The data shows the total number of employee medical leave days taken for on-the-job accidents in the first six months of the year: 12, 6, 15, 9, 18, 12. Use the data for the Exercise. Find the sum of squares of the deviations from the mean.

90

Step-by-step explanation:

The sum of squares of the deviation from mean=sum (x-xbar) ²=?

x

12

6

15

9

18

12

xbar=sumx/n

xbar = (12+6+15+9+18+12) / 6=72/6=12

x x-xbar (x-xbar) ²

12 12-12=0 0

6 6-12=-6 36

15 15-12=3 9

9 9-12=-3 9

18 18-12=6 36

12 12-12=0 0

sum (x-xbar) ²=0+36+9+9+36+0=90

So, the sum of squares of the deviations from the mean is 90.