If for a distribution,Sigma(x-5)=3 and Sigma( | vedclass

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(C) Let $d_i = x_i - 5$. We are given $\Sigma d_i = 3$ and $\Sigma d_i^2 = 43$ with $n = 18$. The variance of a distribution is invariant under change

Vedclass · Accepted Answer

$2.36$

Vedclass · Answer

(C) Let $d_i = x_i - 5$. We are given $\Sigma d_i = 3$ and $\Sigma d_i^2 = 43$ with $n = 18$. The variance of a distribution is invariant under change of origin. Therefore,the variance of $x_i$ is the same as the variance of $d_i$. The formula for variance is $\sigma^2 = \frac{\Sigma d_i^2}{n} - \left( \frac{\Sigma d_i}{n} \right)^2$. Substituting the values: $\sigma^2 = \frac{43}{18} - \left( \frac{3}{18} \right)^2$. $\sigma^2 = \frac{43}{18} - \left( \frac{1}{6} \right)^2 = \frac{43}{18} - \frac{1}{36}$. $\sigma^2 = \frac{86 - 1}{36} = \frac{85}{36} \approx 2.3611$. Thus,the variance is approximately $2.36$.

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

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