Find the mode of the following distribution.
(i)Class interval:
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Frequency:
5
8
7
12
28
20
10
10
(ii)Class interval:
10-15
15-20
20-25
25-30
30-35
35-40
Frequency:
30
45
75
35
25
15
(iii)Class interval: 25-30
30-35
35-40
40-45
45-50
50-60
Frequency:
25
34
50
42
29
15
Answers
SOLUTION:
MODE:
Mode is that value of the given observation which occurs most frequently.
Mode = l + (f1- f0/2f1- f0 - f2) ×h
l = lower limit of the modal class
h = size of the class intervals
f1 = frequency of the modal class
f0 = frequency of the class preceding the modal class
f2 = frequency of the class succeed in the modal class.
(i) Here the maximum frequency is 28 and the class corresponding to this frequency is 40 - 50 . So the modal class is 40 - 50.
From the table : l = 40, h = 50 – 40 = 10, f1 = 28, f0 = 12, f2 = 20
Mode = l +[ (f1- f0) / 2f1- f0 - f2)] ×h
= 40 + [(28−12)/ (2 × 28–12−20) ×10
= 40 + [16 / (56 - 32)] × 10
= 40 + 16× 10 / 24
= 40 + 160/ 24
= 40 + 6.67
= 46.67
Hence, the MODE is 46.67 .
(ii) Here the maximum frequency is 75, and the class corresponding to this frequency is 20 – 25 .So the modal class is 20 - 25 .
From the table : l = 20, h = 25 – 20 = 5, f1 = 75, f0 = 45, f2 = 35
Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h
= 20 + (75 – 45) /( 2×75–45−35) ×5
= 20 + [(30 ) / (150 - 80)]×5
= 20 +( 30/70) × 5
= 20 + 150/ 70
= 20 + 2.14
= 22.14
Hence, the MODE is 22.14 .
(iii) Here the maximum frequency is 50 and the class corresponding to this frequency is 35 – 40. So the modal class is 35 – 40.
From the table : l = 35, h = 40 – 35 = 5, f1 = 50, f0 = 34, f2 = 42
Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h
= 35 + [[(50–34) / (2 × 50–34−42)]×5
= 35 + [(16)/ (100 - 76)] × 5
= 35 + [(16 × 5 )/ 24]
= 35 + 80/24
= 35 + 3.33
= 38.33
Hence, the MODE is 38.33 .
HOPE THIS ANSWER WILL HELP YOU…
(ii) The mode of the given data = 22.14
(iii) The mode of the given data= 38.33