已知随机变量X～N（2，σ2）（σ＞0），若X在（0，2）内取值的概率为0.3，则X在（4，+∞）内的概率为

\u5728\u67d0\u9879\u6d4b\u91cf\u4e2d\uff0c\u6d4b\u91cf\u7ed3\u679c\u03be\u670d\u4ece\u6b63\u6001\u5206\u5e03N\uff082\uff0c\u03c32\uff09\uff08\u03c3\uff1e0\uff09\uff0c\u82e5\u03be\u5728\uff080\uff0c2\uff09\u5185\u53d6\u503c\u7684\u6982\u7387\u4e3a0.4\uff0c\u5219\u03be\u5728

\u2235\u03be\u670d\u4ece\u6b63\u6001\u5206\u5e03N\uff082\uff0c\u03c32\uff09\u2234\u66f2\u7ebf\u7684\u5bf9\u79f0\u8f74\u662f\u76f4\u7ebfx=2\uff0c\u2235\u03be\u5728\uff080\uff0c2\uff09\u5185\u53d6\u503c\u7684\u6982\u7387\u4e3a0.4\uff0c\u2234\u6839\u636e\u6b63\u6001\u66f2\u7ebf\u7684\u6027\u8d28\u77e5\u5728\uff080\uff0c+\u221e\uff09\u5185\u53d6\u503c\u7684\u6982\u7387\u4e3a0.4+0.5=0.9\uff0e\u6545\u9009\uff1aD\uff0e

\u968f\u673a\u53d8\u91cf\u03be\u670d\u4ece\u6b63\u6001\u5206\u5e03N\uff081\uff0c\u03b42\uff09\uff0c\u2234\u66f2\u7ebf\u5173\u4e8ex=1\u5bf9\u79f0\uff0c\u2234P\uff080\uff1c\u03be\uff1c1\uff09=P\uff081\uff1c\u03be\uff1c2\uff09=0.4\uff0c\u2234P\uff080\uff1c\u03be\uff1c2\uff09=2P\uff080\uff1c\u03be\uff1c1\uff09=0.8\uff0c\u6545\u9009D\uff0e

由随机变量X～N（2，σ²）知，X的均值为2，其图象关于x=2对称，
故X在（4，+∞）内的概率为P=

1

2

（1-0.3×2）=0.2
故答案为：0.2

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