Question

Find the mean, mode and median of the following frequency distribution. [CBSE 2010]

Class	0−10	10−20	20−30	30−40	40−50	50−60	60−70
Frequency	4	4	7	10	12	8	5

Answer 1

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

Class	Frequency (fi)	Class mark (xi)	fixi
0−10	4	5	20
10−20	4	15	60
20−30	7	25	175
30−40	10	35	350
40−50	12	45	540
50−60	8	55	440
60−70	5	65	325
Total	∑fi = 50		∑fixi = 1910

$$\begin{gathered} \mathrm{Mean}=\frac{{\displaystyle \sum _i}{f}_i{x}_i}{{\displaystyle \sum _i}{f}_i} \\ =\frac{1910}{50} \\ =38.2 \end{gathered}$$

Thus, mean of the given data is 38.2.

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

Class	Frequency (fi)	Cumulative frequency (cf)
0−10	4	4
10−20	4	8
20−30	7	15
30−40	10	25
40−50	12	37
50−60	8	45
60−70	5	50
Total	N = ∑fi = 50

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

The cumulative frequency just greater than 25 is 37, and the corresponding class is 40−50.

Thus, the median class is 40−50.

∴ l = 40, h = 10, N = 50, f = 12 and cf = 25.

Now,

$$\begin{gathered} \mathrm{Median}=l+\left(\frac{{\displaystyle \frac{N}{2}}-cf}{f}\right)\times h \\ =40+\left(\frac{25-25}{12}\right)\times 10 \\ =40 \end{gathered}$$

Thus, the median is 40.

We know that,
Mode = 3(median) − 2(mean)
= 3 × 40 − 2 × 38.2
= 120 − 76.4
= 43.6

Hence, Mean = 38.2, Median = 40 and Mode = 43.6