Given, Mean = 66 and Mode = 60.
Using, Mode ≈ 3Median – 2Mean
60 ≈ 3Median – 2(66)
3 Median ≈ 60 +132
Therefore, Median ≈ \({192\over 3}≈64\)

Modal class is 16 -20 since it has the maximum frequency.
l = 15.5, f = 32, f₁ = 16, f₂ = 24, c = 20.5–15.5 = 5
Mode \(=l+{\left(f-f_1\over 2f-f_1-f_2\right)}\times c\)
\(=15.5+\left(32-16\over64-16-24\right)\times 5\)
\(=15.5+\left(16\over24\right)\times = 15.5 + 3.33 =18.83\)

Since the median is 24 and median class is 20 – 30
l = 20 N = 55 + x, m = 30, c = 10, f = x
Median = \(l+{\left({N\over2}-m\right)\over f}\times c\)
\(24=20+{\left({55+x\over2}-3\right)\over x}\times 10\)
\(4={5x-25\over x}\)  (after simplification)
4x = 5x – 25
5x – 4x = 25
x = 25

Median class = \(\left(N\over 2\right)^{th}\) value=\(\left(200\over 2\right)^{th}\) value
= 100^th value
Median class = 2000 – 3000
\({N\over 2}=100\ l=2000\)
m = 74, c = 1000, f = 54
Median = \(l+{\left({n\over2}-m\right)\over f}\times c\)
\(=2000+\left(100-74\over54\right)\times1000\)
\(=2000+({26\over54})\times1000=2000+481.5\)
= 2481.5

Let us arrange the marks in ascending order and prepare the following data:

Here N = 41
Median = size of \(\left(N+1\over 2\right)^{th}\) value = size of \(\left(41+1\over2\right)^{th}\) value = size of 21^st value.
If the 41 students were arranged in order (of height), the 21^st student would be the middle most one, since there are 20 students on either side of him/her. We therefore need to find the height against the 21^st student. 15 students (see cumulative frequency) have height less than or equal to 154 cm. 22 students have height less than or equal to 155 cm. This means that the 21^st student has a height 155 cm.
Therefore, Median = 155 cm