---|---|---|---|---|
\(P\) | \(q\) | \(0.4\) | \(0.1\) | \(0.2\) |
若离散型随机变量\(Y\)满足\(Y=3X+1\),则下列结果正确的有( )
A.\(q=0.2\) \(\qquad \qquad\) B.\(EX=2\),\(DX=1.4\) \(\qquad \qquad\) C.\(EX=2\),\(DX=1.8\) \(\qquad \qquad\) D.\(EY=7\),\(DY=16.2\)
解析 由离散型随机变量\(X\)的分布列的性质得:\(q=1-0.4-0.1-0.2-0.2=0.1\),
\(E(X)=0×0.1+1×0.4+2×0.1+3×0.2+4×0.2=2\),
\(D(X)=(0-2)^2 \times 0.1+(1-2)^2 \times 0.4+(2-2)^2 \times 0.1+(3-2)^2 \times 0.2\)\(+(4-2)^2 \times 0.2=1.8\),
\(\because\)离散型随机变量\(Y\)满足\(Y=3X+1\),
\(\therefore E(Y)=3E(X)+1=7\),\(D(Y)=9D(X)=16.2\).
故选:\(CD\).
点拨 \(E(a X+b)=aE(X)+b\), \(D(a X+b)=a^2 D(X)\),注意理解\(a\),\(b\)对期望与方差变化的影响,不要混淆这两个性质.
【典题2】 (多选)已知随机变量\(\xi\)的分布列为
\(\xi\) | \(1\) | \(2\) | \(3\) |
---|---|---|---|
\(P\) | \(a\) | \(b\) | \(a+2b\) |
当\(b\)增大时,( )
A. \(E(\xi )\)减小 \(\qquad \qquad \qquad \qquad\) B. \(E(\xi )\)增大 \(\qquad \qquad \qquad \qquad\) C. \(D(\xi )\)减小 \(\qquad \qquad \qquad \qquad\) D. \(D(\xi )\)增大
解析 由随机变量\(\xi\)的分布列可得,\(a+b+a+2b=1\),故 \(\alpha a=\dfrac{1-3 b}{2}\),
所以\(E(\xi )=a+2b+3(a+2b)=4a+8b=2b+2\),
显然\(E(\xi )\)随\(b\)的增大而增大,
\(E\left(\xi^2\right)=1^2 \cdot a+2^2 \cdot b+3^2 \cdot(a+2 b)=a+4 b+9(a+2 b)=10 a+22 b\),
\(D(\xi)=E\left(\xi^2\right)-[E(\xi)]^2=10 a+22 b-(2 b+2)^2\)\(=5+7 b-4 b^2-8 b-4=-4 b^2-b+1\),
对称轴\(b=-\dfrac{1}{8}\),
\(\because b>0\),\(\therefore D(\xi )\)随\(b\)的增大而减小.
故选:\(BC\).
点拨 求方差利用\(D(\xi )=E(\xi ^2 )-[E(\xi )]^2\)有时候能够化简计算.
【巩固练习】
1.如果\(X\)是离散型随机变量,\(E(X)=6\),\(D(X)=0.5\),\(X_1=2X-5\),那么\(E(X_1)\)和\(D(X_1)\)分别是( )
A.\(E(X_1)=12\),\(D(X_1 )=1\) \(\qquad \qquad \qquad \qquad\) B.\(E(X_1)=7\),\(D(X_1)=1\)
C.\(E(X_1)=12\),\(D(X_1)=2\) \(\qquad \qquad \qquad \qquad\) D.\(E(X_1)=7\),\(D(X_1)=2\)
- 随机变量\(X\)的分布列如表所示,若\(E(X)=\dfrac{1}{3}\),则\(D(3X-2)=\)( )
\(X\) | \(-1\) | \(0\) | \(1\) |
---|---|---|---|
\(P\) | \(\dfrac{1}{6}\) | \(a\) | \(b\) |
A.\(9\) \(\qquad \qquad \qquad \qquad\) B.\(7\) \(\qquad \qquad \qquad \qquad\) C.\(5\) \(\qquad \qquad \qquad \qquad\)D.\(3\)
3.(多选)若随机变量\(X\)服从两点分布,其中\(P(X=0)=\dfrac{1}{3}\),\(E(X)\)、\(D(X)\)分别为随机变量\(X\)均值与方差,则下列结论正确的是( )
A.\(P(X=1)=E(X)\) \(\qquad \qquad\) B.\(E(3X+2)=4\) \(\qquad \qquad\) C.\(D(3X+2)=4\) \(\qquad \qquad\)D.\(D(X)=\dfrac{4}{9}\)
4.袋中有\(20\)个大小相同的球,其中记上\(0\)号的有\(10\)个,记上\(n\)号的有\(n\)个\((n=1,2,3,4)\).现从袋中任取一球,\(\xi\)表示所取球的标号.
