Prove the following

Question:

If $x_{i}{ }^{\prime}$ s are the mid-points of the class intervals of grouped data, $f_{i}{ }^{\prime}$ s are the corresponding frequencies and $\bar{x}$ is

the mean, then $\Sigma\left(f_{i} x_{i}-\bar{x}\right)$ is equal to

(a) 0                 

(b) -1                          

(c) 1                        

(d) 2

Solution:

(a) $\because$ $\bar{x}=\frac{\Sigma f_{i} x_{i}}{n}$

$\therefore$ $\Sigma\left(f_{i} x_{i}-\bar{x}\right)=\Sigma f_{i} x_{i}-\Sigma \bar{x}$

$=n \bar{x}-n \bar{x}$ $[\because \Sigma \bar{x}=n \bar{x}]$

$=0$

Leave a comment