Solve the Problem $= ($ Lower Limit of Median Class $) + ($

Solve the problem.
-When data are summarized in a frequency distribution, the median can be found by first identifying the median class (the class that contains the median) . We then assume that the values in that class are evenly
Distributed and we can interpolate. This process can be described by median $= ($ lower limit of median class $) + ($ class width $) \left( \frac { \frac { \mathrm { n } + 1 } { 2 } - ( \mathrm { m } + 1 ) } { \text { frequency of median class } } \right)$
where $\mathrm { n }$ is the sum of all class frequencies and $\mathrm { m }$ is the sum of the class frequencies that precede the median class. Use this procedure to find the median of the frequency distribution below:
$\begin{array}{c|c}\text { Score } & \text { Frequency } \\\hline 50-59 & 21 \\60-69 & 24 \\70-79 & 22 \\80-89 & 16 \\90-99 & 17\end{array}$

A) $72.5$
B) $72.0$
C) $71.8$
D) $74.5$