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Mode using Grouping Method Example-6 ( Enter your problem )
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  7. Example-6

7. Example-6





Calculate Mode using Grouping Method from the following grouped data
ClassFrequency
0 - 104
10 - 2016
20 - 3015
30 - 4020
40 - 507
50 - 605


Solution:
Mode using Grouping Method :
1) In column-I, write the original frequency
2) In column-II, combine the frequency two by two, starting from top
3) In column-III, combine the frequency two by two, starting from second
4) In column-IV, combine the frequency three by three, starting from top
5) In column-V, combine the frequency three by three, starting from second
6) In column-VI, combine the frequency three by three, starting from third

Grouping Table :
ClassI
Frequency
II
(1+2)
III
(2+3)
IV
(1+2+3)
V
(2+3+4)
VI
(3+4+5)
0-104
4+16=20
10-20164+16+15=35
16+15=31
20-301516+15+20=51
15+20=35
30-402015+20+7=42
20+7=27
40-50720+7+5=32
7+5=12
50-605

1) 20 is the maximum value in the column-I and it is an individual frequency of Class 30-40. Therefore, we have tick(✓) this Class.
2) 35 is the maximum value in the column-II and it is the sum of 15 and 20; i.e., of Class 20-30 and 30-40. Therefore, we have tick(✓) this both Class.
3) 31 is the maximum value in the column-III and it is the sum of 16 and 15; i.e., of Class 10-20 and 20-30. Therefore, we have tick(✓) this both Class.
4) 35 is the maximum value in the column-IV and it is the sum of 4, 16, and 15; i.e, of Class 0-10, 10-20 and 20-30. Therefore, we have tick(✓) this three Class.
5) 51 is the maximum value in the column-V and it is the sum of 16, 15, and 20; i.e, of Class 10-20, 20-30 and 30-40. Therefore, we have tick(✓) this three Class.
6) 42 is the maximum value in the column-VI and it is the sum of 15, 20, and 7; i.e, of Class 20-30, 30-40 and 40-50. Therefore, we have tick(✓) this three Class.

Analysis Table :
ColumnIIIIIIIVVVITotal
Max Frequency203531355142
0-10✓1
10-20✓✓✓3
20-30✓✓✓✓✓5
30-40✓✓✓✓4
40-50✓1
50-60-

Since 20-30 has the maximum ticks 5. So mode class is 20-30.
`:.` The mode class is `20 - 30`.

`:. L = `lower boundary point of mode class `=20`

`:. f_1 = ` frequency of the mode class `=15`

`:. f_0 = ` frequency of the preceding class `=16`

`:. f_2 = ` frequency of the succedding class `=20`

`:. c = ` class length of mode class `=10`

`Z=L+(f_1-f_0)/((f_1-f_0)+(f_1-f_2)) * c`

Here `f_1-f_0=-1 < 0` and `f_1-f_2=-5 < 0`

So we can not apply the above formula

Since, `f_2=20 > f_0=16`

Thus, Mode `Z=L+f_2/(f_2+f_0) * c`

`=20+20/(20+16) * 10`

`=20+20/36 * 10`

`=20+5.5556`

`=25.5556`




This material is intended as a summary. Use your textbook for detail explanation.
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