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
Please Login to comment...