If $x_1, x_2, x_3 \ldots x_n$ are $n$ observations such that $\Sigma\left(x_i+2\right)^2=28 n$ and…
If $x_1, x_2, x_3 \ldots x_n$ are $n$ observations such that $\Sigma\left(x_i+2\right)^2=28 n$ and $\Sigma\left(x_i-2\right)^2=12 n$, then the variance is:
12
14
16
20
Solution
$\begin{aligned} & \text { Given that } \sum\left(x_i+2\right)^2=28 n \\ & \Rightarrow \sum x_i^2+4 \sum x_i+4 n=28 n \\ & \Rightarrow \sum x_i^2+4 \sum x_i=24 n ....(i)\end{aligned}$
Similarly
$\sum\left(x_i-2\right)^2=12 n \Rightarrow \sum x_i^2-4 \sum x_i=8 n$ On solving equation (i) and (ii), we get
$\sum x_i^2=16 n \text { and } \sum x_i=2 n$
$\therefore$ Variance $=\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2$
$=\frac{16 n}{n}-\left(\frac{2 n}{n}\right)^2=16-4=12$