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For a frequency distribution mean deviat...

For a frequency distribution mean deviation from mean in computed by `MdotDdot=(sumf)/(sumf|d|)` (b) `MdotDdot=(sumd)/(sumf)` (c) `MdotDdot=(sumfd)/(sumf)` (d) `MdotDdot=(sumf|d|)/(sumf)`

A

M.D. `=(sum d_(1))/(sum f_(i))`

B

`M.D. =(sum f_(i)d_(i))/(sum f_(i))`

C

`M.D.=(sum f_(i) |d_(i)|)/(sum f_(i))`

D

`M.D.=(sum f_(i))/(sum f_(i) |d_(i)|)`

Text Solution

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The correct Answer is:
C
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