mean and standard devition
asked Aug 4, 2014 in Statistics Answers by anonymous

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If I understand this correctly the summation of each datum minus 50 is 123.5 and the summation of the same difference squared is 238.4 and there are 100 data in the set. So I would say that 50 is a weighted factor which is subtracted from each datum. The actual sum of the data would then be 100*50+123.5=5123.5 and the mean is 5123.5/100=51.235. The variance is the sum of the squares of the differences of each datum and the mean, but we have the sum of the squares of the differences of each datum and 50, which is not the mean.

Let's call S1 summation(x-50)^2 and S2 summation(x-51.235)^2. S1=238.4, so we need S2 to calculate the variance. The square root of the variance will be the standard deviation. Variance=S2/100 and SD=sqrt(S2)/10. If x1, x2, x3, ..., x100 represent the dataset:

S1=238.4=(x1-50)^2+(x2-50)^2+...+(x100-50)^2=(x1^2+x2^2+...+x100^2)-100(x1+x2+...+x100)+100*50^2.

S2=(x1^2+...+x100^2)-102.47(x1+...+x100)+100*51.235^2.

S2-S1=100(51.235^2-50^2)-2.47(x1+...+x100)=12502.5225-2.47(x1+...+x100)=12502.1225-2.47*5123.5=-152.5225.

Therefore, S2=S1-152.5225=238.4-152.5225=85.8775.

The variance is 0.858775 and SD=0.9267.

answered Oct 30, 2015 by Rod Top Rated User (430,620 points)
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