Lesson variability comparison

In September, students missed the following number of lessons:
Girls: 2; 0; 7; 8; 2; 2; 4;
Boys: 4; 4; 0; 3; 10; 3; 8;
Compare the variability of both statistical files.

Final Answer:

r₁ =  7.3878
r₂ =  9.6735

Step-by-step explanation:

$$\begin{gathered} {\mu }_1=\frac{2+0+7+8+2+2+4}{7}=\frac{25}{7}=3\frac{4}{7}\doteq 3.5714 \\ {\mu }_2=\frac{4+4+0+3+10+3+8}{7}=\frac{32}{7}=4\frac{4}{7}\doteq 4.5714 \\ {e}_1=(2-{\mu }_1{)}^2+(0-{\mu }_1{)}^2+(7-{\mu }_1{)}^2+(8-{\mu }_1{)}^2+(2-{\mu }_1{)}^2+(2-{\mu }_1{)}^2+(4-{\mu }_1{)}^2=(2-3.5714{)}^2+(0-3.5714{)}^2+(7-3.5714{)}^2+(8-3.5714{)}^2+(2-3.5714{)}^2+(2-3.5714{)}^2+(4-3.5714{)}^2=\frac{362}{7}=51\frac{5}{7}\doteq 51.7143 \\ {r}_1=\frac{e_1}{7}=\frac{51.7143}{7}=7.3878 \end{gathered}$$

$$\begin{gathered} e_2=(4-{\mu }_2{)}^2+(4-{\mu }_2{)}^2+(0-{\mu }_2{)}^2+(3-{\mu }_2{)}^2+(10-{\mu }_2{)}^2+(3-{\mu }_2{)}^2+(8-{\mu }_2{)}^2=(4-4.5714{)}^2+(4-4.5714{)}^2+(0-4.5714{)}^2+(3-4.5714{)}^2+(10-4.5714{)}^2+(3-4.5714{)}^2+(8-4.5714{)}^2=\frac{474}{7}=67\frac{5}{7}\doteq 67.7143 \\ {r}_2=\frac{e_2}{7}=\frac{67.7143}{7}=\frac{474}{49}=9\frac{33}{49}\doteq 9.6735 \\ {r}_2=r_1<r_2=9.6735 \end{gathered}$$

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