我试图创建一个新列,名为duration_probablity
,它可以获取值在6到12小时之间的概率。P(6 < Origin_Duration ≤ 12)
dput(df)
structure(list(CRD_NUM = c(1000120005478330, 1000130009109199,
1000140001635234, 1000140002374747, 1000140003618308, 1000140007236959,
1000140015078086, 1000140026268650, 1000140027281272, 1000148000012215
), Origin_Duration = c("10:48:38", "07:41:34", "11:16:41", "09:19:35",
"17:09:19", "08:59:05", "11:27:28", "12:17:41", "10:45:42", "12:19:05"
)), .Names = c("CRD_NUM", "Origin_Duration"), class = c("data.table",
"data.frame"), row.names = c(NA, -10L))
CRD_NUM Origin_Duration
1: 1000120005478330 10:48:38
2: 1000130009109199 07:41:34
3: 1000140001635234 11:16:41
4: 1000140002374747 09:19:35
5: 1000140003618308 17:09:19
6: 1000140007236959 08:59:05
7: 1000140015078086 11:27:28
8: 1000140026268650 12:17:41
9: 1000140027281272 10:45:42
10: 1000148000012215 12:19:05
我不确定如何在R中实现这种操作。我正在尝试获取标准正态分布的累积分布函数。计算通勤者在某个站点逗留时间介于6到12小时之间的概率。例如,输出结果可能为11:16:41的持续时间为0.96。
我的累积分布函数应该类似于-
P(6
P(6 <X≤ 12) = Φ((12−μ)/σ)−Φ((6−μ)/σ)
@SotosP(6 <X≤ 12) = Φ((12−μ)/σ)−Φ((6−μ)/σ)
- Prasanna Nandakumarpnorm
,但是你的公式中缺少一些东西:数据与分布参数之间的联系。 - Vincent Guillemot