The data shows the total number of employee medical leave days taken for on-the-job accidents in the first six months of the yea
r: 12, 6, 15, 9, 18, 12. Use the data for the Exercise. Find the sum of squares of the deviations from the mean.
1 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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