The following table gives the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
Marks obtained (in per cent) | 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.
(a) To convert the given frequency distribution into continuous form, adjust the end - limits of each class.
MARKS OBTAINED (in percent) | NUMBER OF STUDENTS (fi) |
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) To find median class,
Assume Σfi = N = Sum of frequencies,
fi = frequency
and Cf = cumulative frequency
MARKS OBTAINED (in percent) | NUMBER OF STUDENTS (fi) | Cf |
10.5 - 20.5 | 141 | 141 |
20.5 - 30.5 | 221 | 141 + 221 = 362 |
30.5 - 40.5 | 439 | 362 + 439 = 801 |
40.5 - 50.5 | 529 | 801 + 529 = 1330 |
50.5 - 60.5 | 495 | 1330 + 495 = 1825 |
60.5 - 70.5 | 322 | 1825 + 322 = 2147 |
70.5 - 80.5 | 153 | 2147 + 153 = 2300 |
TOTAL | 2300 |
So, N = 2300
⇒ N/2 = 2300/2 = 1150
The cumulative frequency just greater than (N/2 = ) 1150 is 1825, so the corresponding median class is 50.5 - 60.5.
∴ median class = 50.5 - 60.5
The class mark of 50.5 - 60.5 is 
(c) For modal class,
Here, the maximum class frequency is 529.
The class corresponding to this frequency is the modal class. ⇒ modal class = 40.5 - 50.5
The cumulative frequency corresponding to the modal class is 1330