Standard Deviation

Variation (dispersion) is the property of deviation of values from the average. The degree of variation is indicated by the measures of variation. There are various measures of variation and the commonly used ones are:
1) Range
2) Mean deviation
3) Standard Deviation and
4) Quartile deviation.
The range is based only on the lowest and the highest values. Quartile deviation is based only on the quartiles, and not based on values. Mean deviation is based on values nut it is not convenient for mathematical analysis. So, we consider standard deviation which is based on all the values.
The standard deviation of a set of values is the positive square root of the mean of the standard deviations of the values from their arithmetic mean. It is denoted by $\sigma$ (sigma).
The measures of dispersion are statistical devices to measure the variability or the dispersion in a series. They tell us the extent to which the values of the series differ between each other or from their average.
The Measures of Central tendency are the average of the original values so they are known as the averages of the first order. The measures of dispersion are only averages of deviations taken from the average. Therefore they are known as the averages of the second order.

Standard Deviation is a measure of dispersion in statistics. It gives an idea of how the individual data in a dataset is dispersed from the mean.

For example : The mean of 5 and 6 is 5.5. Also the mean of 1 and 10 is also 5.5

The data points 5 and 6 are closer to 5.5 than data points 1 and 10. This means that the standard deviation of the data set 1,10 is more than that of the data set 5,6. Thus the standard deviation gives an idea of the dispersion of the data from the mean

A related term is the variance. Variance is the square of the stanadrd deviation.
Standard deviation is defined as the square root of the mean of the squares of the deviations of all the values of a series taken from the arithmetic mean. It is also known as the root mean square deviation. The symbol used for standard deviation is $\sigma$

1. The minimum value of standard deviation is 0. i.e. it cannot be negative.
2. When the items in a series are more dispersed from the mean, then the standard deviation is also large.
Merits of standard deviation
1. It is based on all observations in a distribution
2. It is capable of further algebraic treatment.
3. We can find out measures like coefficient of variation, combined standard deviation extra.
Demerits of Standard deviation
1. It is difficult to calculate
2. It gives more importance to bigger values.

Definition Of Variance:

The square of standard deviation. A measure of the degree of spread among a set of values and also the measure of the tendency of individual values vary from the mean value.

Formulae:

Standard Deviation Population Standard Deviation

S = `sqrt((Sigma(x-M)^2)/(n-1))` S = `sqrt((Sigma(x-M)^2)/(n))`

Where `Sigma` = sum of

`x ` = individual score

`M` = Mean of all scores

`n` =sample size (Number of scores)

Explanation of Parameters:

`x` - Normally in mathematics, letters such as `x` stand for a single number.The letter `x` stands for an entire list of numbers-every single number in the list whose standard deviation we are trying to find. We use each of these numbers to calculate the standard deviation, although we just write them as one letter in the formula. In some textbooks we see this written as `x_(i)` , to show that it stands for a list of numbers` x_(1)` ,` x_(2)` ,` x_(3)` ,` x_(4)` etc.;

`M ` - It is nothing but the arithmetic mean of all the numbers in the list i.e; add them together and divide by `n`

`n` - `n` stands for how many numbers there are in the list.

`S` - This is other version of `Sigma` - the upper case letter. It has a standard meaning in mathematics as Add up a list of numbers. They use the Greek letter `S` because it represents sum, i.e; add together.

To calculate standard deviation, we follow certain steps.
Steps in the standard deviation calculation are: -
1. Calculate the arithmetic Mean
2. Find the deviation of each item from the mean
3. Square these deviations and add them
4. We get $\sum (x-\bar{x})^{2}$
5. Divide this sum by the total number of items
6. Take the square root of this. This will give the standard deviation.

