Let m be the mid-point be the upper class limit of a continuous frequency distribution The lower class limit of the class is

2 m + L

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2 m - L

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m - L

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m + 2 L

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Solution

The correct option is B $2 m - L$
Mid point= $\frac{u p p e r l i m i t + l o w e r l i m i t}{2}$
$\Rightarrow m = \frac{L + l o w e r l i m i t}{2}$
$\Rightarrow L o w e r l i m i t = 2 m - L$

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Let m be the mid-point be the upper class limit of a continuous frequency distribution The lower class limit of the class is

The correct option is B 2m−LMid point=upperlimit+lowerlimit2⇒m=L+lowerlimit2⇒Lowerlimit=2m−L

The correct option is B $2 m - L$
Mid point= $\frac{u p p e r l i m i t + l o w e r l i m i t}{2}$
$\Rightarrow m = \frac{L + l o w e r l i m i t}{2}$
$\Rightarrow L o w e r l i m i t = 2 m - L$