Arithmetic Mean- Solved Examples of Continuous Series, Cumulative Series and Inclusive Class-Interval Series

Arithmetic Mean [Continuous Series]

In continuous series, the value of each individual frequency distribution is unknown, therefore an assumption is made to make them precise and the assumption is that the frequency of the class intervals is concentrated at the center and on this basis, the midpoint of each class interval has to be found out. In continuous series, the mean can be calculated by any of the following methods:

1-    Direct Method

2-    Short cut Method

3-    Step Deviation Method

Direct Method: The following steps are given below for calculating arithmetic mean in continuous series-

Steps:

1. Find out the mid value of each group or class. The mid value is obtained by adding the lower limit and upper limit of the class and dividing the total by 2. For example, in a class interval [10-20] the mid value is

2. Multiply the mid value of each class by the frequency of the class [f].

3. Add up all the products- [∑fm]

4. Total all the frequency N or [∑f].

5. Then, divide [∑fm] by N [total number of frequencies].

Formula:

Question 1[A]: From the following data find out the mean from direct method in continuous series.

Age group [CI]	No. of persons [f]
0-10	5
10-20	15
20-20	25
30-40	8
40-50	7

Solution:

Age group [CI]	No. of persons [f]	Mid-point [m]	fm
0-10	5	5	25
10-20	15	15	225
20-20	25	25	625
30-40	8	35	280
40-50	7	45	315
	N=60		∑fm = 1470

Short cut Method:

Steps:

1- Find the mid value of each class or group [m].

2- Assume any one of the mid values as an assumed mean [A].

3- Find out the deviations of the mid value of each from the assumed mean [d=m-a].

4- Multiply the deviations of each class by its frequency [fd].

5- Add up all the fd..[∑fd].

6- Total all the frequency N or [∑f].

7- Apply the formula: 

Question 1[B]: Short-cut method in continuous series.

Solution:

Age group [CI]	No. of persons [f]	Mid value [m]	A=25 d=m-a	fd
0-10	5	5	-20	-100
10-20	15	15	-10	-150
20-20	25	25	0	0
30-40	8	35	10	80
40-50	7	45	20	140
	N=60			∑fd = -30

Step Deviation Method: In this method, after finding deviation from assumed mean, it is divided by common factor. Scaling down the deviation by a step will reduce the calculation to minimum. Following steps should be taken for this method:

1- Find the mid value of each class or group [m].

2- Assume any one of the mid values as an assumed mean [A].

3- Find out the deviations of the mid value of each from the assumed mean [d=m-a].

4- Deviations (d) are divided by a common factor (c) to get d’.

5- Multiply d’ of each class by frequency ‘f’ (fd’).

6- Sum up all the fd’... (∑fd')

7- Apply the formula: 

Question 1[C] Step deviation method in continuous series.

Solution:

Age group [CI]	No. of persons [f]	Mid value [m]	A=25 d=m-a	C=10 d‘=d/10	fd’
0-10	5	5	-20	-2	-10
10-20	15	15	-10	-1	-15
20-20	25	25	0	0	0
30-40	8	35	10	1	8
40-50	7	45	20	2	14
	N=60				∑fd’ = -3

Question 2[A]: From the following data find out the mean profits through direct method.

Profits per shop[Rs]	Numbers of shops
100-200	10
200-300	18
300-400	20
400-500	26
500-600	30
600-700	28
700-800	18

 Solution:

Profits per shop [Rs]	Numbers of shops	Mid-point [m]	fm
100-200	10	150	1,500
200-300	18	250	4,500
300-400	20	350	7,000
400-500	26	450	11,700
500-600	30	550	16,500
600-700	28	650	18,200
700-800	18	750	13,500
	N=150		∑fm = 72,900

Question 2[B]: Short-cut method in continuous series.

Solution:

Profits per shop [Rs]	Numbers of shops [f]	Mid-point [m]	A=450 d=m-a	fd
100-200	10	150	-300	-3,000
200-300	18	250	-200	-3,600
300-400	20	350	-100	-2,000
400-500	26	450	0	0
500-600	30	550	100	3,000
600-700	28	650	200	5,600
700-800	18	750	300	5,400
	N = 150			∑fd = 5,400

Question 2[C]: From the following data find out the mean profits through Step-Deviation method.

Solution:

Profits per shop [Rs]	Numbers of shops [f]	Mid-point [m]	A=450 d=m-a	C = 100 d‘ = d/100	fd’
100-200	10	150	-300	-3	-30
200-300	18	250	-200	-3	-36
300-400	20	350	-100	-1	-20
400-500	26	450	0	0	0
500-600	30	550	100	1	30
600-700	28	650	200	2	56
700-800	18	750	300	3	54
	N = 150				∑fd’ = 54

Question 3[A]: From the following data find out the mean through direct method.

Class interval	Frequency [f]
0-5	6
5-10	6
10-15	11
15-20	6
20-25	7
25-30	8
30-35	10
35-40	11

 Solution:

Class interval	Frequency [f]	Mid-point [m]	fm
0-5	6	2.5	15
5-10	6	7.5	45
10-15	11	12.5	137.5
15-20	6	17.5	105
20-25	7	22.5	157.5
25-30	8	27.5	220
30-35	10	32.5	325
35-40	11	37.5	412.5
	N=65		∑fm = 1417.5

Question 3[B]: From the following data find out the mean through short-cut method.

