Question

Find the mean, median and mode of the following data:

Class	0−10	10−20	20−30	30−40	40−50	50−60	60−70
Frequency	6	8	10	15	5	4	2

Answer 1

Here the maximum class frequency is 15, and the class corresponding to this frequency is 30−40. So, the modal class is 30−40.

Now,

Modal class = 30−40, lower limit (l) of modal class = 30, class size (h) = 10,

frequency (f₁) of the modal class = 15,

frequency (f₀) of class preceding the modal class = 10,

frequency (f₂) of class succeeding the modal class = 5.

Now, let us substitute these values in the formula:

$$\begin{gathered} \mathrm{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h \\ =30+\left(\frac{15-10}{30-10-5}\right)\times 10 \\ =30+\left(\frac{50}{15}\right) \\ =33.33 \end{gathered}$$

Hence, the mode is 33.33.

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

Class	Frequency (fi)	Class mark (xi)	fixi
0−10	6	5	30
10−20	8	15	120
20−30	10	25	250
30−40	15	35	525
40−50	5	45	225
50−60	4	55	220
60−70	2	65	130
Total	∑ fi = 50		∑ fixi = 1500

$$\begin{gathered} \mathrm{Mean}=\frac{{\displaystyle \sum _i}{f}_i{x}_i}{{\displaystyle \sum _i}{f}_i} \\ =\frac{1500}{50} \\ =30 \end{gathered}$$

Thus, mean of the given data is 30.

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

Class	Frequency (fi)	Cumulative frequency (cf)
0−10	6	6
10−20	8	14
20−30	10	24
30−40	15	39
40−50	5	44
50−60	4	48
60−70	2	50
Total	N = ∑ fi = 50

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

The cumulative frequency just greater than 25 is 39, and the corresponding class is 30−40.

Thus, the median class is 30−40.

∴ l = 30, h = 10, N = 50, f = 15 and cf = 24.

Now,

$$\begin{gathered} \mathrm{Median}=l+\left(\frac{{\displaystyle \frac{N}{2}}-cf}{f}\right)\times h \\ =30+\left(\frac{25-24}{15}\right)\times 10 \\ =30+\frac{10}{15} \\ =30.67 \end{gathered}$$

Thus, the median is 30.67.