Arithmetic Mean Formula of Grouped Data Using Assumed Mean Method | Solved Examples, Important Questions, FAQs

Arithmetic Mean for Grouped Data Using Assumed Mean Method | Class 11 Notes

Written by Neeraj Anand | Published by ANAND TECHNICAL PUBLISHERS

Introduction:

The Assumed Mean Method is a simplified way of calculating the Arithmetic Mean for grouped data, especially useful when dealing with large datasets or complex calculations. This method minimizes computational effort by using a value close to the actual mean as a reference point, reducing the complexity of the arithmetic operations.

For Class 11 students preparing for Board Exams, JEE Mains, and Advanced, mastering this shortcut technique will save time and improve accuracy in problem-solving.

When to Use the Assumed Mean Method:

• When class intervals are large
• When the actual mean is likely to be close to a particular class mark
• To simplify complex calculations and reduce the chances of errors

Arithmetic Mean for Grouped Data Using Assumed Mean Method

In statistics, the assumed mean method is used to calculate mean or arithmetic mean of a grouped data. If the given data is large, then this method is recommended rather than a direct method for calculating mean. This method helps in reducing the calculations and results in small numerical values. This method depends on estimating the mean and rounding to an easy value to calculate with. Again this value is subtracted from all the sample values. When the samples are converted into equal size ranges or class intervals, a central class is chosen and the computations are performed.

Assumed Mean Method simplifies the calculation of the mean by using an assumed mean (a). The formula for the Assumed Mean Method is:

x̄ = a + ∑ƒ_id_i /∑ƒ_i

Where,

• a is assumed mean,
• ƒ_i is frequency of i^th class,
• d_i = x_i – a is derivation of i^th class,
• ∑ƒ_i = n is total number of observations
• x_i is class mark and is equal to (upper class limit + lower class limit)/2

Steps to Find Mean using Assumed Mean Method

For process for calculating mean by using assumed mean method, are discussed below:

Step 1: For each class interval, we have to calculate the class mark by using the formula,

x_i = (upper limit + lower limit)/2

Step 2: Choose an approximate and suitable value of mean, and denote it by “a” .

Step 3: Calculate the deviations using the following formula,

d_i = (x_i – a) for each i

Step 4: Calculate the product by,

(ƒ_i × d_i ), for each i

Step 5: Find total frequency

n = ∑ƒ_i

Step 6: Finally, calculate the mean by using the following formula,

x̄ = a + ∑ ƒ_id_i/∑ƒ_i

Let’s consider an Example

Class Interval	10-25	25-40	40-55	55-70	70-85	85-100
Number of Students	2	3	7	6	6	6

Here we can choose either a = 47.5 or 62.5. Now, let first step is to choose a = 47.5.

The second step is to find the difference (d_i) between each x_i and the assumed mean “a”.

The third step is to find the product of d_i with the corresponding f_i.

Class Interval	Number of students (fi)	Class Mark (xi)	di = xi – 47.5	fidi
10-25	2	17.5	-30	-60
25-40	3	32.5	-15	-45
40-55	7	47.5	0	0
55-70	6	62.5	15	90
70-85	6	77.5	30	180
85-100	6	92.5	45	270
Total	Σfi = 30			Σfidi = 435.

Hence, the mean of the deviations obtained is,

\(\begin{array}{l}\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}\end{array} \)

As, the relationship between \(\begin{array}{l}\bar{d}\end{array} \) and \(\begin{array}{l}\bar{x}\end{array} \) is \(\begin{array}{l}\bar{x}\end{array} \) = a + \(\begin{array}{l}\bar{d}\end{array} \)

We can write,

\(\begin{array}{l}\bar{x}=a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}\end{array} \)

Now, substitute the values of a, Σf_i , and Σf_idi in the above formula to get the mean,

Therefore, x̄ = 47.5 + (435/30)

x̄ = 47.5 + 14.5

x̄ = 62.

Therefore, the mean of the marks obtained by the class 10 students is 62.

