Question

Let $m$ be the mid-point and $l$ be the lower class limit of a class in a continuous frequency distribution' the upper-class limit of the class is?

Answer 1

Determine the upper limit:

Let upper limit $=\mathrm{U}$

Given lower limit $=\mathrm{l}$ and mid point$=\mathrm{m}$

We know that,

$$\begin{gathered} \mathbf{Mid}\mathbf{}\mathbf{point}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{Upper}\mathbf{}\mathbf{limit}\mathbf{}\mathbf{+}\mathbf{}\mathbf{Lower}\mathbf{}\mathbf{limit}}{\mathbf{2}} \\ \Rightarrow 2\times \mathrm{m}=\mathrm{Upper}\mathrm{limit}+\mathrm{l} \\ \Rightarrow \mathrm{Upper}\mathrm{limit}=2\mathrm{m}-\mathrm{l} \end{gathered}$$

Hence, upper limit is $\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{l}$