Method 1:

Ex:1 An example to find the standard deviation of 1,2,3,4,5

Sol:

Step 1: Calculate mean and deviation

`x`	`M`	`(x-M)`	`(x-M)^2`
1	3	-2	4
2	3	-1	1
3	3	0	0
4	3	1	1
5	3	2	4

Step 2: Find the sum of ` (x-M)^2`

4+1+0+1+4=10

Step 3: `n` =5, the total number of values. Find ` n-1` = 5-1 = 4.

Step 4: Now find the standard deviation from , `S` = `sqrt((Sigma(x-M)^2)/(n-1))` = 1.58113.

Method 2:

Ex:2 To find the standard deviation of 1,2,3,4,5.

Sol: Firstly, square each of the scores

`x`	`x^2`
1	1
2	4
3	9
4	16
5	25

Using the formula

S=`sqrt((Sigma(x^2)- (Sigma(x)*Sigma(x/n)))/(n-1))`

= `sqrt[(55-(15*(15/5)))/(5-1)]`

=`sqrt((55-(225/5))/4)`

=`sqrt((55-45)/4)`

=`sqrt [10/4]`

=`sqrt(2.5)`

`S` = 1.58113.

The standard deviation of the above numbers is given by `S` =1.58113.

Population Deviation Method :

Ex: 3 To find the population standard deviation of the following numbers 1,2,3,4,5.

Sol: Perform the steps 1 and 2 as seen in the above example.

Step 3: Now find the population standard deviation using the formula, S = `sqrt((Sigma(x-M)^2)/(n))`

`sqrt(10/5)` = 1.414

The Population Standard Deviation = 1.414

We can find the standard deviation by finding the variance. Given below are the steps to be followed for solving the problem.

Step 1: Calculate arithmetic mean of the list of numbers.

Step 2: We subtract the mean value from all the numbers in the list.

Step 3: In order to avoid the negative numbers from the part of calculation, square the negative numbers.

Step 4: Add the squared differences to get a single number.

Step 5: Now find the variance to get the standard deviation.

Step 6: From the variance we can find the standard deviation by finding the `sqrt(variance)`

Here are some standard deviation example problems:

Problem: Calculate the standard deviation of the numbers 4,1,0,2,1,5,2,2,3,4,3.

Step 1:` x` = (4+1+0+2+1+5+2+2+3+4+3)/10 = 25/10 = 2.5

Step 2: We subtract the mean value from numbers given in the problem

1.5,-1.5,-2.5,-0.5,-1.5,2.5,-0.5,0.5,1.5,0.5

Step 3: Squaring all the numbers to avoid the negative numbers.

2.25,2.25,6.25,0.25,0.25,2.25,6.25,0.25,0.25,2.25,0.25

Step 4: Add the squared differences to get a single number

2.25+2.25+6.25+0.25+0.25+2.25+6.25+0.25+0.25+2.25+0.25 = 22.5

Step 5: Variance `S^2` = 22.5/10 = 2.25

Step 6: standard deviation = `S` = `sqrt(2.25)` = 1.5

Lets consider few examples of standard deviation
Example 1:
Find the standard deviation of the following values
2, 3, 4, 5, 6
Solution
We first find the arithmetic mean of the values. Then we find the deviation of each item from the mean. Find the squares of the deviations and add them. Then divide the sum by the number of items in the series and take the square root
The calculations are shown in the table below

X	$(x_{1}-\bar{x})$	$(x_{1}-\bar{x})^{2}$
2	2-4 = -2	4
3	3-4 = -1	1
4	4-4 = 0	0
5	5-4 = 1	1
6	6-4 = 2	4
		10

