Question

The following table gives the marks obtained by 50 students in a class test:

Marks	11−15	16−20	21−25	26−30	31−35	36−40	41−45	46−50
Number of students	2	3	6	7	14	12	4	2

Calculate the mean, median and mode for the above data.

Answer 1

The frequency distribution into the continuous form is as follows:

Marks	10.5−15.5	15.5−20.5	20.5−25.5	25.5−30.5	30.5−35.5	35.5−40.5	40.5−45.5	45.5−50.5
Number of students	2	3	6	7	14	12	4	2

Here the maximum class frequency is 14, and the class corresponding to this frequency is 30.5−35.5. So, the modal class is 30.5−35.5.

Now,

Modal class = 30.5−35.5, lower limit (l) of modal class = 30.5, class size (h) = 5,

frequency (f₁) of the modal class = 14,

frequency (f₀) of class preceding the modal class = 7,

frequency (f₂) of class succeeding the modal class = 12.

Now, let us substitute these values in the formula:

$$\begin{gathered} \mathrm{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h \\ =30.5+\left(\frac{14-7}{28-7-12}\right)\times 5 \\ =30.5+\left(\frac{35}{9}\right) \\ =34.38 \end{gathered}$$

Hence, the mode is 34.38.

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

Class	Frequency (fi)	Class mark (xi)	fixi
10.5−15.5	2	13	26
15.5−20.5	3	18	54
20.5−25.5	6	23	138
25.5−30.5	7	28	196
30.5−35.5	14	33	462
35.5−40.5	12	38	456
40.5−45.5	4	43	172
45.5−50.5	2	48	96
Total	∑ fi = 50		∑ fixi = 1600

$$\begin{gathered} \mathrm{Mean}=\frac{{\displaystyle \sum _i}{f}_i{x}_i}{{\displaystyle \sum _i}{f}_i} \\ =\frac{1600}{50} \\ =32 \end{gathered}$$

Thus, mean of the given data is 32.

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

Class	Frequency (fi)	Cumulative frequency (cf)
10.5−15.5	2	2
15.5−20.5	3	5
20.5−25.5	6	11
25.5−30.5	7	18
30.5−35.5	14	32
35.5−40.5	12	44
40.5−45.5	4	48
45.5−50.5	2	50
Total	N = ∑ fi = 50

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

The cumulative frequency just greater than 25 is 32, and the corresponding class is 30.5−35.5.

Thus, the median class is 30.5−35.5.

∴ l = 30.5, h = 5, N = 50, f = 14 and cf = 18.

Now,

$$\begin{gathered} \mathrm{Median}=l+\left(\frac{{\displaystyle \frac{N}{2}}-cf}{f}\right)\times h \\ =30.5+\left(\frac{25-18}{14}\right)\times 5 \\ =30.5+\frac{35}{14} \\ =33 \end{gathered}$$

Thus, the median is 33.