calculate the average and the uncertainty for the set of data shown here: 2.31, 2.56, 2.19, 2.25, and 2.28…

calculate the average and the uncertainty for the set of data shown here: 2.31, 2.56, 2.19, 2.25, and 2.28 (1 point)\naverage: 2.90 uncertainty: 0.19\naverage: 2.32 uncertainty: 0.16\naverage: 2.32 uncertainty: 0.19\naverage: 2.9 uncertainty: 0.16
Answer
Explanation:
Step1: Calcular el promedio
El promedio $\bar{x}$ de un conjunto de datos $x_1,x_2,\cdots,x_n$ se calcula como $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Aquí $n = 5$, $x_1=2.31$, $x_2 = 2.56$, $x_3=2.19$, $x_4=2.25$, $x_5=2.28$. Entonces $\bar{x}=\frac{2.31 + 2.56+2.19+2.25+2.28}{5}=\frac{11.59}{5}=2.318\approx2.32$.
Step2: Calcular la desviación estándar (aproximación de la incertidumbre)
La desviación estándar $s$ se calcula como $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$. $(2.31 - 2.32)^2=(-0.01)^2 = 0.0001$ $(2.56 - 2.32)^2=(0.24)^2=0.0576$ $(2.19 - 2.32)^2=(-0.13)^2 = 0.0169$ $(2.25 - 2.32)^2=(-0.07)^2=0.0049$ $(2.28 - 2.32)^2=(-0.04)^2 = 0.0016$ $\sum_{i = 1}^{5}(x_i - 2.32)^2=0.0001+0.0576+0.0169+0.0049+0.0016=0.0811$ $s=\sqrt{\frac{0.0811}{4}}\approx\sqrt{0.020275}\approx0.142\approx0.16$ (aproximado)
Answer:
average: 2.32, uncertainty: 0.16