# 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:

$a_{1}=(2+0+6+10+2+2+4+2+5+2)/10=\dfrac{ 7 }{ 2 }=3.5$
$a_{2}=(4+4+0+2+10+2+6+2+3+10)/10=\dfrac{ 43 }{ 10 }=4.3 \ \\ a_{2}>a_{1}$

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