Measures of Dispersion

Definition: If we carefully observed a series of data, we find that the observation very among themselves and
also form the average. While studying a dispersion it is equally important to know how the variables are
clustered around or scattered a way from the point of central tendency. If two distribution centers round the
same point viz the mean, yet differ in variously mean such variation is called dispersion or spread or scatter or
variability.
Dispersion is absolute measured:(i)Range (ii) Quartile deviation (iii)Mean deviation (iv)Standard deviation.

Range: The range is the difference between two extreme observations of the distribution. If A and B are the
greatest and smallest observations respectively in a distribution then its range is A-B.
1
Quartile deviation: Quartile deviation or semi-inter quartile range Q is given by Q  (Q3  Q1 ) .
2
Where Q1 and Q3 are the first and third quartiles of the distribution respectively.

Mean deviation: For the frequency distribution xi | f i , i  1, 2, 3,, n , the mean deviation from the
n
1
average A (usually mean, median or mode) is given by Mean deviation=
N
 f i xi  A , where  f i  N .
i 1

n
1
Thus, Mean deviation from mean=
N
 f i xi  x
i 1

1 n
Mean deviation from median=  f i xi  median
N i1

1 n
Mean deviation from mode=  f i xi  mod e
N i 1
Standard deviation: Standard deviation, usually denoted by the Greek letter  (sigma), is the square root of
the arithmetic mean of the squares of the deviations of the given values from their arithmetic mean. For the

1 n
frequency distribution xi | f i , i  1, 2, 3,, n  , standard deviation,    f i ( xi  x ) 2 .
N i 1

where x is the arithmetic mean of the distribution and  fi  N.

1 n
Variance: The square of standard deviation is called the variance and is given by  2   ( xi  x ) 2 .
n i1
n
For the frequency distribution xi | f i , i  1, 2, 3,, n  , variance,  2  1  f i ( xi  x ) 2 .
N i 1

where x is the arithmetic mean of the distribution and  fi  N.

1
Root mean square deviation: Let, xi , i  1, 2, 3,, n  be n observations and A be any arbitrary number,
1 n
Root mean square deviation is denoted by ‘S’ is given by S   ( xi  A) 2 .
n i 1

1 n
For the frequency distribution xi | fi , i  1, 2,3,, n , S  f i ( xi  A) 2 . where  fi  N .
N i 1

Mean square deviation: The square of the root mean square deviation is called the mean square deviation
1 n
and denoted by ‘S2’ is given by S 2  
n i 1
( xi  A) 2 .

n
For the frequency distribution xi | fi , i  1, 2,3,, n ,variance, S 2  1  fi (xi  A)2 , where  f i  N.
N i1

Coefficient of dispersion: The coefficient of dispersion (C.D) based on different measures of dispersion are
A B
as follows: (i) Based upon range, C.D  , A and B are the greatest and the smallest terms in the series.
A B
Meandeviation
(ii) Based upon mean deviation: C.D  ,
A varage from which it is calculate

Q3  Q1
(iii) Based upon quartile deviation: C.D  ,
Q3  Q1

(iv) Based upon standard deviation: C.D  .
x
Coefficient of variation: 100 times the coefficient of dispersion based upon standard deviation is called

coefficient of variation, i.e, C.V  100 .
x
Question-1: Which measures is the best measures of dispersion?
Answer: Among the four absolute measures of dispersion, standard deviation is the best measures of dispersion,
because it posses almost all the requisites of a good measure of dispersion.
(i) It is rigidly defined.
(ii) It is based on all the observations even if one of the observations is changed standard deviation changed.
However, range, Quartile deviation does not possess this property.
(iii) It is easy to understand and easy to calculate.
(iv) It is amenable to algebraically treatment.
(v) It is affected least by fluctuation of sampling.

Question-2: Determine the relation between standard deviation and root mean square deviation i.e,  and S .
1 n
Solution: By definition we have S2  
N i1
fi (xi  A)2 ,

2
 1 n
 
N i1
fi (xi  x  x  A)2 ,


1 n

N i1
 
fi (xi  x)2  (x  A)2  2(xi  x)(x  A) ,

1 n 2 1
n
1 n
 i i
N i1
f ( x  x ) 2
 ( x  A)  i N  fi (xi  x)(x  A),
N i1
f  2
i1

1 n
  fi (xi  x)2  (x  A)2  0, [Since sum of the deviation of A.M=0]
N i1

or, S 2   2  (x  A)2

or, S 2   2  d 2 which is the required relation. where d  x  A .

