As we are aware that statistics is the collection of data where we find represent them graphically to study it,. Also we find the mean, median, mode of the given data which will help us to study more about the given data. There are different methods of calculating the mean, median and mode, and each time we arrive at approximately the same result whatever be the method we adopt for.

The mean or average does not tell the full story. It is hardly full representative of a mass unless we know the manner in which the individual items scatter around it. A further description of the series is necessary if we have to gauge how representative the average is. Thus the measures of central tendency must be supported and supplemented by some other measures called Dispersion.

The basic dispersions are range and quartile deviation. As they are not based on all the observations and also they do not exhibit any scatter of the observations from an average and completely ignore the composition of the series we have the Mean Deviation and the Standard Deviation. In this section let us study about the Standard Deviation.

Standard deviation is the square root of the arithmetic average of the squares of the deviations measured from the mean.

Thus in the calculation of standard deviation, first arithmetic average is calculated and the deviation of various items from the arithmetic average are squared, The squared deviations are totaled and the sum is divided by the number of items.
The square root of the resulting figure is the standard deviation of the series, The standard deviation is conventionally represented by the Greek letter Sigma $\sigma$.

Symbolically, $\sigma$ = $\sqrt{\frac{\sum d^{\;2}}{N}}$
Where $\sigma$ stands for the standard deviation, $\sum d^{\;2}$ for the sum of the squares of the deviations measured from the arithmetic average and N for the number of items.

How to Get Standard Deviation?

Steps for computation of Standard Deviation:
1. Compute the arithmetic Mean.
2. Compute the deviation ( X - $\overline{X}$ ) of each observation from arithmetic mean.
3. Square each deviation obtained in step 2.
4. Find the sum of the squared deviations in step 3.
5. Divide the sum obtained in Step 4 by N ( total frequency )
6. Take the positive square root of the value obtained in Step 5.
7. The resulting value gives the standard deviation of the distribution.
In case of frequency distribution, the standard deviation is given by,

$\sigma$ = $\sqrt{\frac{1}{N}\sum f(X-\overline{X})^{2}}$,

where X is the value of the variable or the mid-value of the class (in case of grouped or continuous frequency distributions):
f is the corresponding frequency of the value X
N = $\sum$ $f$, is the total frequency and $\overline{X}$ = $\frac{1}{N}$ $\sum$ $f X$, is the arithmetic mean of the distribution.

Standard Deviation Table:

$X$
Frequency ( f )
$d = X - A$
(A - Assumed Mean )
$f d$
$f d^{2}$

Where Mean = A + $\frac{\sum\;f\;d}{N}$
$\sigma^{2}$ = $\frac{1}{N}$ $\sum$ f d2

$\sigma$ = $\sqrt{\sigma^{2}}$

Population Standard Deviation

The Population Standard Deviation is given by the formula,
$\sigma$ = $\sqrt{\frac{\sum(X-M)^{2}}{N}}$,
where $\sum$ - sum of
X - the individual scores
M - the mean of all the scores
N - total number of scores (total frequency)
Standard Deviation of the Mean is calculated using the formula,

$\sigma_{\overline{x}}$ = $\frac{s}{\sqrt{N}}$,
where, x - is the individual data
s - the standard deviation
and N - total number of data (also called the total frequency)

Whenever we calculate the standard deviation of the samples, for the purpose of quality control, we use this method to calculate the standard deviation.
For the given data we first calculate the the mean and the Standard Deviation.
We draw the bell curve as shown below and then divide it accordingly as
$\overline{X}$ - 3 $\sigma$, $\overline{X}$ - 2 $\sigma$, $\overline{X}$ -  $\sigma$, $\overline{X}$ +  $\sigma$, $\overline{X}$ +  2 $\sigma$, $\overline{X}$ + 3 $\sigma$

By using Empirical rule, we can estimate approximately what percent of data values like between the given standard deviation of the mean.