Assumed Mean Method Solved Examples

Example.1 : Find mean using assumed mean method for following data:

Class Interval	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Frequency (f)	3	7	12	15	8	5

Solution:

Step 1: Calculate mid-point for each class-interval:

Class Interval	Midpoint (xi​)	Frequency (f)
10 – 20	15	3
20 – 30	25	7
30 – 40	35	12
40 – 50	45	15
50 – 60	55	8
60 – 70	65	5

Step 2: Select a midpoint close to the center of the data. Let’s assume a = 45.

Step 3: For each class interval, calculate the deviation from the assumed mean using d_i = (x_i – a) for each i.

Step 4: Multiply each deviation by the corresponding frequency.

Class Interval	Midpoint (xi​)	Frequency (fi)	Deviation (di​)	fi​ × di​
10 – 20	15	3	15 – 45 = -30	3 × (-30) = -90
20 – 30	25	7	25 – 45 = -20	7 × (20) = -140
30 – 40	35	12	35 – 45 = -10	12 × (-10) = -120
40 – 50	45	15	45 – 45 = 0	15 × 0 =0
50 – 60	55	8	55 – 45 = 10	8 × 10 =80
60 – 70	65	5	65 – 45 = 20	5 × 20 = 100
		∑ƒi = 50		∑ƒidi = -170

Step 5: Calculate the mean using formula: x̄ = a + ∑ ƒ_id_i/∑ƒ_i

x̄ = 45 + (-170)/50

⇒ x̄ = 45 – 3.4 = 41.6

Thus, mean of given dataset is 41.6.

Example 2: The following table gives information about the marks obtained by 110 students in an examination.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	12	28	32	25	13

Find the mean marks of the students using the assumed mean method.

Solution:

Class (CI)	Frequency (fi)	Class mark (xi)	di = xi – a	fidi
0-10	12	5	5 – 25 = – 20	-240
10-20	28	15	15 – 25 = – 10	-280
20-30	32	25 = a	25-25 = 0	0
30-40	25	35	35-25 = 10	250
40-50	13	45	45-25 = 20	260
Total	Σfi =110			Σfidi = -10

Assumed mean = a = 25

Mean of the data: x̄ = a + ∑ ƒ_id_i/∑ƒ_i

= 25 + (-10/ 110)

= 25 -( 1/11)

= (275-1)/11

= 274/11

=24.9

Hence, the mean marks of the students are 24.9.

Example 3: The table below gives information about the percentage distribution of female employees in a company of various branches and a number of departments.

Percentage of female employees	Number of departments
5-15	1
15-25	2
25-35	4
35-45	4
45-55	7
55-65	11
65-75	6

Find the mean percentage of female employees by the assumed mean method.

Solution:

Percentage of female employees (CI)	Number of departments (fi)	Class mark (xi)	di = xi – a	fidi
5-15	1	10	-30	-30
15-25	2	20	-20	-40
25-35	4	30	-10	-40
35-45	4	40 = a	0	0
45-55	7	50	10	70
55-65	11	60	20	220
65-75	6	70	30	180
Total	Σfi =35			Σfidi = 360

Assumed mean = a = 40

Mean = a+ (Σf_id_i /Σf_i)

=40+ (360/35)

= 40+(72/7)

= 40 + 10.28

=50.28 (approx)

Hence, the mean percentage of female employees is 50.28.

Example 4: The following table gives information about the marks obtained by 120 students in an examination.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	14	30	34	27	15

Find the mean marks of the students using the assumed mean method.

Solution:

Let assume mean of the given data be 25 i.e., a = 25.

