Find out mode, given the following information:

Size 6−10 11−15 16−20 21−25 26−30

Frequency 20 30 50 40 10

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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.
$Value of Adjustment = \frac{Value of lower limit of one class - Value of upper limit of the preceeding class}{2}$
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.

Size Exclusive Class Interval Frequency
(f)

6 − 10
11 − 15
16 − 20
21 − 25
26 − 30 5.5 − 10.5
10.5 − 15.5
15.5 − 20.5
20.5 − 25.5
25.5 − 30.5 20
30 f₀
50 f₁
40 f₂
10

By inspection, we can say that the modal class is (15.5 – 20.5) as it has the highest frequency of 50.
$Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 15.5 + \frac{50 - 30}{2 (50) - 30 - 40} \times 5 or, Z = 15.5 + \frac{20}{100 - 70} \times 5 or, Z = 15.5 + \frac{100}{30} or, Z = 15.5 + 3.33 = 18.33 Hence, mode is 18.33$

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Class-Interval	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Frequency	5	7	15	18	16	9	6	3

Size	0−4	4−6	6−10	10−15	15−20	20−30	30−35	35−40
Frequency	2	4	3	5	2	20	6	8

Marks	0−10	10−20	20−30	30−40	40−50	50−60
Frequency	6	8	16	25	12	8

Size	10−20	20−30	30−40	40−50
Frequency	42	25	58	40

Class interval	1-5	6-10	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
Frequency	3	8	13	18	28	20	13	8	6	4

Size	Exclusive Class Interval	Frequency (f)
6 − 10 11 − 15 16 − 20 21 − 25 26 − 30	5.5 − 10.5 10.5 − 15.5 15.5 − 20.5 20.5 − 25.5 25.5 − 30.5	20 30 f₀ 50 f₁ 40 f₂ 10