Question

Let L be the lower class boundary of a class in a frequency distribution and m be the midpoint of the class. Which one of the following is the upper class boundary of the class?
(a) $m+\frac{(m+L)}{2}$
(b) $L+\frac{m+L}{2}$
(c) $2m-L$
(d) $m-2L$

Answer 1

(c) 2m$-$L

$$\begin{gathered} \mathrm{Mid}\mathrm{value}=\frac{\mathrm{Lower}\mathrm{limit}+\mathrm{Upper}\mathrm{limit}}{2} \\ \Rightarrow m=\frac{L+U}{2} \\ \Rightarrow U=2m-L \\ \therefore \mathrm{Upper}\mathrm{class}\mathrm{boundary}\mathrm{of}\mathrm{the}\mathrm{class}=2m-L \end{gathered}$$