Respuesta :
Solution
Given the following scores below
[tex]31,22,24,15,23[/tex]To find the sum of squares of the deviations,
Firstly, we find the mean, the formula to find the mean of ungrouped data is
[tex]\begin{gathered} \bar{x}=\frac{\sum_X}{n} \\ Where\text{ n is the number of terms} \end{gathered}[/tex]Where n = 5
Substitute the given data into the formula to find the mean of ungrouped data
[tex]\begin{gathered} \bar{x}=\frac{31+22+24+15+23}{5} \\ \bar{x}=\frac{115}{5}=23 \\ \bar{x}=23 \end{gathered}[/tex]Secondly, we subtract the mean from each of the values given as shown below
[tex]\begin{gathered} 31-23=8 \\ 22-23=-1 \\ 24-23=1 \\ 15-23=-8 \\ 23-23=0 \end{gathered}[/tex]Then, to find the sum of squares of the deviations
We add the square of each of the deviations deduced above
[tex]\begin{gathered} =(8)^2+(-1)^2+(1)^2+(-8)^2+(0)^2 \\ =64+1+1+64+0 \\ =130 \end{gathered}[/tex]Hence, the sum of squares is 130
