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.

Result

a1 =  3.5
a2 =  4.3
s1 =  2.729
s2 =  3.226

Solution:

a1=(2+0+6+10+2+2+4+2+5+2)/10=72=3.5a_{1}=(2+0+6+10+2+2+4+2+5+2)/10=\dfrac{ 7 }{ 2 }=3.5
a2=(4+4+0+2+10+2+6+2+3+10)/10=4310=4.3 a2>a1a_{2}=(4+4+0+2+10+2+6+2+3+10)/10=\dfrac{ 43 }{ 10 }=4.3 \ \\ a_{2}>a_{1}
 b1=(2−a1)2+(0−a1)2+(6−a1)2+(10−a1)2+(2−a1)2+(2−a1)2+(4−a1)2+(2−a1)2+(5−a1)2+(2−a1)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=1492=74.5  r1=1/10⋅ b1=1/10⋅ 74.5=14920=7.45  s1=r1=7.45≐2.7295≐2.729 \ \\ 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 }=\dfrac{ 149 }{ 2 }=74.5 \ \\ \ \\ r_{1}=1/10 \cdot \ b_{1}=1/10 \cdot \ 74.5=\dfrac{ 149 }{ 20 }=7.45 \ \\ \ \\ s_{1}=\sqrt{ r_{1} }=\sqrt{ 7.45 } \doteq 2.7295 \doteq 2.729
 b2=(4−a2)2+(4−a2)2+(0−a2)2+(2−a2)2+(10−a2)2+(2−a2)2+(6−a2)2+(2−a2)2+(3−a2)2+(10−a2)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=104110=104.1  r2=1/10⋅ b2=1/10⋅ 104.1=1041100=10.41  s2=r2=10.41≐3.2265≐3.226  s2>s1 \ \\ 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=\dfrac{ 1041 }{ 10 }=104.1 \ \\ \ \\ r_{2}=1/10 \cdot \ b_{2}=1/10 \cdot \ 104.1=\dfrac{ 1041 }{ 100 }=10.41 \ \\ \ \\ s_{2}=\sqrt{ r_{2} }=\sqrt{ 10.41 } \doteq 3.2265 \doteq 3.226 \ \\ \ \\ s_{2}>s_{1}

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