(1)求\(\xi\)的均值和方差;
(2)若\(η=a\xi +b\),\(E(η)=1\),\(D(η)=11\),试求\(a\),\(b\)的值.
参考答案
-
答案 \(D\)
解析 \(E(X_1)=2E(X)-5=12-5=7\),\(D(X_1)=4D(X)=4×0.5=2\). -
答案 \(C\)
解析 \(\because E(X)=\dfrac{1}{3}\),
\(\therefore\)由随机变量\(X\)的分布列得: \(\left\{\begin{array}{l} \dfrac{1}{6}+a+b=1 \\ -\dfrac{1}{6}+b=\dfrac{1}{3} \end{array}\right.\),
解得\(a=\dfrac{1}{3}\),\(b=\dfrac{1}{2}\),
\(\therefore D(X)=\left(-1-\dfrac{1}{3}\right)^2 \times \dfrac{1}{6}+\left(0-\dfrac{1}{3}\right)^2 \times \dfrac{1}{3}+\left(1-\dfrac{1}{3}\right)^2 \times \dfrac{1}{2}=\dfrac{5}{9}\).
\(\therefore D(3 X-2)=9 D(X)=9 \times \dfrac{5}{9}=5\).
故选:\(C\). -
答案 \(AB\)
解析 随机变量\(X\)服从两点分布,其中\(P(X=0)=\dfrac{1}{3}\),
\(\therefore P(X=1)=\dfrac{2}{3}\),
\(E(X)=0×\dfrac{1}{3}+1×\dfrac{2}{3}=\dfrac{2}{3}\),
\(D(X)=\left(0-\dfrac{2}{3}\right)^2 \times \dfrac{1}{3}+\left(1-\dfrac{2}{3}\right)^2 \times \dfrac{2}{3}=\dfrac{2}{9}\),
在\(A\)中,\(P(X=1)=E(X)\),故\(A\)正确;
在\(B\)中,\(E(3X+2)=3E(X)+2=3×\dfrac{2}{3}+2=4\),故\(B\)正确;
在\(C\)中,\(D(3X+2)=9D(X)=9\times \dfrac{2}{9}=2\),故\(C\)错误;
在\(D\)中,\(D(X)=\dfrac{2}{9}\),故\(D\)错误.
故选:\(AB\). -
答案 (1)\(E(\xi )=1.5\),\(D(\xi )=2.75\).(2) \(a=2\),\(b=-2\)或\(a=-2\),\(b=4\)
解析 (1)由题意得,\(\xi\)的所有可能取值为\(0\),\(1\),\(2\),\(3\),\(4\),
\(P(\xi =0)=\dfrac{10}{20}=\dfrac{1}{2}\),\(P(\xi =1)=\dfrac{1}{20}\),\(P(\xi =2)=\dfrac{2}{20}=\dfrac{1}{10}\),
\(P(\xi =3)=\dfrac{3}{20}\),\(P(\xi =4)=\dfrac{4}{20}=\dfrac{1}{5}\).
