正态分布范围内列的概率

从你的问题来看,不清楚你是否已经知道均值和方差。我将讨论这两种情况。此外,我假设你们有理由相信持续时间实际上是正态分布的。

如果给定了预先指定的均值和方差。说 mu = 11 和 sigma = 3 P(6 < X ≤ 12) = P(X ≤ 12) - P(X ≤ 6) . 基R函数 pnorm() 能够计算:

mu    <- 11
sigma <- 3
pnorm(12, mu, sigma) - pnorm(6, mu, sigma)
# 0.5827683

P(6 < X < 12) student t-distribution df$Origin_Duration 从字符到某个时间类型:

df$Origin_Duration <- as.POSIXct(df$Origin_Duration, format = "%H:%M:%S")

mu          <- mean(df$Origin_Duration)       # "2017-09-04 11:12:28 CEST"
df$demeaned <- df$Origin_Duration - mu
sigma       <- var(df$demeaned)^0.5           # 153.68 

我们将使用 pt 计算概率的函数 P(X ≤ 12) - P(X ≤ 6) . 为此,我们需要一个标准化/比例化/标准化版本的 12 和 6 . 也就是说,我们必须减去平均值,然后除以标准差:

x6  <- as.numeric(difftime("2017-09-04 06:00:00", mu), unit = "mins")/sigma
x12 <- as.numeric(difftime("2017-09-04 12:00:00", mu), unit = "mins")/sigma

deg_fr <- length(df$demeaned)-1

p_x_smaller_than12 <- pt( x12, df = deg_fr )    #  0.6178973
p_x_smaller_than6  <- pt( x6,  df = deg_fr )    #  0.03627651
p_x_smaller_than12 - p_x_smaller_than6
# [1] 0.5816208

在回复注释时添加:未知参数,所有条目:

# scale gives the distance from the mean in terms of standard deviations:
df$scaled <- scale(df$Origin_Duration)

pt(df$scaled, df = deg_fr)
# [1,] 0.4400575
# [2,] 0.1015886
# [3,] 0.5106114
# [4,] 0.2406431
# [5,] 0.9773264
# [6,] 0.2039751
# [7,] 0.5377728
# [8,] 0.6593331
# [9,] 0.4327620
# [10,] 0.6625280