Question

If $x_i$'s are the midpoints of the class intervals of grouped data, $f_i$’s are the corresponding frequencies and $x$ is the mean, then $\Sigma (f_i{x}_i–x)$ is equal to:

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is A

$0$

Find the summation

We know that, Mean $\left(x\right)=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{observations}}{\mathrm{Number}\mathrm{of}\mathrm{observations}}$

$\Rightarrow$ $x=\frac{(f_1{x}_1+f_2{x}_2+\dots ..+f_n{x}_n)}{f_1+f_2+\dots \dots +f_n}$

$\Rightarrow$ $x=\frac{\Sigma f_i{x}_i}{\Sigma f_i}$ and $\Sigma f_i=n$

$\Rightarrow$ $x=\frac{\Sigma f_i{x}_i}{n}$

$\Rightarrow$ $nx=\Sigma f_i{x}_i$ ……………$\left(i\right)$

Now, $\Sigma (f_i{x}_i–x)=(f_1{x}_1–x)+(f_2{x}_2–x)+\dots ..+(f_n{x}_n–x)$

$\Rightarrow$ $\Sigma (f_i{x}_i–x)=(f_1{x}_1+f_2{x}_2+\dots ..+f_n{x}_n)–(x+x+\dots .n\mathrm{times})$

$\Rightarrow$ $\Sigma (f_i{x}_i–x)=\Sigma f_i{x}_i–nx$

$\Rightarrow$ $\Sigma (f_i{x}_i–x)=nx–nx$ ……………(from $\left(i\right)$)

$\Rightarrow$ $\Sigma (f_i{x}_i–x)=0$

Hence, $\Sigma (f_i{x}_i–x)=0$, so, option(A) is the correct answer.