Question

Let m be the midpoint and L be the upper limit of a class in a continuous frequency distribution. What is the lower class limit of the class?

Answer 1

Hey dude, your answer is here,

Let x and y be the lower and upper class limit of a continuous frequency distribution


Now, mid-point of a class = (x+y) /2=m (given)

X+y=2m=x+l=2m

[ therefore y=l=upper class limit (given)]

X=2m-l

Hence, the lower class limit of the class is 2m-l

Thank you friend

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]