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 result:

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

Solution:

a1=(2+0+6+10+2+2+4+2+5+2)/10=72=3.5
a2=(4+4+0+2+10+2+6+2+3+10)/10=4310=4.3 a2>a1
 b1=(2a1)2+(0a1)2+(6a1)2+(10a1)2+(2a1)2+(2a1)2+(4a1)2+(2a1)2+(5a1)2+(2a1)2=(23.5)2+(03.5)2+(63.5)2+(103.5)2+(23.5)2+(23.5)2+(43.5)2+(23.5)2+(53.5)2+(23.5)2=1492=74.5  r1=1/10 b1=1/10 74.5=14920=7.45  s1=r1=7.45=2.7295
 b2=(4a2)2+(4a2)2+(0a2)2+(2a2)2+(10a2)2+(2a2)2+(6a2)2+(2a2)2+(3a2)2+(10a2)2=(44.3)2+(44.3)2+(04.3)2+(24.3)2+(104.3)2+(24.3)2+(64.3)2+(24.3)2+(34.3)2+(104.3)2=104110=104.1  r2=1/10 b2=1/10 104.1=1041100=10.41  s2=r2=10.413.2265=3.2265  s2>s1



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