设随机变量X , Y 相互独立, 且X~b(10,0.3) , Y~b(10,0.4) , 则 E(2X-Y)~2= 设随机变量x~b(10,0.2),y~e(0.5),且x与y...
\u8bbeX~e\uff080.5\uff09\uff0cY~b\uff0810,0.3\uff09\uff0c\u4e14X\uff0cY\u76f8\u4e92\u72ec\u7acb\uff0c\u5219D\uff08X-2Y\uff09=\u4f60\u597d\uff01\u7531\u4e8eX~e(0.5)\uff0cY~b(10,0.3)\uff0c\u7531\u671f\u671b\u516c\u5f0f\u5f97DX=1/0.5=2\uff0cDY=10*0.3=3\uff0c\u53c8X\uff0cY\u76f8\u4e92\u72ec\u7acb\uff0c\u6240\u4ee5D(X-2Y)=DX+4DY=2+4*3=14\u3002\u7ecf\u6d4e\u6570\u5b66\u56e2\u961f\u5e2e\u4f60\u89e3\u7b54\uff0c\u8bf7\u53ca\u65f6\u91c7\u7eb3\u3002\u8c22\u8c22\uff01
\u4f60\u597d\uff01\u968f\u673a\u53d8\u91cfX~B(10,0.2),Y~e(0.5),\u5219\u6709EX=10*0.2=2\uff0cEY=1/0.5=2\uff0c\u6240\u4ee5E(4X-3Y)=4EX-3EY=2\u3002\u7ecf\u6d4e\u6570\u5b66\u56e2\u961f\u5e2e\u4f60\u89e3\u7b54\uff0c\u8bf7\u53ca\u65f6\u91c7\u7eb3\u3002\u8c22\u8c22\uff01
答案是B,可以反用方差计算公式来计算平方的期望。请采纳,谢谢!
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