Question

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)
average: 2.90 uncertainty: 0.19
average: 2.32 uncertainty: 0.16
average: 2.32 uncertainty: 0.19
average: 2.9 uncertainty: 0.16

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