Question

Find the mean, mode and median of the following data: [CBSE 2008]

Class	0−20	20−40	40−60	60−80	80−100	100−120	120−140
Frequency	6	8	10	12	6	5	3

Answer 1

To find the mean let us put the data in the table given below:

Class	Frequency (fi)	Class mark (xi)	fixi
0−20	6	10	60
20−40	8	30	240
40−60	10	50	500
60−80	12	70	840
80−100	6	90	540
100−120	5	110	550
120−140	3	130	390
Total	∑fi = 50		∑fixi = 3120

$$\begin{gathered} \mathrm{Mean}=\frac{{\displaystyle \sum _i}{f}_i{x}_i}{{\displaystyle \sum _i}{f}_i} \\ =\frac{3120}{50} \\ =62.4 \end{gathered}$$

Thus, mean of the given data is 62.4.

Now, to find the median let us put the data in the table given below:

Class	Frequency (fi)	Cumulative frequency (cf)
0−20	6	6
20−40	8	14
40−60	10	24
60−80	12	36
80−100	6	42
100−120	5	47
120−140	3	50
Total	N = ∑fi = 50

Now, N = 50 $\Rightarrow \frac{N}{2}=25$.

The cumulative frequency just greater than 25 is 36, and the corresponding class is 60−80.

Thus, the median class is 60−80.

∴ l = 60, h = 20, N = 50, f = 12 and cf = 24.

Now,

$$\begin{gathered} \mathrm{Median}=l+\left(\frac{{\displaystyle \frac{N}{2}}-cf}{f}\right)\times h \\ =60+\left(\frac{25-24}{12}\right)\times 20 \\ =60+1.67 \\ =61.67 \end{gathered}$$

Thus, the median is 61.67.

We know that,
Mode = 3(median) − 2(mean)
= 3 × 61.67 − 2 × 62.4
= 185.01 − 124.8
= 60.21

Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21.