Arithmetic Mean of Grouped Data Using Step Deviation Method | Class 11 Notes
Written by Neeraj Anand | Published by ANAND TECHNICAL PUBLISHERS
Introduction:
The Step Deviation Method is an advanced technique for calculating the Arithmetic Mean of grouped data. It simplifies the process by using deviations from an assumed mean, divided by a common factor (class width). This method is particularly useful when data values are large or complex.
For Class 11 students preparing for Board Exams, JEE Mains, and Advanced, mastering the Step Deviation Method provides a faster and more efficient way to compute the arithmetic mean.
Advantages of the Step Deviation Method:
✔️ Reduces calculations by working with smaller numbers.
✔️ Ideal for large data sets with uniform class widths.
✔️ Useful in Board Exams and JEE Mains for quick and accurate calculations.
Table of Contents
Derivation of Formula for Arithmetic Mean by Step Deviation Method
The general formula for mean in statistics is:
Mean = Σfixi / Σfi
Where,
Σfixi: the weighted sum of elements and
Σfi: the number of elements
In the case of grouped data, assume that the frequency in each class is centered at its class-mark. If there are n classes and fi denotes the frequency and xi denotes the class-mark of the ith class the mean is given by,
Mean = Σfixi / Σfi
When the number of classes is large or the value of fi and xi is large, an approximate (assumed) mean is taken near the middle, represented by A and deviation (di) is taken into consideration. Then mean is given by,
Mean = A + Σfidi / Σfi
In the problems where the width of all classes is the same, then further simplify the calculations of the mean by computing the coded mean, i.e. the mean of u1, u2, u3, …..un where,
ui = (xi – A) / h
Where h = Class Interval or Class Width
Then the mean is given by the formula,
Mean = A + h x (Σfiui / Σfi)
This method of finding the mean is called the Step Deviation Method.
Step Deviation Method
Consider the Example as given below.
Class Interval | 10-25 | 25-40 | 40-55 | 55-70 | 70-85 | 85-100 |
Number of Students | 2 | 3 | 7 | 6 | 6 | 6 |
In the step deviation method, we will add one more column to the table to find the mean, which is ui = (xi – a)/h
Where “a” is the assumed mean and “h” is the class size, which is equal to 15 (i.e) width of the class interval.
Class Interval | Number of students (fi) | Class Mark (xi) | di = xi – 47.5 di = xi – a | ui =(xi – a)/h (h=15) | fiui |
10-25 | 2 | 17.5 | -30 | -2 | -4 |
25-40 | 3 | 32.5 | -15 | -1 | -3 |
40-55 | 7 | 47.5 | 0 | 0 | 0 |
55-70 | 6 | 62.5 | 15 | 1 | 6 |
70-85 | 6 | 77.5 | 30 | 2 | 12 |
85-100 | 6 | 92.5 | 45 | 3 | 18 |
Total | Σfi = 30 | Σfiui = 29 |
Therefore, we obtained
\(\begin{array}{l}\bar{u}=\frac{\sum f_{i}u_{i}}{\sum f_{i}}\end{array} \)
The relation between \(\begin{array}{l}\bar{u}\end{array} \) and \(\begin{array}{l}\bar{x}\end{array} \) is:
\(\begin{array}{l}\bar{x}=a+h\frac{\sum f_{i}u_{i}}{\sum f_{i}}\end{array} \)
Now, substitute the values of a, h,Σfi , and Σfiui in the above formula to get the mean,
x̄ = 47.5 + 15(29/30)
x̄ = 47.5 + 15(0.967)
x̄= 47.5+ 14.5
x̄ = 62
Hence, the mean of the marks scored by the students = 62.
Difference between Direct, Assumed Mean and Step Deviation Method
Direct Method, Assumed Mean Method, and Step Deviation Method are three different techniques for calculating the mean (average) of a dataset. Each method has its own way of simplifying the calculations. Some of the common differences among these methods are:
Direct Method | Assumed Mean Method | Step Deviation Method |
---|---|---|
Best method for simple problems and small data set. | Best method for large data sets with potentially large values. | Best method for large data sets with uniform class intervals. |
This method is straightforward and easy to understand. | It simplifies large arithmetic calculations using the assumed mean. | Further simplifies calculations and arithmetic complexity by working with smaller numbers. |
In this method we calculate midpoints(xi) of each class intervals and multiply them with corresponding frequency(ƒ), i.e., ƒixi . | In this method we calculate deviation(di) of each data point and multiply them with corresponding frequency(ƒ), i.e., ƒidi . | In this method we calculate step deviation(ui) of each class intervals and multiply them with corresponding frequency(ƒ), i.e., ƒiui . |
The formula of this method isMean = ∑ƒixi / ∑ƒi | Formula for assumed mean method isMean = a + ∑ ƒidi / ∑ ƒi | Formula for step deviation method isMean = a + h × ∑ ƒiui / ∑ ƒi |
Solved Examples For Arithmetic Mean of Grouped Data Using Step Deviation Method
Question 1: Find the mean for the following frequency distribution?
Class Intervals | 84-90 | 90-96 | 96-102 | 102-108 | 108-114 |
Frequency | 8 | 12 | 15 | 10 | 5 |
Solution:
Applying the Standard Deviation Method,
We take the assumed mean to A = 99, and here the width of each class(h) = 6
Classes | Class-mark(xi) | ui = (xi – A) / h | frequency(fi) | fiui |
---|---|---|---|---|
84-90 | 87 | -2 | 8 | -16 |
90-96 | 93 | -1 | 12 | -12 |
96-102 | 99 | 0 | 15 | 0 |
102-108 | 105 | 1 | 10 | 10 |
108-114 | 111 | 2 | 5 | 10 |
Total | 50 | -8 |
Mean = A + h x (Σfiui / Σfi)
= 99 + 6 x (-8/50)
= 99 – 0.96
= 98.04
Question 2: Find the mean for the following frequency distribution?
