Solution:

Here, we have \[N\text{ }=\text{ }140,\]

So, \[N/2\text{ }=\text{ }140/\text{ }2\text{ }=\text{ }70\]

The cumulative frequency just greater than \[N/\text{ }2\text{ }is\text{ }98\]then median class is \[55\text{ }\text{ }65\]such that \[L\text{ }=\text{ }55,\text{ }f\text{ }=\text{ }40,\text{ }F\text{ }=\text{ }58,\text{ }h\text{ }=\text{ }65\text{ }\text{ }55\text{ }=\text{ }10\]

\[=\text{ }55\text{ }+\text{ }3\text{ }=\text{ }58\]