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:

a1​=(2+0+6+10+2+2+4+2+5+2)/10=3.5
a2​=(4+4+0+2+10+2+6+2+3+10)/10=1043​=4103​=4.3 a2​>a1​
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=2149​=7421​=74.5  r1​=1/10⋅ b1​=1/10⋅ 74.5=20149​=7209​=7.45  s1​=r1​​=7.45​=2.7295
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=101041​=104101​=104.1  r2​=1/10⋅ b2​=1/10⋅ 104.1=1001041​=1010041​=10.41  s2​=r2​​=10.41​≐3.2265  s2​>s1​

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