**Standard deviation**is the positive square root of the arithmetic mean of the squares of the deviations of the given observation from their arithmetic mean.

Among all the methods of finding out dispersion,

**standard deviation**is regarded as the best. It is free from the defects which the earlier methods range, quartile deviation and mean deviation suffer. Its value is based upon each and every item of the series and it also take into account algebraic signs.**Standard deviation**is also known as ‘**Root Mean-Square Deviation**’ because it is the square root of the arithmetic mean of the square of the deviations. It is denoted by the**σ**(sigma).###
__Individual Series__**:**

__Individual Series__

The standard deviation of the set of n observation is
given by

###
__Discrete
and Continuous Series__**:**

__Discrete and Continuous Series__

For a discrete and continuous frequency distribution, the
standard deviation is given by

Where N is the total number
of observations of the given data. In case of continuous distribution x is
taken as the mid-value of the corresponding class.

If the values of the observations and their corresponding
frequencies are large, then the calculation of standard deviation is tedious
and time consuming. In such cases, we can calculate the standard deviation by
taking deviation as follows:

###
__Individual
Series__**:**

__Individual Series__

When the values of the observations are very large in the
individual series, the standard deviation can be calculated by using the
following formula,

Where d=x–A and A is
assumed mean.

###
__Discrete
or Continuous Series__**:**

__Discrete or Continuous Series__

When the values of the observations and the frequencies are
large in case of discrete or continuous series, the calculation of standard deviation
can be made easy by using the following formula.

Where d=x–A and A is
assumed mean.

When the value of variables or mid-values have some common
factor, the calculation of the standard deviation can be made still easier by
taking the deviation as follows:

###
**Coefficient
of Standard Deviation**

The relative measure or the coefficient of standard deviation is
given by the formula,

*Workout Examples*

*Example 1: The weekly expenditure of 5 families in rupees are given below:*Family |
A |
B |
C |
D |
E |

Expenditure
(Rs.) |
300 |
325 |
350 |
375 |
425 |

*Calculate*

*the standard deviation and its coefficient by:*

*a.*

*Direct method*

*b.*

*Deviation method*

*Solution:*

*Here,*

*a.*

*Direct method*Families |
Expenditure |
x^{2} |

ABCDE |
300325350375425 |
90000105625122500140625180625 |

∑x = 1775 |
∑x^{2} = 639375 |

*b.*

*Deviation method*Family |
Exp. (x) |
d = x – A(350) |
d^{2} |

ABCDE |
300325350375425 |
-50-2502575 |
250062506255625 |

25 |
9375 |

*Example 2: Find the standard deviation and its coefficient from the following frequency distribution.*

Marks |
20 |
24 |
30 |
35 |
38 |
40 |

No. of students |
8 |
7 |
10 |
12 |
6 |
3 |

*Solution:*

*Here,*

Marks |
No. of students |
fx |
fx^{2} |

202430353840 |
87101263 |
160168300420228120 |
3200403290001470086644800 |

N = 46 |
∑fx = 1396 |
∑fx^{2} = 44396 |

*Example 3: Following are the marks obtained by students in a test exam.*

Marks |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
80-90 |

No. of students |
4 |
6 |
10 |
17 |
11 |
9 |
3 |

*Calculate the standard deviation and its coefficient by*

*a.*

*Direct method*

*b.*

*Deviation method*

*Solution:*

*Here,*

*a.*

*Direct method*

Marks |
No. of std(f) |
Mid value(x) |
fx |
fx^{2} |

20-3030-4040-5050-6060-7070-8080-90 |
4610171193 |
25354555657585 |
100210450935715675255 |
250073502025051425464755062521675 |

N = 60 |
∑fx = 3340 |
∑fx^{2} = 200300 |

*b.*

*Deviation method*Marks |
f |
Mid value(x) |
d=(x-55)/10 |
fd |
fd^{2} |

20-3030-4040-5050-6060-7070-8080-90 |
4610171193 |
25354555657585 |
-3-2-10123 |
-12-12-10011189 |
3624100113627 |

N = 60 |
∑fd = 4 |
∑fd^{2} = 144 |

*Example 4: Find the variance and its coefficient from the following data.*

Age in yrs |
10 |
15 |
20 |
25 |
30 |
35 |

No. of people |
5 |
7 |
15 |
25 |
10 |
8 |

*Solution:*

*Here,*

Age (x) |
No. of people (f) |
fx |
fx^{2} |

101520253035 |
571525108 |
50105300625300280 |
500157560001562590009800 |

N = 70 |
∑fx = 1660 |
∑fx^{2} = 42500 |

*You can comment your questions or problems regarding*

**standard deviation**here.
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