The following example illustrates the graph of the standard deviation.
Draw the standard deviation graph where the mean height of 10 students in a class is 163 cm and the standard deviation 3 cm

We have $\overline{X}$ = 3 and $\sigma$ = 163 cm
$\overline{X}$ - $\sigma$    = 163 - 3 =  160
$\overline{X}$ - 2 $\sigma$ = 163 - 2 (3) = 163 - 6 = 157
$\overline{X}$ - 3 $\sigma$ = 163 - 3 (3) = 163 - 9 = 154
$\overline{X}$ + $\sigma$  = 163 + 3 = 166
$\overline{X}$ - $\sigma$   = 163 + 2(3) = 163 + 6 = 169
$\overline{X}$ - $\sigma$   = 163 + 3(3) = 163 + 9 = 172
The following represent the the mean and the standard deviation of the given information.

Graphing Standard Deviation
Observation from the Graph:
From the above graph we can observe that about 68 % of the scores are between 160 and 166,
                                                              about 95 % of the scores are between 157 and 169
                                                      and about 99.7% of the scores are between 154 and 172.

The relative Standard Deviation is also called the Coefficient of Variation, which is the ratio of the Standard Deviation and the Mean expressed as percentage.
Relative Standard Deviation = $\frac{\sigma}{N}$ x 100We can calculate the Relative Standard Deviation for the given collection of data (whose graph is given above) once we calculate the Mean ($\overline{X}$)and the Standard Deviation($\sigma$).

Solved Example

Question: Find the coefficient of variation, given that the mean height of 10 students is 160 cm and the Standard Deviation is 3 cm.
Solution:
 
From the above formula, we have $\sigma$ = 3 and Mean M = 160

                                     Relative Standard Deviation = $\frac{\sigma}{N}$ x 100

                                                                             = $\frac{3}{160}$ x 100

                               Relative Standard Deviation = 1.875

 


Solved Examples

Question 1: Calculate the Standard Deviation for the following distribution:
Values  
      10           
   20          
     30          
     40         
    50       
     60       
      70           
 Frequency            1
      4      10
     20
    17
       8
        4

Solution:
 
  Values ( x )    
Frequency ( f)   
    f x                      
     d = x - A      
    f d            
       f d2                     
    10
       1
     10       - 33.75
      - 33.75
       1139.0625
    20
       4
     80
      - 23.75
      -  95
     2256.25
    30
     10
    300   
      - 13.75
     - 137.5
     1890.625
    40
     20
    800
         -3.75
          -75
          281.25
    50
     17
    850
      6.25
       106.25
     664.0625
    60
      8
    480
       16.25
       130
     2112.5
    70
      4
    280
       26.25
       105
     2756.25
                    
     N = 64                 
 $\sum$ f x = 2800   
                        
                    
$\sum$ f d2 = 11100   

                                     Mean, A =  $\frac{2800}{64}$ = 43.75

                  Standard Deviation $\sigma$ = $\sqrt{\frac{\sum fd^{2}}{N}}$

                                                            = $\sqrt{\frac{11100}{64}}$

                                                            = $\sqrt{173.4375}$

                                           $\sigma$  = 13.16956

 

Question 2: Evaluate the Population Standard Deviation of the following observations:
  X : 1, 2, 4, 6, 8, 9
Solution:
 
The mean of the given data = $\frac{1+2+4+6+8+9}{6}$ = 5
We have M = 5
The data can be tabulated as follows,

          X              
      X - M        
    (X - M )2               
          1
         -4
          16
          2
         -3
           9
          4
         -1
           1
          6
          1
           1
          8
          3
           9
         9
          4
          16
              52
         
                 Variance  = $\frac{52}{6}$ = 8.67

Standard Deviation = $\sqrt{8.67}$ = 2.944
 

1. Show that the standard deviation of the natural numbers 1, 2, 3 ,4 and 5 is $\sqrt{2}$.

2. The mean of 10 items is 50 and the Standard Deviation is 14. Find the sum of the squares of all the terms.

3. The following table shows the number of persons in different age groups in a town. Calculate the Standard Deviation of the given data.

Age (years)
20 - 25
25 - 30
30 - 35
35 - 40
40 - 45
45 - 50
No. of Persons 170 110
80
45
40
35