Class	Frequency	Class mark (xi)	di = (xi – a)	ƒidi
0-10	14	5	5 – 25 = -20	-280
10-20	30	15	15 – 25 = -10	-300
20-30	34	25 = a	25 – 25 = 0	0
30-40	27	35	35 – 25 = 10	270
40-50	15	45	45 – 25 = 20	300
	∑ƒi = 120			∑ƒidi = -10

The formula for mean,

x̄ = a + ∑ƒ_id_i/∑ƒ_i

⇒ x̄ = 25 + (-10/120)

⇒ x̄ = 25 – 1/12

⇒ x̄ = (300-1)/12

⇒ x̄ = 299/12

⇒ x̄ = 24.91

Therefore, the mean marks of the students are 24.91 .

Example 5: A group of students surveyed as a part of their environmental awareness

Number of plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of houses	1	2	2	4	3	2	6

The program in which they collected the following data of plants in 20 homes in a area. Find the mean number of plants per household using the assumed mean method.

Solution:

No. of Plants	No. of houses/ Frequency (ƒi)	Class mark (xi)	di = (xi – a)	ƒidi
0-2	1	1	-6	-6
2-4	2	3	-4	-8
4-6	2	5	-2	-4
6-8	4	7 = a	0	0
8-10	3	9	2	6
10-12	2	11	4	8
12-14	6	13	6	36
	∑ ƒi = 20			∑ƒidi = 32

x̄ = a+ (Σƒ_id_i /Σƒ_i)

⇒ x̄ = 7+(32/20)

⇒ x̄ = 7+(8/5)

⇒ x̄ = 8.6

∴ Mean number of plants per household = 8.6

Practice Questions on Assumed Mean Method

Problem 1: The following table contains information on the grades received by 160 students in an examination. Find the mean marks of the students using the assumed mean method.

Class	0-20	20-40	40-60	60-80	80-100
Frequency	20	30	45	35	30

Problem 2: Find the mean of the following data using the assumed mean method formula.

Marks	Number of students
0-10	5
10-20	3
20-30	4
30-40	3
40-50	3
50-60	4
60-70	7
70-80	9
80-90	7

Problem 3: Find the arithmetic mean using the assumed-mean method:

Class interval	100-120	120-140	140-160	160-180	180-200
Frequency	20	30	15	10	5

Problem 4: The following table contains, distribution of daily wages of 60 worker of a factory.

Daily Wages	100-120	120-140	140-160	160-180	180-200
Number of Workers	15	12	10	13	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Problem 5: Find the mean of the following data using assumed mean method.

Class Interval	Frequency
0-10	8
10-20	12
20-30	10
30-40	12

Problem 6: Find the mean of the following data by assumed mean method.

CI	20 – 60	60 – 100	100 – 150	150 – 250	250 – 350	350 – 4507
f	7	5	16	12	2	3

Problem 7: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored	Number of batsmen
3000 – 4000	4
4000 – 5000	18
5000 – 6000	9
6000 – 7000	7
7000 – 8000	6
8000 – 9000	3
9000 – 10000	1
10000 – 11000	1

Find the mean of the data.

Problem.8: Find the mean of the following data using the assumed mean method formula.

Marks	Number of students
0 – 10	5
10 – 20	3
20 – 30	4
30 – 40	3
40 – 50	3
50 – 60	4
60 – 70	7
70 – 80	9
80 – 90	7
90 – 100	8

FAQS

Q1

What are the three methods used to find the mean of grouped data?

The three methods used to find the mean of the grouped data are:
Direct method
Assumed mean method
Step deviation method

Q2

What is meant by class-mark?

The classmark is also called the midpoint of the class intervals, which can be found by taking the average of its upper-class limit and lower-class limit.
Class Mark = (upper class limit + lower class limit)/2

Q3

Can we get the same mean value for both the grouped and ungrouped data?

The mean value of grouped data slightly differs from the ungrouped data because of the midpoint assumption.

Q4

How to choose the value of “a” in the assumed mean method?

In the assumed mean method, the value of “a” can be chosen which lies in the centre of x₁, x₂, . . ., x_n.

📥 Download PDF Notes:

Download detailed Class 11 Statistics Notes on the Arithmetic Mean Using Assumed Mean Method from ANAND CLASSES for Board exams and JEE Mains/Advanced preparation.