Question

Which of the following can be an alternate representation of variance, where $x_i$ being the midpoint of class intervals, $f_i$ being frequency of class interval and $\overline{x}$ the mean,N = ${\sum }_{i=1}^n{f}_i$

• A.
  
• B.
  
• C.
  
• D.

Answer 1

Answer: (A), (B)

The correct options are
A

B

Variance ${\sigma }^2$ = $\frac{1}{N}\left({\sum }_{i=1}^n{f}_i\right(x_i-\overline{x}{)}^2$ = $\frac{1}{N}{\sum }_{i=1}^n{f}_i(x_i^2+{\overline{x}}^2-2\overline{x}{x}_i)$

$=\frac{1}{N}[{\sum }_{i=1}^n{f}_i{x}_i^2+{\overline{x}}^2{\sum }_{i=1}^n{f}_i-2\overline{x}{\sum }_{i=1}^n{f}_i{x}_i{f}_i]$

$=\frac{1}{N}[{\sum }_{i=1}^{n}fi{x}_i+{\overline{x}}^{2}N-2\overline{x}N\overline{x}]$

Because, $\frac{1}{N}{\sum }_{i=1}^n{x}_{i}fi=\overline{x}$

= $\frac{1}{N}{\sum }_{i=1}^n{f}_i{x}_i^2+{\overline{x}}^2-2{\overline{x}}^2$ = $\frac{1}{N}{\sum }_{i=1}^n{f}_i{x}_i^2-{\overline{x}}^2$

= $\frac{1}{N}{\sum }_{i=1}^n{f}_i{x}_i^2$ - $(\frac{{\sum }_{i=1}^n{f}_i{x}_i{)}^2}{N}{)}^2$

= $\frac{1}{N^2}[N{\sum }_{i=1}^n{f}_i{x}_i^2-({\sum }_{i=1}^n{f}_i{x}_i{)}^2]$

$\therefore$ option a and b are correct.