Q. If $\sum_{i = 1}^{18}{\left( x_{i} - 8 \right) = 9\ and\ }\sum_{i = 1}^{18}{\left( x_{i} - 8 \right)^{2} = 45}$, then the standard derivation of $x_{1},x_{2},\ldots,x_{18}$ is
A) $\frac{4}{9}$
B) $\frac{9}{4}$
C) $\frac{3}{2}$
D) None of these
Explanation
(c)
Let $d_{i} = x_{i} - 8$
$\therefore\ \sigma_{x}^{2} = \sigma_{d}^{2} = \frac{1}{18}\sum_{}^{}{d_{i}^{2} -}\left( \frac{1}{8}\sum_{}^{}d_{i} \right)^{2}$
$= \frac{1}{18} \times 45 - \left( \frac{9}{18} \right)^{2} = \frac{5}{2} - \frac{1}{4} = \frac{9}{4}$
$\Rightarrow \ \sigma_{x}^{2} = \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.