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:

a2=(4+4+0+2+10+2+6+2+3+10)/10=4310=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=7412=74.5  r1=1/10 b1=1/10 74.5=14920=7920=7.45  s1=r1=7.45=2.7295

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