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