The variance of the data set 10.4, 10.3, 11.7, 11.1, 8.0, 4.4, 2.6, 1.8, 2.5, 4.4, 7.3, 9.5 is obtained to be 12.4 approximately.
Variance is the average of the squared deviation of the data values from their mean.
Thus, if [tex]x_1, x_2, \cdots, x_n[/tex] are n data values of the considered data set, then we get variance of that data set as:
[tex]\sigma^2 = \dfrac{\sum_{i=1}^n(x_i - \overline{x})^2}{n}[/tex]
where [tex]\overline{x}[/tex] is the mean of the data set's observations.
The mean of the considered data set is:
[tex]\overline{x} = \dfrac{\sum_{i=1}^n(x_i)}{n}\\\\\overline{x} = \dfrac{10.4+10.3+11.7+11.1+8.0+4.4+2.6+1.8+2.5+4.4+7.3+9.5}{12} \\\\\overline{x} = \dfrac{84}{12} = 7[/tex]
Thus, the variance of the considered data set is:
[tex]\sigma^2 = \dfrac{\sum_{i=1}^n(x_i - \overline{x})^2}{n}\\\\\sigma^2 = \dfrac{(10.4-7)^2 + (10.3-7)^2 + \cdots + (9.5 -7)^2}{12} \\\\\sigma^2 = \dfrac{11.56 + 10.89 + \cdots + 6.25}{12} \approx 12.4[/tex]
Thus, the variance of the data set 10.4, 10.3, 11.7, 11.1, 8.0, 4.4, 2.6, 1.8, 2.5, 4.4, 7.3, 9.5 is obtained to be 12.4 approximately.
Learn more about variance here:
https://brainly.com/question/3699980
