Step deviation Method for Finding the Mean with Examples
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, Letâ€™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 = Î£f_{i}x_{i }/ Î£f_{i}
Where,
Î£f_{i}x_{i}: the weighted sum of elements and
Î£f_{i}: the number of elements
In the case of grouped data, assume that the frequency in each class is centered at its classmark. If there are n classes and f_{i} denotes the frequency and y_{i} denotes the classmark of the i_{th} class the mean is given by,
Mean = Î£f_{i}y_{i }/ Î£f_{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 + Î£f_{i}d_{i }/ Î£f_{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 (Î£f_{i}u_{i }/ Î£f_{i})
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 
8490 
9096 
96102 
102108 
108114 
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 
Classmark(y_{i}) 
u_{i }= (y_{i }– A) / c 
frequency(f_{i}) 
f_{i}u_{i} 

8490 
87 
2 
8 
16 
9096 
93 
1 
12 
12 
96102 
99 
0 
15 
0 
102108 
105 
1 
10 
10 
108114 
111 
2 
5 
10 
Total 


50 
8 
Mean = A + c x (Î£f_{i}u_{i }/ Î£f_{i})
= 99 + 6 x (8/50)
= 99 – 0.96
= 98.04
Question 2: Find the mean for the following frequency distribution?
Class Intervals 
2030 
3040 
4050 
5060 
6070 
7080 
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 
Classmark(y_{i}) 
u_{i }= (y_{i }– A) / c 
frequency(f_{i}) 
f_{i}u_{i} 

2030 
25 
2 
10 
20 
3040 
35 
1 
6 
6 
4050 
45 
0 
8 
0 
5060 
55 
1 
12 
12 
6070 
65 
2 
5 
10 
7080 
75 
3 
9 
27 
Total 


50 
23 
Mean = A + c x (Î£f_{i}u_{i }/ Î£f_{i})
= 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 
8085 
8590 
9095 
95100 
100105 
105110 
110115 
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  Classmark(y_{i})  u_{i }= (y_{i }– A) / c  frequency(f_{i})  f_{i}u_{i} 

8085 
82.5 
3 
5 
15 
8590 
87.5 
2 
8 
16 
9095 
92.5 
1 
10 
10 
95100 
97.5 
0 
12 
0 
100105 
102.5 
1 
8 
8 
105110 
107.5 
2 
4 
8 
110115 
112.5 
3 
3 
9 
Total 


50 
16 
Mean = A + c x (Î£f_{i}u_{i }/ Î£f_{i})
= 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 
05 
510 
1015 
1520 
2025 
2530 
3035 
3540 
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 
Classmark(y_{i}) 
u_{i }= (y_{i }– A) / c 
frequency(f_{i}) 
f_{i}u_{i} 

05 
2.5 
3 
3 
9 
510 
7.5 
2 
7 
14 
1015 
12.5 
1 
15 
15 
1520 
17.5 
0 
24 
0 
2025 
22.5 
1 
16 
16 
2530 
27.5 
2 
8 
16 
3035 
32.5 
3 
5 
15 
3540 
37.5 
4 
2 
8 
Total 


80 
17 
Mean = A + c x (Î£f_{i}u_{i }/ Î£f_{i})
= 17.5 + 5 x (17/80)
= 17.5 + 1.06
= 18.56
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