Solution:

Class interval	Frequency [f]	Mid-point [m]	A=22.5 d=m-a	fd
0-5	6	2.5	-20	-120
5-10	6	7.5	-15	-90
10-15	11	12.5	-10	-110
15-20	6	17.5	-5	-30
20-25	7	22.5	0	0
25-30	8	27.5	5	40
30-35	10	32.5	10	100
35-40	11	37.5	15	165
	N=65			∑fd = -45

Question 3[C] From the following data find out the mean through step-deviation method.

Solution:

Class interval	Frequency [f]	Mid-point [m]	A=22.5 d=m-a	C = 5 d‘ = d/5	fd’
0-5	6	2.5	-20	-4	-24
5-10	6	7.5	-15	-3	-18
10-15	11	12.5	-10	-2	-22
15-20	6	17.5	-5	-1	-6
20-25	7	22.5	0	0	0
25-30	8	27.5	5	1	8
30-35	10	32.5	10	2	20
35-40	11	37.5	15	3	33
	N=65				∑fd’ = -9

Arithmetic Mean [Cumulative Series]

Cumulative series can be either more than type or less than type. In the more than type, the frequencies are cumulated upwards so that the first-class interval has the highest frequency and it goes on declining in subsequent classes. In case of less than type, the frequencies are cumulated downwards so that the first-class interval has the lowest frequency and it goes on inclining in subsequent classes. In cumulative series, the data are first converted into a simple series. Then, the method of calculation of mean is same as done in the continuous series.

Question 4. Calculate the mean of the marks scored by the students in an examination by using direct method.

Percentage Marks	Number of Students
Less than 10	4
Less than 20	16
Less than 30	20
Less than 40	65
Less than 50	85
Less than 60	97
Less than 70	100

Solution: 

Marks Class-interval	Frequency       [f]	Mid-point     [m]	fm
0-10	4	5	20
10-20	16-4= 12	15	180
20-30	20-16=4	25	100
30-40	65-20=45	35	1575
40-50	85-65=20	45	900
50-60	97-85=12	55	660
60-70	100-97=3	65	195
	N=100		∑fm = 3630

Note: Solve this question by using short-cut and step deviation method.

Question 5. Calculate the average wage paid to workers by using short-cut method.

Wages [ Rs]	No of workers
More than   75	150
More than   85	140
More than   95	115
More than 105	95
More than 115	70
More than 125	60
More than 135	40
More than 145	25

Solution:

Wages [ Rs] Class-interval	No of workers [f]	Mid-point       [m]	d=m-A A=110	fd
75-85	150-140=10	80	-30	-300
85-95	140-115=25	90	-20	-500
95-105	115-95= 20	100	-10	-200
105-115	95-70=    25	110	0	0
115-125	70-60=    10	120	10	100
125-135	60-40=    20	130	20	400
135-145	40-25=    15	140	30	450
145-155	25	150	40	1000
	N= 150			∑fd =950

Note: Solve this question by using direct method and step deviation method.

Inclusive Class Intervals

In the inclusive method, the upper limit of a class interval is included in the class itself [ 0-9, 10-19, 20-29 etc]. While in exclusive method, the upper limit of the class interval is excluded from the class and is included in the next class [0-10, 10- 20, 20-30 etc].

It is not necessary to convert the inclusive class intervals into an exclusive class interval by adjusting class limits while calculating mean in continuous series. It is also not necessary to re-arrange the series in an ascending or descending order.

Question 6: Find mean of the following data.

Class-interval	Frequency
50-59	1
40-49	3
30-39	9
20-29	10
10-19	15
0-9	2

 Solution:

Class-interval	Frequency	Mid-point[m]	d=m-A A=24.5	d’=d/C C=10	fd’
50-59	1	54.5	30	3	3
40-49	3	44.5	20	2	6
30-39	9	34.5	10	1	9
20-29	10	24.5	0	0	0
10-19	15	14.5	-10	-1	-15
0-9	2	4.5	-20	-2	-4
	N=40				∑fd’ = -1

Note: Solve this question by using direct method and short cut method.

Question 7: Calculate the arithmetic mean from the following data:

Value	Frequency
0-4	328
5-9	350
10-19	720
20-29	664
30-39	598
40-49	524
50-59	378
60-69	244

Solution:
To find the mean from the above data, first make one combined class interval for 0-4 and 5-9 and add both the frequencies [328+350=678]. It will be 0-9 and frequency falls in this class will be 678. Remaining class interval and their respective frequency will be the same as given in the data.

Value	Frequency	Mid-point[m]	d= m-A A=34.5	d’=d/C  C=10	fd’
0-9	678	4.5	-30	-3	-2034
10-19	720	14.5	-20	-2	-1440
20-29	664	24.5	-10	-1	-  664
30-39	598	34.5	0	0	0
40-49	524	44.5	10	1	524
50-59	378	54.5	20	2	756
60-69	244	64.5	30	3	732
	N=3806				∑fd’ = -2126

Dr. Swati Gupta

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