Question

The following table, construct the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.

Marks obtained (in percent)	11 – 20	21 – 30	31 – 40	41 – 50	51 – 60	61 – 70	71 – 80
Number of Students	141	221	439	529	495	322	153

(a) Convert the given frequency distribution into the continuous form.
(b) Find the median class and write its class mark.
(c) Find the modal class and write its cumulative frequency.

Answer 1

(a) The frequency distribution into the continuous form is as follows:

Marks obtained (in per cent)	Number of students (f)
10.5-20.5	141
20.5-30.5	221
30.5-40.5	439
40.5-50.5	529
50.5-60.5	495
60.5-70.5	322
70.5-80.5	153

(b)Now, to find the median class let us put the data in the tale given below:

Marks obtained (in per cent)	Number of students (f)	Cumulative frequency (cf)
10.5-20.5	141	141
20.5-30.5	221	362
30.5-40.5	439	801
40.5-50.5	529	1330
50.5-60.5	495	1825
60.5-70.5	322	2147
70.5-80.5	153	2300

Now, 𝑁 = 2300
⟹`N/2 = 1150`
The cumulative frequency just greater than 1150 is 1330, and the corresponding class is 40.5-50.5.
Thus, the median class is 40.5-50.5
Now, class mark =`( "Upper class limit + Lower class lilit")/2`

`(40.5 + 50.5 )/2 = 91/2 = 45.5`
Thus, class mark of the median class is 45.5
(c)Here the maximum class frequency is 529, and the class corresponding to this frequency is 40.5-50.5
.So, the modal class is 40.5-50.5 and its cumulative frequency is 1330.