If for a distribution $\Sigma(x-5)=3$,$\Sigma(x-5)^{2}=43$ and the total number of items is $18$,find the mean and standard deviation.
Given: $n=18$,$\Sigma(x-5)=3$,and $\Sigma(x-5)^{2}=43$.
Let $d = x-5$. Then $\Sigma d = 3$ and $\Sigma d^2 = 43$.
The mean of $d$ is $\bar{d} = \frac{\Sigma d}{n} = \frac{3}{18} = \frac{1}{6} \approx 0.167$.
The mean of $x$ is $\bar{x} = 5 + \bar{d} = 5 + 0.167 = 5.167$.
The standard deviation is independent of the origin,so $SD(x) = SD(d)$.
$SD(d) = \sqrt{\frac{\Sigma d^2}{n} - (\bar{d})^2} = \sqrt{\frac{43}{18} - (\frac{1}{6})^2} = \sqrt{2.3889 - 0.0278} = \sqrt{2.3611} \approx 1.537$.
Let $d = x-5$. Then $\Sigma d = 3$ and $\Sigma d^2 = 43$.
The mean of $d$ is $\bar{d} = \frac{\Sigma d}{n} = \frac{3}{18} = \frac{1}{6} \approx 0.167$.
The mean of $x$ is $\bar{x} = 5 + \bar{d} = 5 + 0.167 = 5.167$.
The standard deviation is independent of the origin,so $SD(x) = SD(d)$.
$SD(d) = \sqrt{\frac{\Sigma d^2}{n} - (\bar{d})^2} = \sqrt{\frac{43}{18} - (\frac{1}{6})^2} = \sqrt{2.3889 - 0.0278} = \sqrt{2.3611} \approx 1.537$.
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