MCQ
Let m be the midpoint and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
- A2m - u
- B2m + u
- Cm - u
- Dm + u
✓
Answer
- 2m - u
Solution:
Given:
Mid value = m
Upper limit = u
We know,
$\frac{\text{Upper}\ \text{limit}+\text{Lower}\ \text{limit}}{2}=\text{Mid}\ \text{value}$
$\Rightarrow\frac{\text{Lower}\ \text{limit}+\text{u}}{2}=\text{m}$
⇒ Lower limit + u = 2m
⇒ Lower limit = 2m - u
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