Obviously S2 will be least when d=0 i.e, x  A .
Hence mean square deviation and consequently root mean square deviation is least when the
deviations are taken from A  x i.e, standard deviation is the least value of root mean square deviation.
n2  1
Question-3: Prove that the variance of the first n natural number is .
12

n2 1
OR, Prove that the standard deviation of the first n natural number is .
12
Proof: Let x1 , x2 , x3 , xn be n observations with corresponding the first n natural number 1, 2, 3, - - - -, n.

1 n
We know that, Arithmetic mean, x   xi
n i 1

1
 ( x1  x2  x3    xn )
n
1
 (1  2  3    n)
n
1 n(n  1)

n 2
n 1

2

Again,
1 n 2 1 2

 xi  n x1  x22  x32    xn2
n i1

1 2
 (1  2 2  32    n 2 )
n
1 n(n  1)(2n  1)

n 6
(n  1)(2n  1)

6

3
 1 n
We know, V ( x)   xi  x 2
n i 1

1 n 2
 
n i 1
xi  x 2

2
(n  1)(2n  1)  n  1 
  
6  2 
 n  1  2n  1 n  1
   
 2  3 2 

 n  1   4n  2  3n  3 
  
 2  6 
1
 (n  1)(n  1)
12
n2 1
V ( x) 
12

1 n
OR, We know,    xi  x 2
n i 1

1 n 2
 
n i 1
xi  x 2

2
(n  1)(2n  1)  n  1 
  
6  2 

 n  1  2n  1 n  1
    
 2  3 2 

 n  1  4n  2  3n  3 
   
 2  6 

1
 (n  1)(n  1)
12

n2 1
  (Proved).
12

H.W. Question-4: Prove that S 

 x 2    x  2 2
N  N   x x .
 

4
Problem-5: Calculate the mean and standard deviation for the following data:
Size of item 6 7 8 9 10 11 12
Frequency 3 6 9 13 8 5 4

Solution: Compute the following table:

Size of item (x) Frequency (f) d=x-a fd f d2
6 3 -3 -9 27
7 6 -2 -12 24
8 9 -1 -9 9
9 13 0 0 0
10 8 1 8 8
11 5 2 10 20
12 4 3 12 36
 f  N  48.  fd  0.  fd 2  124
Assumed mean=9,

Mean= x  a 
 fd  9  0  9 Ans.
N

Standard deviation= S .D 
 f (x  x)2 .
f
2


 fd 2    fd 
 f   f 
124
  1.6 Ans.
48
Problem-6: Calculate the mean and standard deviation for the following data:
Age(in year) 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of people 3 61 132 153 140 51 2

Solution: From the given data compute the following table:

Age No. of people (f) x d=(x-a)/h d2 fd f d2
20-30 3 25 -3 9 -9 27
30-40 61 35 -2 4 -122 244
40-50 132 45 -1 1 -132 132
50-60 153 55 0 0 0 0
60-70 140 65 1 1 140 140
70-80 51 75 2 4 102 204
80-90 2 85 3 9 6 18

f  N  542. fd 15  fd 2  765

5
 Assumed mean=55,

Mean= x  a 
 fd  h  55    15  10  54.7232 Ans.
N  542 
2

Standard deviation= S .D 
 fd 2    fd 
 f   f 
2
765   15 
    1.188 Ans.
542  542 
H.W.Problem-7: Find the mean deviation and standard deviation for the following data:
(i) 12, 6, 7, 3, 15, 10, 18, 5 Ans. M.D=4.25, S.D=4.87
(ii) 9, 3, 8, 8, 9, 8, 9, 18. Ans. M.D=2.25, S.D=3.87
H.W.Problem-8: Find the mean deviation and standard deviation of the heights of the 100 students of
RUET:
Hight (inches) 60-62 63-65 66-68 69-71 72-74
No. of students 5 18 42 27 8
Ans: M.D=2.26 inches, S.D= 2.92 inches.

H.W.Problem-9: Find the mean and standard deviation for the wage distribution of the 65 employees:
Wages $250.00-$259.99 260.00-269.99 270.00-279.99 280.00-289.99 290.00-299.99 300.00-309.99310.00-319.99
No. of Employees 8 10 16 14 10 5 2
Ans: A. M=$ 279.77, S.D= $ 15.60

H.W.Problem-10: Compute the standard deviation for the following frequency distribution:
Class interval 0-4 4-8 8-12 12-16
Frequency 4 8 2 1
Ans: S.D= 3.266 .

H.W.Problem-11: The following table shows the marks obtained by 100 candidates in an examination.
Compute the mean, media and standard deviation:
Marks 1-10 11-20 21-30 31-40 41-50 51-60
No. of candidates 3 16 26 31 16 8
Ans: Mean=32, Median=32.61, S.D= 12.36 .