Solution
Given the following scores below
![31,22,24,15,23](https://img.qammunity.org/2023/formulas/mathematics/high-school/bvrje1ujikm2l1irdf3ma2n33q2sxsd2pe.png)
To find the sum of squares of the deviations,
Firstly, we find the mean, the formula to find the mean of ungrouped data is
![\begin{gathered} \bar{x}=(\sum_X)/(n) \\ Where\text{ n is the number of terms} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/iay491uhy9o7gd28w84stoiahwv3zkvf7r.png)
Where n = 5
Substitute the given data into the formula to find the mean of ungrouped data
![\begin{gathered} \bar{x}=(31+22+24+15+23)/(5) \\ \bar{x}=(115)/(5)=23 \\ \bar{x}=23 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/bhjkyaifh3doiluz73vpcs9xaygezsezzi.png)
Secondly, we subtract the mean from each of the values given as shown below
![\begin{gathered} 31-23=8 \\ 22-23=-1 \\ 24-23=1 \\ 15-23=-8 \\ 23-23=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vm26kyyk13zgqzwnztj0jxte0zsynhpu2q.png)
Then, to find the sum of squares of the deviations
We add the square of each of the deviations deduced above
![\begin{gathered} =(8)^2+(-1)^2+(1)^2+(-8)^2+(0)^2 \\ =64+1+1+64+0 \\ =130 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/fcs4wc8k7a0mwmz18o1wua4hgvd6hhbpd8.png)
Hence, the sum of squares is 130