鸟 · 2020年01月08日
问题如下：
For a normally distributed random variable, the probability that random varible is greater than three standard deviations excess the mean is:
选项：
A.0.0013.
B.0.0228.
C.0.9987.
解释：
A is correct.
1 - F(3) = 1 - 0.9987 = 0.0013.
题目是作对了，但这题好像不用查表也可以吧，3明显大于2.58 ，结果应该是小于1%，可以这么理解么
NO.PZ2015120604000106 问题如下 For a normally stributeranm variable, the probability thranm varible is greater ththree stanrviations excess the meis: A.0.0013. B.0.0228. C.0.9987. A is correct.1 - F(3) = 1 - 0.9987 = 0.0013. 如标题
NO.PZ2015120604000106问题如下For a normally stributeranm variable, the probability thranm varible is greater ththree stanrviations excess the meis:A.0.0013.B.0.0228.C.0.9987.A is correct.1 - F(3) = 1 - 0.9987 = 0.0013.这里Z大于等于3，3是怎么算出来的？
NO.PZ2015120604000106问题如下For a normally stributeranm variable, the probability thranm varible is greater ththree stanrviations excess the meis:A.0.0013.B.0.0228.C.0.9987.A is correct.1 - F(3) = 1 - 0.9987 = 0.0013.为什么就等于是求X-μ≥3σ的概率？X-μ是什么
NO.PZ2015120604000106 问题如下 For a normally stributeranm variable, the probability thranm varible is greater ththree stanrviations excess the meis: A.0.0013. B.0.0228. C.0.9987. A is correct.1 - F(3) = 1 - 0.9987 = 0.0013. '题目问服从正态分布的随机变量落在超过均值3个标准差的部分的概率为多少。也就是要求的就是X-μ≥3σ的概率\",老师为什么题干可以解读出要求的是X-μ≥3σ的概率，能一下这个不等式是怎么得出来的吗，运用的是哪个知识点？
NO.PZ2015120604000106 0.0228. 0.9987. A is correct. 1 - F(3) = 1 - 0.9987 = 0.0013. 根据正态分布标准化的公式（X-μ）/σ进行转化，相当于求P(Z≥3)。———解析中这句话没有理解
