Question

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Let ‘m’ be the mid-point (class mark) and ‘l’ be the upper limit of a class in a continuous frequency distribution, then express value of lower limit in terms of l and m

Answer 1

It is a question of Statistics.

We use the lower limit and upper limit to find out the Class mid-point.

This method is adopt usually to find out the MEAN of the grouped data.

Now as mentioned in the question

m = midpoint

l   = upper limit

Let

v   = Lower limit

So we have

Mid-point = (lower-limit + upper limit ) / 2

               m = ( v+ l ) / 2

             2m = v + l

         2m - l = v

Rearranging, we get

             v = 2m - l   which is the value of lower limit in terms of I and m .

Answer 2

$hey\: mate \:here \: is \: ur\: required\: answer \\ \\ let \: x = lower \: class\: limit \\ \\ let \:y = upper\: class\: limit\\ \\ let\: m = midpoint \\ \\ we\: know\: that \\ \\ \frac{(x + l)}{2} = m \\ 2m \: = \: x + l \\ x = 2m - l$

hence this is the required answer

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