import math

#网上找来的正态分布的分布函数，返回z>u的概率，z服从标准正态分布
def st_norm(u):
    x=abs(u)/math.sqrt(2)
    T=(0.0705230784,0.0422820123,0.0092705272,
    0.0001520143,0.0002765672,0.0000430638)
    E=1-pow((1+sum([a*pow(x,(i+1))
    for i,a in enumerate(T)])),-16)
    p=0.5-0.5*E if u else 0.5+0.5*E
    if u>=0:
        return p
    else:
        return 1-st_norm(-u)