故\(\xi\)的分布列为
\(\xi\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
---|---|---|---|---|---|
\(P\) | \(\dfrac{1}{2}\) | \(\dfrac{1}{20}\) | \(\dfrac{1}{10}\) | \(\dfrac{3}{20}\) | \(\dfrac{1}{5}\) |
所以\(E(\xi )=0\times \dfrac{1}{2}+1\times \dfrac{1}{2}0+2\times \dfrac{1}{10}+3\times \dfrac{3}{20}+4\times \dfrac{1}{5}=1.5\),
\(D(\xi)=(0-1.5)^2 \times \dfrac{1}{2}+(1-1.5)^2 \times \dfrac{1}{20}+(2-1.5)^2 \times \dfrac{1}{10}\) \(+(3-1.5)^2 \times \dfrac{3}{20}+(4-1.5)^2 \times \dfrac{1}{5}=2.75\)
(2)由\(D(a\xi +b)=a^2 D(\xi )=11\),\(E(a\xi +b)=aE(\xi )+b=1\),及\(E(\xi )=1.5\),\(D(\xi )=2.75\),
得\(2.75a^2=11\),\(1.5a+b=1\),
解得\(a=2\),\(b=-2\)或\(a=-2\),\(b=4\).
分层练习
【A组---基础题】
1.有甲、乙两种水稻,测得每种水稻各\(10\)株的分蘖数据,计算出样本方差分别为\(D\left(X_{\text {甲 }}\right)=11\)_, _\(D\left(X_{\text {乙 }}\right)=3.4\).由此可以估计( )
A.甲种水稻比乙种水稻分蘖整齐 \(\qquad \qquad \qquad \qquad\) B.乙种水稻比甲种水稻分蘖整齐
C.甲、乙两种水稻分蘖整齐程度相同 \(\qquad \qquad \qquad \qquad\) D.甲、乙两种水稻分蘖整齐程度不能比较
2.已知\(X\)的分布列为
\(\xi\) | \(1\) | \(2\) | \(3\) | \(4\) |
---|---|---|---|---|
\(P\) | \(\dfrac{1}{4}\) | \(\dfrac{1}{6}\) | \(\dfrac{1}{6}\) | \(\dfrac{1}{4}\) |
则\(D(X)\)的值为( )
A.\(\dfrac{29}{12}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{121}{144}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{179}{144}\) \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{17}{12}\)
3.设随机变量\(\xi\)的分布列是
\(\xi\) | \(0\) | \(1\) | \(2\) |
---|---|---|---|
\(P\) | \(\dfrac{1-p}{2}\) | \(\dfrac{1}{2}\) | \(\dfrac{p}{2}\) |
则当\(p\in [0,1]\)时,\(D(\xi )\)的最大值为( )
A.\(\dfrac{1}{4}\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(\dfrac{3}{4}\) \(\qquad \qquad \qquad \qquad\) D.\(1\)
4.设\(0<a<1\),离散型随机变量\(X\)的分布列是
\(X\) | \(0\) | \(1\) | \(2\) |
---|---|---|---|
\(P\) | \(\dfrac{1-a}{2}\) | \(\dfrac{1}{2}\) | \(\dfrac{p}{2}\) |
则当\(a\)在 \(\left(0, \dfrac{2}{3}\right)\)内增大时( )
A.\(D(X)\)增大 \(\qquad \qquad\) B.\(D(X)\)减小 \(\qquad \qquad\)C.\(D(X)\)先减小后增大 \(\qquad \qquad\) D.\(D(X)\)先增大后减小
5.(多选)已知\(A=B=\{1,2,3\}\),分别从集合\(A\),\(B\)中各随机取一个数\(a\),\(b\),得到平面上一个点\(P(a,b)\),事件“点\(P(a,b)\)恰好落在直线\(x+y=n\)上”对应的随机变量为\(X\),\(P(X=n)=P_n\),\(X\)的数学期望和方差分别为\(E(X)\),\(D(X)\),则( )
A.\(P_4=2P_2\) \(\qquad \qquad\) B.\(P(3≤X≤5)=\dfrac{7}{9}\)\(\qquad \qquad\) C.\(E(X)=4\) \(\qquad \qquad\) D.\(D(X)=\dfrac{4}{3}\)
6.(多选)袋内有大小完全相同的\(2\)个黑球和\(3\)个白球,从中不放回地每次任取\(1\)个小球,直至取到白球后停止取球,则( )
A.抽取\(2\)次后停止取球的概率为\(\dfrac{3}{5}\)
B.停止取球时,取出的白球个数不少于黑球的概率为\(\dfrac{9}{10}\)
C.取球次数\(\xi\)的期望为\(2\)
D.取球次数\(\xi\)的方差为\(\dfrac{9}{20}\)
7.已知离散型随机变量\(\xi\)的分布列如下:
\(\xi\) | \(1\) | \(3\) | \(5\) |
---|---|---|---|
\(P\) | \(0.5\) | \(m\) | \(0.2\) |
则其方差\(D(\xi )=\) \(\underline{\quad \quad}\) .
