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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.

Answer :

Answer:

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.

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