If for a distribution, $\Sigma(x-5)=3$ and $\Sigma(x-5)^2=43$ and total number of observations is 18 , then the variance of the distribution is

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Aicharya · Accepted Answer

Given $\Sigma(x-5)=3$ and $\Sigma(x-5)^2=43$ Number of observations (n) = 18 To Find Variance = ? $\begin{aligned} \because 	ext { Variance } & =\frac{\Sigma\left(x_i-\bar{x}ight)^2}{n} \ & =\frac{\Sigma(x-5)^2}{18}=\frac{43}{18}=2.36\end{aligned}$

If for a distribution, $\Sigma(x-5)=3$ and $\Sigma(x-5)^2=43$ and total number of observations is 18 , then…

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