8.已知随机变量\xi 的分布列是
\(\xi\) | \(-1\) | \(0\) | \(1\) |
---|---|---|---|
\(P\) | \(\dfrac{1}{2}\) | \(\dfrac{1-p}{2}\) | \(\dfrac{p}{2}\) |
随机变量\(η\)的分布列是
\(X\) | \(1\) | \(2\) | \(3\) |
---|---|---|---|
\(P\) | \(\dfrac{1}{2}\) | \(\dfrac{1-p}{2}\) | \(\dfrac{p}{2}\) |
则当\(p\)在\((0,1)\)内增大时,下列选项中正确的是 \(\underline{\quad \quad}\).
①\(E(\xi )=E(η)\);\(\qquad \qquad\) ②\(D(\xi )=D(η)\);\(\qquad \qquad\) ③\(E(\xi )\)增大; \(\qquad \qquad\) ④\(D(η)\)先增大后减小.
9.有\(10\)张卡片,其中\(8\)张标有数字\(2\),\(2\)张标有数字\(5\),从中随机地抽取\(3\)张卡片,设这\(3\)张卡片上的数字之和为\(\xi\).
(1)求\(E(\xi )\)和\(D(\xi )\);\(\qquad \qquad\) (2)若\(X=3\xi -2\),求\(E(X)\),\(D(X)\).
参考答案
-
答案 \(B\)
解析 \(\because D\left(X_{\text {甲 }}\right)>D\left(X_{\text {乙 }}\right)\),\(\therefore\)乙种水稻比甲种水稻分蘖整齐. -
答案 \(C\)
解析 \(E(X)=1\times \dfrac{1}{4}+2\times \dfrac{1}{3}+3\times \dfrac{1}{6}+4\times \dfrac{1}{4}=\dfrac{29}{12}\),
\(\therefore D(X)=\left(1-\dfrac{29}{12}\right)^2 \times \dfrac{1}{4}+\left(2-\dfrac{29}{12}\right)^2 \times \dfrac{1}{3}+\left(3-\dfrac{29}{12}\right)^2 \times \dfrac{1}{6}\)\(+\left(4-\dfrac{29}{12}\right)^2 \times \dfrac{1}{4}=\dfrac{179}{144}\). -
答案 \(B\)
解析 由随机变量\(\xi\)的分布列得:\(E(\xi )=\dfrac{1}{2}+p\),
\(\because p\in [0,1]\),
\(\therefore D(\xi)=\left(-\dfrac{1}{2}-p\right)^2 \times \dfrac{1-p}{2}+\left(\dfrac{1}{2}-p\right)^2 \times \dfrac{1}{2}+\left(\dfrac{3}{2}-p\right)^2 \times \dfrac{p}{2}\)
\(=-p^2+p+\dfrac{1}{4}=-\left(p-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\),
\(\therefore\)当\(p=\dfrac{1}{2}\)时,\(D(\xi )\)取最大值为\(\dfrac{1}{2}\).
