Step Deviation Method for Finding the Mean with Examples (original) (raw)
Last Updated : 23 Jul, 2025
**Statistics is a discipline of mathematics that uses quantified models and representations to gather, review, analyze, and draw conclusions from data. The most commonly used statistical measures are mean, median, and mode. **Variance and standard deviation are measures of dispersion in statistics and various measures of concentration including quartiles, quintiles, deciles, and percentiles. Statistics is a way more beyond the topics mentioned, but here we stop for the "Mean" by Step Deviation method. In general, there are 3 types of mean:
- Arithmetic mean
- Geometric mean
- Harmonic mean
This article is about the Arithmetic mean by Step Deviation method. The arithmetic mean, also called the average or average value is the quantity obtained by summing two or more numbers or variables and then dividing by the number of numbers or variables. The arithmetic mean is important in statistics. For example, Let’s say there are only two quantities involved, the arithmetic mean is obtained simply by adding the quantities and dividing by 2. **Mean or Arithmetic Mean is the average of the numbers i.e. a calculated central value for a set of numbers. General Formulae for Mean is,
Mean = Sum of observation / Number Of Observation
**Example:
The marks obtained by 5 students in a class test are 7, 9, 6, 4, 2 out of 10. Find the mean marks for the class?
According to the formula mean marks of the class are:
Average marks = Sum of observation / Number Of Observation
Here average marks = (7 + 9 + 6 + 4 + 2) / 5 = 28 / 5 = 5.6
Hence the mean marks for the class is 5.6
Derivation of Formula for Mean by Step Deviation Method
The general formula for mean in statistics is:
**Mean = Σf i x i ****/ Σf** i
**Where,
**Σf i x i : the weighted sum of elements and
**Σf i : 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 **f i denotes the frequency and **y i denotes the class-mark of the **i th class the mean is given by,
**Mean = Σf i y i ****/ Σf** i
When the number of classes is large or the value of **f i and **y i is large, an approximate (assumed) mean is taken near the middle, represented by **A and deviation ****(d** i ) is taken into consideration. Then mean is given by,
**Mean = A + Σf i d i ****/ Σf** i
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 **u 1 , u 2 , u 3 , .....u n where,
**u i **= (y i **- A) / c
Then the mean is given by the formula,
Mean = A + c x (Σfiui / Σfi)
This method of finding the mean is called the **Step Deviation Method.
Examples
**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(c) = 6
| Classes | Class-mark(yi) | ui = (yi - A) / c | 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 + c 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(c) = 10.
| Classes | Class-mark(yi) | ui = (yi - A) / c | 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 + c 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(c) = 5
| Classes | Class-mark(yi) | ui = (yi - A) / c | 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 + c 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(c) = 5
| Classes | Class-mark(yi) | ui = (yi - A) / c | 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 + c x (Σfiui / Σfi)
= 17.5 + 5 x (17/80)
= 17.5 + 1.06
= 18.56
Practice Examples - Step Deviation Method
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 |
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 |
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 |
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 |
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 |
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 |