# 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:

${a}_{1}=\left(2+0+6+10+2+2+4+2+5+2\right)\mathrm{/}10=\frac{7}{2}=3.5$

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