故选:\(B\). -
答案 \(D\)
解析 由离散型随机变量\(X\)的分布列得:\(E(X)=\dfrac{1}{2}+a\),
\(D(X)=\left(-\dfrac{1}{2}-a\right)^2 \times \dfrac{1-a}{2}+\left(\dfrac{1}{2}-a\right)^2 \times \dfrac{1}{2}+\left(\dfrac{3}{2}-a\right)^2 \times \dfrac{a}{2}\)\(=-a^2+a+\dfrac{1}{4}=-\left(a-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\),
\(\because 0<a<1\),\(\therefore D(X)\)先增大后减小.
故选:\(D\). -
答案 \(BCD\)
解析 由题意得对应的点\(P\)有:\((1,1),(1,2),(1,3),(2,1),(2,2)\),\((2,3),(3,1),(3,2),(3,3)\),
\(\therefore\)对应的\(X\)的可能取值为\(2\),\(3\),\(4\),\(5\),\(6\),
\(P(X=2)=\dfrac{1}{9}\),\(P(X=3)=\dfrac{2}{9}\),\(P(X=4)=\dfrac{3}{9}\),
\(P(X=5)=\dfrac{2}{9}\),\(P(X=6)=\dfrac{1}{9}\),
对于\(A\),\(p_4=P(X=4)=\dfrac{3}{9}=\dfrac{1}{3}≠2P_2=\dfrac{2}{9}\),故\(A\)错误;
对于\(B\),\(P(3≤X≤5)=P(X=3)+P(X=4)+P(X=5)=\dfrac{2}{9}+\dfrac{3}{9}+\dfrac{2}{9}=\dfrac{7}{9}\),
故\(B\)正确;
对于\(C\),\(E(X)=2\times \dfrac{1}{9}+3\times \dfrac{2}{9}+4\times \dfrac{3}{9}+5\times \dfrac{2}{9}+6\times \dfrac{1}{9}=4\),故\(C\)正确;
对于\(D\), \(D(X)=(2-4)^2 \times \dfrac{1}{9}+(3-4)^2 \times \dfrac{2}{9}+(4-4)^2 \times \dfrac{3}{9}+(5-4)^2 \times \dfrac{2}{9}\)\(+(6-4)^2\times \dfrac{1}{9}=\dfrac{4}{3}\),故\(D\)正确.
故选:\(BCD\). -
答案 \(BD\)
解析 设取球次数为\(\xi\),\(\xi\)的可能值是\(1\),\(2\),\(3\),
\(P(\xi =1)=\dfrac{3}{5}\),\(P(\xi =2)=\dfrac{2}{5}\times \dfrac{3}{4}=\dfrac{3}{10}\),\(P(\xi =3)=\dfrac{2}{5}\times \dfrac{1}{4}\times \dfrac{3}{3}=\dfrac{1}{10}\),
随机变量\(\xi\)的分布列为
\(X\) | \(1\) | \(2\) | \(3\) |
---|---|---|---|
\(P\) | \(\dfrac{3}{5}\) | \(\dfrac{3}{10}\) | \(\dfrac{1}{10}\) |
则\(P(\xi =2)=\dfrac{3}{10}\),\(A\)错;
取出的白球个数不少于黑球的概率为\(P(\xi =1)+P(\xi =2)=\dfrac{9}{10}\),故\(B\)正确;
\(E(\xi)=1 \times \dfrac{3}{5}+2 \times \dfrac{3}{10}+3 \times \dfrac{1}{10}=\dfrac{3}{2}\),故\(C\)错误;
\(D(\xi)=\left(1-\dfrac{3}{2}\right)^2 \times \dfrac{3}{5}+\left(2-\dfrac{3}{2}\right)^2 \times \dfrac{3}{10}+\left(3-\dfrac{3}{2}\right)^2 \times \dfrac{1}{10}=\dfrac{9}{20}\),故\(D\)正确.
故选:\(BD\).
- 答案 \(2.44\)
解析 \(\therefore E(\xi )=1\times 0.5+3\times 0.3+5\times 0.2=2.4\).
