Note: These are samples of size n=2 so we divide by n-1 = 2-1 or 1, not n=2
sample means variances
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1,1 (1+1)/2 = 1.0 [(1-1.0)²+(1-1.0)²]/1 = 0.0/1 = 0.0
1,2 (1+2)/2 = 1.5 [(1-1.5)²+(2-1.5)²]/1 = 0.5/1 = 0.5
1,3 (1+3)/2 = 2.0 [(1-2.0)²+(3-2.0)²]/1 = 2.0/1 = 2.0
1,4 (1+4)/2 = 2.5 [(1-2.5)²+(4-2.5)²]/1 = 4.5/1 = 4.5
1,5 (1+5)/2 = 3.0 [(1-3.0)²+(5-3.0)²]/1 = 8.0/1 = 8.0
2,1 (2+1)/2 = 1.5 [(2-1.5)²+(1-1.5)²]/1 = 0.5/1 = 0.5
2,2 (2+2)/2 = 2.0 [(2-2.0)²+(2-2.0)²]/1 = 0.0/1 = 0.0
2,3 (2+3)/2 = 2.5 [(2-2.5)²+(3-2.5)²]/1 = 0.5/1 = 0.5
2,4 (2+4)/2 = 3.0 [(2-3.0)²+(4-3.0)²]/1 = 2.0/1 = 2.0
2,5 (2+5)/2 = 3.5 [(2-3.5)²+(5-3.5)²]/1 = 4.5/1 = 4.5
3,1 (3+1)/2 = 2.0 [(3-2.0)²+(1-2.0)²]/1 = 2.0/1 = 2.0
3,2 (3+2)/2 = 2.5 [(3-2.5)²+(2-2.5)²]/1 = 0.5/1 = 0.5
3,3 (3+3)/2 = 3.0 [(3-3.0)²+(3-3.0)²]/1 = 0.0/1 = 0.0
3,4 (3+4)/2 = 3.5 [(3-3.5)²+(4-3.5)²]/1 = 0.5/1 = 0.5
3,5 (3+5)/2 = 4.0 [(3-4.0)²+(5-4.0)²]/1 = 2.0/1 = 2.0
4,1 (4+1)/2 = 2.5 [(4-2.5)²+(1-2.5)²]/1 = 4.5/1 = 4.5
4,2 (4+2)/2 = 3.0 [(4-3.0)²+(2-3.0)²]/1 = 2.0/1 = 2.0
4,3 (4+3)/2 = 3.5 [(4-3.5)²+(3-3.5)²]/1 = 0.5/1 = 0.5
4,4 (4+4)/2 = 4.0 [(4-4.0)²+(4-4.0)²]/1 = 0.0/1 = 0.0
4,5 (4+5)/2 = 4.5 [(4-4.5)²+(5-4.5)²]/1 = 0.5/1 = 0.5
5,1 (5+1)/2 = 3.0 [(5-3.0)²+(1-3.0)²]/1 = 8.0/1 = 8.0
5,2 (5+2)/2 = 3.5 [(5-3.5)²+(2-3.5)²]/1 = 4.5/1 = 4.5
5,3 (5+3)/2 = 4.0 [(5-4.0)²+(3-4.0)²]/1 = 2.0/1 = 2.0
5,4 (5+4)/2 = 4.5 [(5-4.5)²+(4-4.5)²]/1 = 0.5/1 = 0.5
5,5 (5+5)/2 = 5.0 [(5-5.0)²+(5-5.0)²]/1 = 0.0/1 = 0.0
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sum = 50.0
There are 25 samples, so mean of variances = 50.0/25 = 2
Population: 1,2,3,4,5
The mean of the population is m = (1+2+3+4+5)/5 = 15/5 = 3
Note: Since this is a population, we will divide by n=5, (not n-1):
x (x-m)²
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1 (1-3)² = 4
2 (2-3)² = 1
3 (3-3)² = 0
4 (4-3)² = 1
5 (5-3)² = 4
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5)10
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2 = variance of population
So they are the same, both 2.
Edwin