7个数据的A类不确定度计算方法 - 详细步骤与公式

首先计算平均值：

$$\bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{189.8+189.4+189.6+189.7+190.0+189.5+189.7}{7} = 189.7~\text{mm}$$

然后计算每个数据与平均值的偏差：

$$\Delta x_i = x_i - \bar{x}$$

分别为：

$$\Delta x_1 = 0.1~\text{mm},~~\Delta x_2 = -0.3~\text{mm},~~\Delta x_3 = -0.1~\text{mm},~~\Delta x_4 = 0~\text{mm},~~\Delta x_5 = 0.3~\text{mm},~~\Delta x_6 = -0.2~\text{mm},~~\Delta x_7 = 0~\text{mm}$$

下面计算偏差的平均值：

$$\bar{\Delta x} = \frac{1}{n}\sum_{i=1}^{n}\Delta x_i = \frac{0.1-0.3-0.1+0+0.3-0.2+0}{7} = 0~\text{mm}$$

然后计算每个偏差的平方：

$$\Delta x_i^2 = (\Delta x_i)^2$$

分别为：

$$\Delta x_1^2 = 0.01~\text{mm}^2,~~\Delta x_2^2 = 0.09~\text{mm}^2,~~\Delta x_3^2 = 0.01~\text{mm}^2,~~\Delta x_4^2 = 0~\text{mm}^2,~~\Delta x_5^2 = 0.09~\text{mm}^2,~~\Delta x_6^2 = 0.04~\text{mm}^2,~~\Delta x_7^2 = 0~\text{mm}^2$$

最后计算偏差平方的平均值：

$$\bar{\Delta x^2} = \frac{1}{n}\sum_{i=1}^{n}\Delta x_i^2 = \frac{0.01+0.09+0.01+0+0.09+0.04+0}{7} = 0.03~\text{mm}^2$$

于是A类不确定度为：

$$u_A = \sqrt{\bar{\Delta x^2}} = \sqrt{0.03} \approx 0.17~\text{mm}$