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In any discrete series (when all values are not same) the relationship between $M.D.$ about mean and $S.D.$ is
$M.D. = S.D.$
$M.D.\ge S.D.$
$M.D. < S.D.$
$M.D. \le S.D.$
Solution
(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 $ $M.D.$
Similar Questions
If the variance of the following frequency distribution is $50$ then $x$ is equal to:
| Class | $10-20$ | $20-30$ | $30-40$ |
| Frequency | $2$ | $x$ | $2$ |
