Variability

In the table bellow the number of wrong produced goods in two shifts:

morning shift: 2; 0; 6; 10; 2; 2; 4; 2; 5; 2;
afternoon shift: 4; 4; 0; 2; 10; 2; 6; 2; 3; 10;

Compare the variability in both shifts, compare the average number of wrong goods on both shifts as well as other variability and positioning rates.

Correct answer:

a₁ =  3.5
a₂ =  4.3
s₁ =  2.7295
s₂ =  3.2265

Step-by-step explanation:

$a_1=(2+0+6+10+2+2+4+2+5+2)\mathrm{/}10={\displaystyle \frac{7}{2}}=3\frac{1}{2}=3.5$

$$\begin{gathered} a_2=(4+4+0+2+10+2+6+2+3+10)\mathrm{/}10={\displaystyle \frac{43}{10}}=4\frac{3}{10}=4.3 \\ {a}_2>a_1 \end{gathered}$$

$$\begin{gathered} {b}_1=(2-a_1{)}^2+(0-a_1{)}^2+(6-a_1{)}^2+(10-a_1{)}^2+(2-a_1{)}^2+(2-a_1{)}^2+(4-a_1{)}^2+(2-a_1{)}^2+(5-a_1{)}^2+(2-a_1{)}^2=(2-3.5{)}^2+(0-3.5{)}^2+(6-3.5{)}^2+(10-3.5{)}^2+(2-3.5{)}^2+(2-3.5{)}^2+(4-3.5{)}^2+(2-3.5{)}^2+(5-3.5{)}^2+(2-3.5{)}^2={\displaystyle \frac{149}{2}}=74\frac{1}{2}=74.5 \\ {r}_1=1\mathrm{/}10\cdot b_1=1\mathrm{/}10\cdot 74.5={\displaystyle \frac{149}{20}}=7\frac{9}{20}=7.45 \\ {s}_1=\sqrt{r_1}=\sqrt{7.45}=2.7295 \end{gathered}$$

$$\begin{gathered} {b}_2=(4-a_2{)}^2+(4-a_2{)}^2+(0-a_2{)}^2+(2-a_2{)}^2+(10-a_2{)}^2+(2-a_2{)}^2+(6-a_2{)}^2+(2-a_2{)}^2+(3-a_2{)}^2+(10-a_2{)}^2=(4-4.3{)}^2+(4-4.3{)}^2+(0-4.3{)}^2+(2-4.3{)}^2+(10-4.3{)}^2+(2-4.3{)}^2+(6-4.3{)}^2+(2-4.3{)}^2+(3-4.3{)}^2+(10-4.3{)}^2={\displaystyle \frac{1041}{10}}=104\frac{1}{10}=104.1 \\ {r}_2=1\mathrm{/}10\cdot b_2=1\mathrm{/}10\cdot 104.1={\displaystyle \frac{1041}{100}}=10\frac{41}{100}=10.41 \\ {s}_2=\sqrt{r_2}=\sqrt{10.41}\doteq 3.2265=3.2265 \\ {s}_2>s_1 \end{gathered}$$

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