Find out the modal value from the following data:

Marks 0−9 10−19 20−29 30−39 40−49 50−59

No. of Students 3 7 15 25 10 4

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Solution

Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula.

$V a l u e o f A d j u s t m e n t = \frac{L o w e r l i m i t o f o n e c l a s s - U p p e r l i m i t o f t h e p r e c e e d i n g c l a s s}{2}$ Value of lower limit of one class − Value of upper limit of the preceeding class2
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.

Inclusive
Class Interval Exclusive
Class Interval Frequency
(f)

0 − 9
10 − 19
20 − 29
30 − 39
40 − 49
50 − 59 -0.5 − 9.5
9.5 − 19.5
19.5 − 29.5
29.5 − 39.5
39.5 − 49.5
49.5 − 59.5 3
7
15 f₀
25 f₁
10 f₂
4

By inspection, we can say that the modal class is (29.5 – 39.5) as it has the highest frequency of 25.
$Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 29.5 + \frac{25 - 15}{2 (25) - 15 - 10} \times 10 or, Z = 29.5 + \frac{10}{50 - 25} \times 10 or, Z = 29.5 + \frac{100}{25} or, Z = 29.5 + 4 = 33.5$

Hence, mode is 33.5.

Suggest Corrections

Marks (more than)	0-9	10-19	20-29	30-39	40-49	50-59
No of students	2	5	15	25	10	3

Marks	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Number of Students	7	10	20	13	12	19	14	9

Marks	0−9	10−19	20−29	30−39	40−49	50−59
No. of Students	3	7	15	25	10	4

Inclusive Class Interval	Exclusive Class Interval	Frequency (f)
0 − 9 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59	-0.5 − 9.5 9.5 − 19.5 19.5 − 29.5 29.5 − 39.5 39.5 − 49.5 49.5 − 59.5	3 7 15 f₀ 25 f₁ 10 f₂ 4

Marks	0−10	10−20	20−30	30−40	40−50
No. of Students	4	6	10	20	10

Class Interval	60−69	50−59	40−49	30−39	20−29	10−19
Frequency	13	15	21	20	19	12