如何使用 ruby 将 z-score 转换为概率?
例子:
z_score = 0
probability should be 0.5
z_score = 1.76
probability should be 0.039204
最佳答案
根据这个https://stackoverflow.com/a/16197404/1062711发布,这是根据 z 分数为您提供 p 概率的函数
def getPercent(z)
return 0 if z < -6.5
return 1 if z > 6.5
factk = 1
sum = 0
term = 1
k = 0
loopStop = Math.exp(-23)
while term.abs > loopStop do
term = 0.3989422804 * ((-1)**k) * (z**k) / (2*k+1) / (2**k) * (z**(k+1)) /factk
sum += term
k += 1
factk *= k
end
sum += 0.5
1-sum
end
puts getPercent(1.76)
关于ruby - z 分数到概率,反之亦然,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31875909/