Question

Find the mode, median and mean for the following data: [CBSE 2009]

Marks obtained	25−35	35−45	45−55	55−65	65−75	75−85
Number of students	7	31	33	17	11	1

Answer 1

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

Marks obtained	Number of students (fi)	Class mark (xi)	fixi
25−35	7	30	210
35−45	31	40	1240
45−55	33	50	1650
55−65	17	60	1020
65−75	11	70	770
75−85	1	80	80
Total	∑fi = 100		∑fixi = 4970

$$\begin{gathered} \mathrm{Mean}=\frac{{\displaystyle \sum _i}{f}_i{x}_i}{{\displaystyle \sum _i}{f}_i} \\ =\frac{4970}{100} \\ =49.7 \end{gathered}$$

Thus, mean of the given data is 49.7.

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

Class	Frequency (fi)	Cumulative frequency (cf)
25−35	7	7
35−45	31	38
45−55	33	71
55−65	17	88
65−75	11	99
75−85	1	100
Total	N = ∑fi = 100

Now, N = 100 $\Rightarrow \frac{N}{2}=50$.

The cumulative frequency just greater than 50 is 71, and the corresponding class is 45−55.

Thus, the median class is 45−55.

∴ l = 45, h = 10, N = 100, f = 33 and cf = 38.

Now,

$$\begin{gathered} \mathrm{Median}=l+\left(\frac{{\displaystyle \frac{N}{2}}-cf}{f}\right)\times h \\ =45+\left(\frac{50-38}{33}\right)\times 10 \\ =45+3.64 \\ =48.64 \end{gathered}$$

Thus, the median is 48.64.

We know that,
Mode = 3(median) − 2(mean)
= 3 × 48.64 − 2 × 49.70
= 145.92 − 99.4
= 46.52

Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52.