\(D(\xi)=(1-2.4)^2 \times 0.5+(3-2.4)^2 \times 0.3+(5-2.4)^2 \times 0.2=2.44\). - 答案 ②③
解析 对于①,由随机变量的分布列可知\(\xi =η-2\),
则\(E(\xi )=E(η)-2\),即\(E(\xi )≠E(η)\),即①错误;
对于②,因为\(\xi =η-2\),所以\(D(\xi )=D(η)\),即②正确;
对于③, \(E(\eta)=1 \times \dfrac{1}{2}+2 \times \dfrac{1-p}{2}+3 \times \dfrac{p}{2}=\dfrac{3+p}{2}\),
则\(p\)在\((0,1)\)内增大时,\(E(\xi )\)增大,即③正确;
对于④,由题意可知 \(E(\eta)=1 \times \dfrac{1}{2}+2 \times \dfrac{1-p}{2}+3 \times \dfrac{p}{2}=\dfrac{3+p}{2}\),
则 \(D(\eta)=\left(1-\dfrac{3+p}{2}\right) \times \dfrac{1}{2}+\left(2-\dfrac{3+p}{2}\right)^2 \times \dfrac{1-p}{2}+\left(3-\dfrac{3+p}{2}\right)^2 \times \dfrac{p}{2}\)\(=-\dfrac{1}{4}(p-2)^2+\dfrac{5}{4}\),
即当\(p\)在\((0,1)\)内增大时,\(D(η)\)增大,
即④错误,
故答案为:②③. - 答案 (1) \(E(\xi )=7.8\),\(D(\xi )=3.36\) (2) \(E(X)=21.4\),\(D(X)=30.24\)
解析 (1)\(3\)张卡片上的数字之和\(\xi\)的可能取值为\(6\),\(9\),\(12\).
\(\xi =6\)表示取出的\(3\)张卡片上都标有\(2\),则\(P(\xi=6)=\dfrac{C_8^3}{c_{10}^3}=\dfrac{7}{15}\).
\(\xi =9\)表示取出的\(3\)张卡片上\(2\)张标有\(2\),\(1\)张标有\(5\),则 \(P(\xi=9)=\dfrac{C_8^2 C_2^1}{C_{10}^3}=\dfrac{7}{15}\).
\(\xi =12\)表示取出的\(3\)张卡片上\(2\)张标有\(5\),\(1\)张标有\(2\),则 \(P(\xi=12)=\dfrac{C_8^1 C_2^2}{C_{10}^3}=\dfrac{1}{15}\).
\(\therefore \xi\)的分布列为
\(X\) | \(6\) | \(9\) | \(12\) |
---|---|---|---|
\(P\) | \(\dfrac{7}{15}\) | \(\dfrac{7}{15}\) | \(\dfrac{1}{15}\) |
\(\therefore E(\xi)=6 \times \dfrac{7}{15}+9 \times \dfrac{7}{15}+12 \times \dfrac{1}{15}=7.8\).
\(D(\xi)=(6-7.8)^2 \times \dfrac{7}{15}+(9-7.8)^2 \times \dfrac{7}{15}+(12-7.8)^2 \times \dfrac{1}{15}=3.36\).
(2)\(\because X=3\xi -2\),\(\therefore E(X)=3E(\xi )-2=3\times 7.8-2=21.4\).
\(D(X)=9D(\xi )=3.36\times 9=30.24\).
