Given a frequency distribution table with marks and frequencies, find the median class, class size, modal class, and mean.

Class size $=$ upper limit $-$ lower limit of any class.

Class size $= 20 - 0 = 20$.

Class size $= 20$

Total frequency $N = 68$.

Median position $= \frac{N+1}{2} = \frac{69}{2} = 34.5$.

Cumulative frequencies:

$$\begin{array}{c|c} \text{Class} & \text{C.F.} \\ \hline 0-20 & 8 \\ 20-40 & 22 \\ 40-60 & 40 \\ 60-80 & 56 \\ 80-100 & 68 \\ \hline \end{array}$$

C.F. first exceeds 34.5 at class $40-60$.

Median class $= 40-60$

Modal class is the class with highest frequency.

Highest $f=18$ for class $40-60$.

Modal class $= 40-60$

Midpoints $x_i$: $10, 30, 50, 70, 90$.

$$\begin{array}{|c|c|c|} \hline \text{Class} & f & x_i & f x_i \\ \hline 0-20 & 8 & 10 & 80 \\ 20-40 & 14 & 30 & 420 \\ 40-60 & 18 & 50 & 900 \\ 60-80 & 16 & 70 & 1120 \\ 80-100 & 12 & 90 & 1080 \\ \hline & 68 & & 3600 \\ \hline \end{array}$$

$\sum f x_i = 80 + 420 + 900 + 1120 + 1080 = 3600$.

Mean $\bar{x} = \frac{\sum f x_i}{N} = \frac{3600}{68}$.

$3600 \div 68 = \frac{3600 \div 4}{68 \div 4} = \frac{900}{17}$.

$\frac{900}{17} = 52 \frac{16}{17}$.

Mean $= \dfrac{900}{17}$
\dfrac{900}{17}