小可爱和大灰狼 · 2020年10月19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.
老师，这道题如果改成要求变量落在距离均值三倍标准差之外的概率应该怎么表述呢？ 这个时候应该把求出来的这个概率✖️2吧？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