【B组---提高题】
1.已知随机变量\(X\)的分布列如表:
\(X\) | \(-1\) | \(0\) | \(1\) |
---|---|---|---|
\(P\) | \(a\) | \(b\) | \(c\) |
其中\(a\),\(b\),\(c>0\).若\(X\)的方差\(DX≤\dfrac{1}{3}\)对所有\(a\in (0,1-b)\)都成立,则( )
A.\(b≤\dfrac{1}{3}\) \(\qquad \qquad \qquad \qquad\) B.\(b≤\dfrac{2}{3}\) \(\qquad \qquad \qquad \qquad\) C.\(b≥\dfrac{1}{3}\) \(\qquad \qquad \qquad \qquad\) D.\(b≥\dfrac{2}{3}\)
参考答案
- 答案 \(D\)
解析 依题意\(a+b+c=1\),故\(c=1-a-b\),
当\(a\in (0,1-b)\)时,故\(EX=-a+c=1-b-2a\),
\(D X=E\left(X^2\right)-E^2(X)=a+c-(c-a)^2=a+c-\left[(c-a)^2+4 a c\right]+4 a c\)
\(=(a+c)-(a+c)^2+4 a[1-b-a]=(1-b)-(1-b)^2+4 a[1-b-a]\),
令\(1-b=t\),则\(t\in (0,1)\)
\(DX=t-t^2+4a(t-a)≤\dfrac{1}{3}\),\(a\in (0,t)\),
故 \(4 a^2-4 a t+t^2-t+\dfrac{1}{3} \geq 0\),在\(a\in (0,t)\)时恒成立,
当\(a=\dfrac{t}{2}\)时\(DX\)有最小值,故 \(4\left(\dfrac{t}{2}\right)^2-4 \dfrac{t}{2} \times t+t^2-t+\dfrac{1}{3} \geq 0\),
故\(t≤\dfrac{1}{3}\),即\(-1-b≤\dfrac{1}{3}\),所以\(b≥\dfrac{2}{3}\),
故选:\(D\).
【C组---拓展题】
1.广雅高一年级和高二年级进行篮球比赛,赛制为\(3\)局\(2\)胜制,若比赛没有平局,且高二队每局获胜的概率都是\(p(0<p<\dfrac{1}{2})\),记比赛的最终局数为随机变量\(X\),则( )
A. \(E(X)=\dfrac{5}{2}\) \(\qquad \qquad \qquad\) B.\(E(X)>\dfrac{21}{8}\) \(\qquad \qquad \qquad\) C.\(D(X)<\dfrac{1}{4}\) \(\qquad \qquad \qquad\) D.\(D(X) \in\left(\dfrac{1}{4}, \dfrac{1}{2}\right)\)
参考答案
- 答案 \(C\)
解析 \(X\)的可能取值为\(2\),\(3\),且 \(P(X=2)=p^2+(1-p)^2=2 p^2-2 p+1\),
\(P(X=3)=1-P(X=2)=2 p-2 p^2\).
\(E(X)=2 \times\left(2 p^2-2 p+1\right)+3\left(2 p-2 p^2\right)=-2 p^2+2 p+2\)\(=-2\left(p-\dfrac{1}{2}\right)^2+\dfrac{5}{2}\),
\(\because 0<p<\dfrac{1}{2}\), \(\therefore E(X)<\dfrac{5}{2}\),排除\(A\),\(B\).
\(E\left(X^2\right)=4 \times\left(2 p^2-2 p+1\right)+9\left(2 p-2 p^2\right)=-10 p^2+10 p+4\),
\(D(X)=E\left(X^2\right)-E^2(X)=-10 p^2+10 p+4-\left(-2 p^2+2 p+2\right)^2\)\(=-4 p^4+8 p^3-6 p^2+2 p\),
\(D^{\prime}(X)=-16 p^3+24 p^2-12 p+2, \quad D^{\prime \prime}(X)=-12(2 p-1)^2 \leqslant 0\),
所以\(D'(X)\)在\(p\in (0,1)\)上单调递减,又\(D^{\prime}\left(\dfrac{1}{2}\right)=0\),
所以当\(p \in\left(0, \dfrac{1}{2}\right)\)时,\(D' (X)>0\),
故\(p \in\left(0, \dfrac{1}{2}\right)\)时,\(D(X)\)单调递增,
所以 \(D(X)<-4 \times\left(\dfrac{1}{2}\right)^4+8 \times\left(\dfrac{1}{2}\right)^3-6 \times\left(\dfrac{1}{2}\right)^2+2 \times \dfrac{1}{2}=\dfrac{1}{4}\).
故选:\(C\).