Mean $\bar{x}=\frac{2+3+4+5+6}{5}=4$
Standard deviation $\sigma =\sqrt{\frac{\sum (x_{1}-\bar{x})^{2}}{n}}=\sqrt{\frac{10}{4}}=\sqrt{2.5}=1.58$
Here you can see that all items are taken into consideration. Also we can see that it give more importance to bigger values since we are squaring the values.
Lets consider another example of standard deviation
Example 2: -
Find the standard deviation of
x: 1 2 3 4
f: 10 4 3 3
Solution
We first find the arithmetic mean of the values. Then we find the deviation of each item from the mean. Find the squares of the deviations, multiply by the corresponding frequencies and add them. Then divide the sum by the number of items in the series and take the square root
The calculations are shown in the table below

x	f	$(x_{1}-\bar{x})$	$(x_{1}-\bar{x})^{2}$	$f(x_{1}-\bar{x})^{2}$
1	10	1-1.95 = -0.95	0.9025	9.025
2	4	2-1.95 = 0.05	0.0025	0.01
3	3	3-1.95 = 1.05	1.1025	3.3075
4	3	4-1.95 = 2.05	4.2025	12.6075
	20		Total	24.95

Mean, $\bar{x}=\frac{\sum fx}{\sum f}=\frac{1*10+2*4+3*3+4*3}{20}=\frac{10+8+9+12}{20}=\frac{39}{20}=1.95$
Standard deviation, $\sigma =\sqrt{\frac{\sum f(x_{1}-\bar{x})^{2}}{N}}=\sqrt{\frac{24.95}{20}}=1.12$
Here also we can see that its based on all the observation and it gives more importance to bigger values.

Lets learn what is standard deviation.
Standard deviation is a measure of variation in a series. When the items in a series are more dispersed from mean, then the standard deviation is large and when the items is the series are less dispersed from mean, then the standard deviation is small. The square of standard deviation is called Variance, which is again a measure of dispersion.
In the case of an Individual series
If x₁,x₂,x₃, ….x_n are n items then the formula for standard deviation is given by
$\sigma =\sqrt{\frac{(x_{1}-\bar{x}+(x_{2}-\bar{x}+.....+(x_{x}-\bar{x}}{n}}$
$=\sqrt{\frac{\sum (x_{1}-\bar{x})^{2}}{n}}$
In the case of discrete and continuous distribution
If x₁,x₂,x₃, ….x_n are n items and f₁,f₂,f₃, ….f_n are the corresponding frequencies, the equation of standard deviation is given by
$\sigma =\sqrt{\frac{f_{1}(x_{1}-\bar{x})^{2}+f_{2}(x_{2}-\bar{x})^{2}+.....+f_{x}(x_{x}-\bar{x})^{2}}{N}}$
$=\sqrt{\frac{\sum f(x_{1}-\bar{x})^{2}}{N}}$, where $N=\sum f$
In a continuous series we take x as the class marks and use the same formula for standard deviation as in the case of discrete distribution.
Lets learn how to find the standard deviation
To find the standard deviation, we first find the arithmetic mean of the values. Then we find the deviation of each item from the mean. Find the squares of the deviations and add them. Then divide the sum by the number of items in the series and take the square root.
These are the important steps in finding the standard deviation.
The mean and standard deviation are the most useful measures. Mean is the measure of central tendency and the standard deviation is the measure of dispersion, which is using the mean. We can say that the mean is the first order average and the standard deviation is the second order average.
Coefficient of variation is a measure that is used to compare the dispersion in two series. Coefficient of variation is the percentage ratio of the standard deviation to the mean Greater the coefficient of variation lesser will be the consistent. Also greater the coefficient of variation, greater will be the variability.
Coefficient of variation = (Standard deviation / Mean) *100
The consistency increases with the decrease in the coefficient of variation.Variance
Variance is another measure of dispersion which is obtained by squaring standard deviation
Variance = (Standard deviation)²
The sample standard deviation is an important measure in Statistics. It plays a very important role in the theory of estimation. Here we find the standard deviation of the sample unlike population standard deviation. The sample standard deviation is denote by s.
The graph of Normal distribution shows the area between the mean and standard deviation. The area between the Mean $\pm$ standard deviation is 68.27 % of the total area. The area between Mean $\pm$ 2 standard deviation is 95.45 % of the total area
area between Mean $\pm$ 3 standard deviation is 98.73 % of the total area.