Calculate the median mark and the 60th percentile mark using linear interpolation from the provided data.

The $40^{th}$ student falls in the 50-59 class, as the cumulative frequency goes from 37 (for 40-49) to 57 (for 50-59).

The lower class boundary ($L$) for the 50-59 class is 49.5.
The cumulative frequency before the median class ($C_f$) is 37.
The frequency of the median class ($f_m$) is 20.
The class width ($c$) is $59.5 - 49.5 = 10$.

b)(ii) The $60^{th}$ percentile mark:

The $48^{th}$ student falls in the 50-59 class, as the cumulative frequency goes from 37 to 57.

The lower class boundary ($L$) for the 50-59 class is 49.5.
The cumulative frequency before this class ($C_f$) is 37.
The frequency of this class ($f_p$) is 20.
The class width ($c$) is 10.

Students scoring 50% or more are those in the classes 50-59, 60-69, 70-79, 80-89, and 90-99.
Number of students passed = $20 + 11 + 7 + 3 + 2 = 43$.

Total number of students = 80.