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

Given, mid point = 'm'. and upper limit = 'l'.

We know that Midpoint = (upper limit + limit class)/2

⇒ m = (l + lower limit)/2

⇒ 2m = l + lower limit

⇒ lower limit = 2m - l.

Therefore, lower limit = 2m - l.

Hope it helps!

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

hope this helps u please mark as BRAINLIEST [/tex]