Question

Find the median.

Marks	No. of students (f)
0-7	3
7-14	4
14-21	7
21-28	11
28-35	0
35-42	16
42-49	9

Answer 1

Median can be referred to as the middlemost number of the series of numbers given. Here we cannot use the basic method of finding the median. We need to use the formula, which has to be remembered, while solving these types of sums having intervals.
Median $= L + \\{ h \times \dfrac{{(\dfrac{N}{2} - cf)}}{f}\\}$
We need to add one more column of cumulative frequency.

Complete answer:
To find: Median

Marks	No. of students (f)	Cumulative frequency (cf)
0-7	3	3
7-14	4	3+4=7
14-21	7	7+7=14
21-28	11	14+11=25
28-35	0	25+0=25
35-42	16	25+16=41
42-49	9	41+9=50

In the above formula,
N is the total number of students. We get it by adding the no. of students’ column.
For this sum, N=50, therefore, $\dfrac{N}{2}$ becomes 25.
In this sum, the selected interval is of 21-28, as cf of this interval is 25 which is equal to $\dfrac{N}{2}$
(Here are 2 intervals with if 25. Whenever there are same cf’s for different intervals, always choose the one which comes first)
Cf is the cumulative frequency, which is given in the table. The value of cf in the formula has to be the cf above the cf of the selected interval. That is if used in the formula has to be the one above cf 25, which is 14.
L is the lower limit of the selected interval. That is 21.
In the formula, h is the height of the interval, which is simply upper limit-lower limit, which is
28-21=7.
And lastly, f is the frequency. The value of f for the selected interval is 11.
We have all the values which we want to put in the formula of median in our hands, now let’s substitute.
Median $= 21 + \\{ 7 \times \dfrac{{(25 - 14)}}{{11}}\\}$ Using all the above statements.
On further solving, we get median as,
Median $= 21 + \dfrac{{77}}{{11}} = 28$

Note:
The last row of the cumulative frequency column has to be N. This can be the way of checking whether you have added the frequencies correctly. Do not confuse mean, median and mode with each other. Median is the middlemost term and has to be somewhat around $\dfrac{N}{2}$. Also, the cf mentioned in the formula is not the cf of the selected interval. This has to be the important thing to remember.