Question

If M = $\frac{x_1+x_2+...+x_{20}}{20}$, then the value of ${\sum }_{i-1}^{20}\frac{(x_i-M)}{20}$ is

• A. $\frac{19M}{20}$
• B. 1
• C. 0
• D. $\frac{1}{20}$

Answer 1

Answer: (A)

$\frac{19M}{20}$

Given that:

$M=\frac{x_1+x_2+x_3+...x_{20}}{20}$

$M={\sum }_{i=1}^{20}\frac{x_i}{20}$

Hence,

${\sum }_{i=1}^{20}\frac{x_i-M}{20}$

$\Rightarrow {\sum }_{i=1}^{20}\frac{x_i}{20}-\frac{M}{20}$

$\Rightarrow M-\frac{M}{20}$

$\Rightarrow \frac{19M}{20}$

Hence, this is the required answer.