Calculate the mean mark in the distribution given below :

Marks	Frequency
1 - 10	7
11 - 20	9
21 - 30	15
31 - 40	8
41 - 50	6
51 - 60	5

Also state :

(i) the median class

(ii) the modal class

The above distribution is discontinuous converting into continuous distribution, we get :

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\dfrac{21 - 20}{2} = \dfrac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Mean = $\dfrac{Σfx}{Σf} = \dfrac{1395}{50}$ = 27.9

Given, total terms = 50.

Since, terms are even.

By formula,

Median = $\dfrac{50}{2}$ = 25th term.

From table,

15.5 th to 25.5 th term marks lies in the range 20.5 - 30.5.

∴ Median class = 21 - 30.

From table class 21 - 30 has the highest frequency.

∴ Modal class = 21 - 30.

Hence, mean = 27.9, median class = 21 - 30 and modal class = 21 - 30.