Question
Let m be the mid-point and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
✓
Answer
- 2m - I
Let the lower limit = k
Mid-point = m
Upper limit = l
Mid-point $=\frac{\text{(upper limit + lower limit)}}{2}$
$\text{m}=\frac{(\text{k+l})}{2}$
$2\text{m}=\text{k}+\text{l}$
$\text{k}=2\text{m}-\text{l}$
Therefore, lower limit = 2m - l
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