MCQ
In any discrete series (when all values are not same) the relationship between $M.D.$ about mean and $S.D.$ is
- A$M.D. = S.D.$
- B$M.D.\ge S.D.$
- C$M.D. < S.D.$
- ✓$M.D. \le S.D.$
✓
Answer
Correct option: D.
$M.D. \le S.D.$
d
(d) Let ${x_i}/{f_i};$ $i = 1,2,......n$ be a frequency distribution.
(d) Let ${x_i}/{f_i};$ $i = 1,2,......n$ be a frequency distribution.
Then,${\rm{S}}{\rm{.D}}{\rm{.}} = \sqrt {\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{{({x_i} - \bar x)}^2}} } $
and ${\rm{M}}{\rm{.D}}{\rm{.}} = \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}|{x_i}} - \bar x|$
Let $|{x_i} - \bar x| = {z_i};i = 1,2,.....n$ .
Then,
$(S.D.)2 -(M.D.)2$ $ = \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}z_i^2 - {{\left( {\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{z_i}} } \right)}^2}} $
$ = \sigma _z^2 \ge 0$==> S. D. $ \ge $ $M.D.$
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