Calculate Mode using Grouping Method from the following grouped data

Class	Frequency
0 - 10	4
10 - 20	2
20 - 30	18
30 - 40	22
40 - 50	21
50 - 60	19
60 - 70	10
70 - 80	3
80 - 90	1

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 :

Class	I Frequency	II (1+2)	III (2+3)	IV (1+2+3)	V (2+3+4)	VI (3+4+5)
0-10	4
		4+2=6
10-20	2			4+2+18=24
			2+18=20
20-30	18				2+18+22=42
		18+22=40
30-40	22					18+22+21=61
			22+21=43
40-50	21			22+21+19=62
		21+19=40
50-60	19				21+19+10=50
			19+10=29
60-70	10					19+10+3=32
		10+3=13
70-80	3			10+3+1=14
			3+1=4
80-90	1

1) 22 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) 40 is the maximum value in the column-II and it is the sum of 18 and 22; i.e., of Class 20-30 and 30-40. Therefore, we have tick(✓) this both Class.
2) 40 is the maximum value in the column-II and it is the sum of 21 and 19; i.e., of Class 40-50 and 50-60. Therefore, we have tick(✓) this both Class.
3) 43 is the maximum value in the column-III and it is the sum of 22 and 21; i.e., of Class 30-40 and 40-50. Therefore, we have tick(✓) this both Class.
4) 62 is the maximum value in the column-IV and it is the sum of 22, 21, and 19; i.e, of Class 30-40, 40-50 and 50-60. Therefore, we have tick(✓) this three Class.
5) 50 is the maximum value in the column-V and it is the sum of 21, 19, and 10; i.e, of Class 40-50, 50-60 and 60-70. Therefore, we have tick(✓) this three Class.
6) 61 is the maximum value in the column-VI and it is the sum of 18, 22, and 21; i.e, of Class 20-30, 30-40 and 40-50. Therefore, we have tick(✓) this three Class.

Analysis Table :

Column	I	II	III	IV	V	VI	Total
Max Frequency	22	40	43	62	50	61
0-10							-
10-20							-
20-30		✓				✓	2
30-40	✓	✓	✓	✓		✓	5
40-50		✓	✓	✓	✓	✓	5
50-60		✓		✓	✓		3
60-70					✓		1
70-80							-
80-90							-

Since 5 is the maximum ticks repeated 2 times, So grouping method fails to give the modal class.

We use formula Mode = 3 Median - 2 Mean

Find Mean, Median

Class `(1)`	Frequency `(f)` `(2)`	Mid value `(x)` `(3)`	`d=(x-A)/h=(x-45)/10` `A=45,h=10` `(4)`	`f*d` `(5)=(2)xx(4)`	`cf` `(7)`
0 - 10	4	5	-4	-16	4
10 - 20	2	15	-3	-6	6
20 - 30	18	25	-2	-36	24
30 - 40	22	35	-1	-22	46
40 - 50	21	45=A	0	0	67
50 - 60	19	55	1	19	86
60 - 70	10	65	2	20	96
70 - 80	3	75	3	9	99
80 - 90	1	85	4	4	100
---	---	---	---	---	---
	`n = 100`	-----	-----	`sum f*d=-28`	-----

Mean `bar x = A + (sum fd)/n * h`

`=45 + (-28)/100 * 10`

`=45 + (-0.28) * 10`

`=45 -2.8`

`=42.2`

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To find Median Class
= value of `(n/2)^(th)` observation

= value of `(100/2)^(th)` observation

= value of `50^(th)` observation

From the column of cumulative frequency `cf`, we find that the `50^(th)` observation lies in the class `40 - 50`.

`:.` The median class is `40 - 50`.

Now,
`:. L = `lower boundary point of median class `=40`

`:. n = `Total frequency `=100`

`:. cf = `Cumulative frequency of the class preceding the median class `=46`

`:. f = `Frequency of the median class `=21`

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

Median `M = L + (n/2 - cf)/f * c`

`=40 + (50 - 46)/21 * 10`

`=40 + (4)/21 * 10`

`=40 + 1.9048`

`=41.9048`

---

We have given Mean (`bar X`) `=42.2`, Median(`M`) `=41.9048`, Mode(`Z`) `=?`

`Z=3 M - 2 bar X`

`Z=3*41.9048-2*42.2`

`Z=125.7144-84.4`

`Z=41.3144`

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