Class Intervals | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency | 10 | 6 | 8 | 12 | 5 | 9 |
Solution:
Applying the Standard Deviation Method,
Construct the table as under, taking assumed mean A = 45, and width of each class(h) = 10.
Classes | Class-mark(xi) | ui = (xi – A) / h | frequency(fi) | fiui |
---|---|---|---|---|
20-30 | 25 | -2 | 10 | -20 |
30-40 | 35 | -1 | 6 | -6 |
40-50 | 45 | 0 | 8 | 0 |
50-60 | 55 | 1 | 12 | 12 |
60-70 | 65 | 2 | 5 | 10 |
70-80 | 75 | 3 | 9 | 27 |
Total | 50 | 23 |
Mean = A + h x (Σfiui / Σfi)
= 45 + 10 x (23/50)
= 45 + 4.6
= 49.6
Question 3: The weight of 50 apples was recorded as given below
Weight in grams | 80-85 | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |
Number of apples | 5 | 8 | 10 | 12 | 8 | 4 | 3 |
Calculate the mean weight, to the nearest gram?
Solution:
Construct the table as under, taking assumed mean A = 97.5. Here width of each class(h) = 5
Classes | Class-mark(xi) | ui = (yi – A) / h | frequency(fi) | fiui |
---|---|---|---|---|
80-85 | 82.5 | -3 | 5 | -15 |
85-90 | 87.5 | -2 | 8 | -16 |
90-95 | 92.5 | -1 | 10 | -10 |
95-100 | 97.5 | 0 | 12 | 0 |
100-105 | 102.5 | 1 | 8 | 8 |
105-110 | 107.5 | 2 | 4 | 8 |
110-115 | 112.5 | 3 | 3 | 9 |
Total | 50 | -16 |
Mean = A + h x (Σfiui / Σfi)
= 97.5 + 5 x (-16/50)
= 97.5 – 1.6
= 95.9
Hence the mean weight to the nearest gram is 96 grams.
Question 4: The following table gives marks scored by students in an examination:
Marks | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
Number of students | 3 | 7 | 15 | 24 | 16 | 8 | 5 | 2 |
Calculate the mean marks correct to 2 decimal places?
Solution:
Construct the table as under, taking assumed mean A = 17.5. Here width of each class(h) = 5
Classes | Class-mark(xi) | ui = (xi – A) / h | frequency(fi) | fiui |
---|---|---|---|---|
0-5 | 2.5 | -3 | 3 | -9 |
5-10 | 7.5 | -2 | 7 | -14 |
10-15 | 12.5 | -1 | 15 | -15 |
15-20 | 17.5 | 0 | 24 | 0 |
20-25 | 22.5 | 1 | 16 | 16 |
25-30 | 27.5 | 2 | 8 | 16 |
30-35 | 32.5 | 3 | 5 | 15 |
35-40 | 37.5 | 4 | 2 | 8 |
Total | 80 | 17 |
Mean = A + h x (Σfiui / Σfi)
= 17.5 + 5 x (17/80)
= 17.5 + 1.06
= 18.56
Practice Problems – Step Deviation Method
Problem.1 : Find the mean for the following frequency distribution:
Class-Interval | Frequency |
---|---|
10-20 | 3 |
20-30 | 5 |
30-40 | 8 |
40-50 | 4 |
50-60 | 2 |
Problem.2. Calculate the mean for the given frequency distribution:
Class Interval | Frequency |
---|---|
5-15 | 7 |
15-25 | 10 |
25-35 | 15 |
35-45 | 6 |
45-55 | 2 |
Problem.3. Calculate the mean for the given frequency distribution:
Class Interval | Frequency |
---|---|
1-5 | 6 |
5-9 | 8 |
9-13 | 10 |
13-17 | 4 |
17-21 | 5 |
Problem.4. Determine the mean for this frequency distribution:
Class Interval | Frequency |
---|---|
2-6 | 3 |
6-10 | 6 |
10-14 | 9 |
14-18 | 7 |
18-22 | 5 |
Problem.5. Find the mean for the given frequency distribution:
Class Interval | Frequency |
---|---|
3-8 | 4 |
8-13 | 10 |
13-18 | 6 |
18-23 | 8 |
23-28 | 2 |
Problem.6. Calculate the mean for the following frequency distribution:
Class Interval | Frequency |
---|---|
0-4 | 5 |
4-8 | 8 |
8-12 | 10 |
12-16 | 7 |
16-20 | 3 |
Frequently Asked Questions (FAQs) on Mean of Grouped Data
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
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
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.
What does “h” mean in the step-deviation method?
In the step-deviation method, “h” represents class size.
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, . . ., xn.
What is the step deviation method for finding the mean?
The step deviation method is a simplified approach for calculating the mean of a frequency distribution when the data is large. It involves using assumed mean and step deviations (class intervals) to reduce the arithmetic calculations.
How do you select the assumed mean (A) in the step deviation method?
The assumed mean (A) is usually chosen from the class midpoints of the distribution. It can be any midpoint, but usually the midpoint of the central class interval is selected to simplify calculations.
What are the step deviations and how are they calculated?
Step deviations di are calculated using the formula
di = (xi – A) / h
, where xi is the class midpoint, A is the assumed mean, and h is the class width.
How do you choose the class width (h) in the step deviation method?
The class width (h) is typically chosen based on the uniform width of the class intervals in the frequency distribution.
What are the advantages of using the step deviation method?
The step deviation method simplifies calculations, especially when dealing with large data sets or data with large values. It reduces the magnitude of the numbers involved, making arithmetic operations easier and less error-prone.
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