.Moment.

Question-1: Define moment (central and raw). Prove that  0  1, 1  0,  2   2 (variance) and

1  x = mean.
Solution: Let, x1 , x2 , x3 ,  , xn be the observations with corresponding frequencies f1 , f 2 , f 3 ,, f n then
n n
1
r-th moment about the mean x is denoted by  r is defined as  r 
N
 f i ( xi  x ) r ; N   f i ; The r-th
i 1 i 1

moment about mean x is known as r-th central moment.

If x1 , x2 , x3 ,  , xn be the observations with corresponding frequencies f1 , f 2 , f 3 ,, f n then r-th

1 n
moment about any point A , usually denoted by  r is defined as  r  
N i 1
f i ( xi  A) r ; The r-th moment

about A is known as r-th raw moment.

n
1
Proof: We know that  r 
N
 f i ( xi  x ) r ;
i 1

1 n 1 n
In particular if r  0 then  0  
N i 1
f i ( x i  x ) 0   f i  1   0  1.
N i 1

1 n 1 n 1 n
if r  1 then 1   i i
N i 1
f ( x  x )   i i N  f i x  x  x  0  1  0.
N i 1
f x 
i 1

n
1
if r  2 then  2 
N
 f i ( xi  x ) 2   2 ,   2   2 (variance).
i 1

1 n
We also know,  r   f i ( xi  A) r ;
N i 1

1 n 1 n 
If A  0 then  r   ii
N i 1
f x r
; when r  1 then 1   f i xi  x ,
N i 1
 1  x = mean. Proved.

Question-2: Define skewness with different measures.

Solution: Literally, skewness means “Lack of symmetry.” Skewness is the degree of a symmetry or departure
from symmetry of a distribution. we study skewness to have an idea about the shape of the curve which we
can draw with the help of the given data.
A distribution is said to be skewed if
(i) Mean, median and mode falls at the different points, i,e. Mean  Median  Mode.
(ii) Quartiles are not equidistant from median.
(iii) The curve drawn with the help of the given data is not symmetrical but stretched more to one side then to
the other.
Various measures of skewness are: (i) S k  M  M d (ii) S k  M  M 0 where M is the mean, Md is the
median and M0 is the mode of the distribution.

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(iii) S k  Q3  M d   M d  Q1 , where Q1 , Q3 are first and third quartiles.
These are the absolute measures of skewness.
Question-3: Describe the co-efficient of skewness.
Solution: For comparing two series, we calculate the relative measures called the coefficients of skewness
which are pure numbers independent of units of measurement. The following are the co-efficient of skewness:

32
(i) Co-efficient of skewness, 1  for symmetrical distribution, 1  0 .
 23
Mean  Mode M  M0
(ii) Prof. Karl Pearson’s Co-efficient of skewness S k  
STANDARD deviation 
3Mean  Median  3M  M d 
If mode is ill defined then S k  
STANDARD deviation 
If mean > median > mode or the tail on graph is longer on the side of higher values the data have positive
skewness.
If mean < median < mode or the tail on graph is longer on the side of smaller values, the data have negative
skewness.
If mean = median = mode, the skewness is zero.

Question-4: Define kurtosis.

Solution: Kurtosis is the degree of peakedness of a distribution. It gives us the idea about the flatness or
peakedness of the curve. This is measured by the co-efficient of kurtosis,  2 or its derivation  2 given as
4
2  ,  2  2  3 .
 22
The curve which is neither flat nor peaked is called normal curve or Mesokurtic and for such a curve  2  3
or  2  0 .
The curve which is flatter than the normal curve is known as Platykurtic and for such a curve  2  3 or,

2  0.

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The curve which is more peaked than the normal curve is known as Leptokurtic and for such a curve  2  3
or,  2  0 .

Question-5: Establish the general relationship between central moments and raw moments. Hence find the
relation between raw and central moments for particular four moments.
Solution: Let, x1 , x2 , x3 ,  , xn be the observations with corresponding frequencies f1 , f 2 , f 3 ,, f n then
n n
1
r-th moment (central) is defined as  r 
N
 f i ( xi  x ) r ; N   f i ;
i 1 i 1

1 n
  f i ( xi  A)  ( x  A)r ;
N i 1
n
1

N
 f i ( xi  A) r   1r ( xi  A) r 1 ( x  A)   2r ( xi  A) r 2 ( x  A) 2    (1) r ( x  A) r 
i 1

1 n 1 n 1 n 1 n
  f i ( xi  A) r   r ( x  A)  f i ( xi  A) r 1   r ( x  A) 2  f i ( xi  A) r 2    (1) r ( x  A) r  f i
N i1 1 N i1  2 N i1 N i1

  r   r 1 r1   r 1 2 r2   r 1 3 r 3    (1) r 1 r

1  2  3
r (r  1) 2 r (r  1)(r  2) 3
 r  r1 r1  1  r2  1 r3    (1) r 1 r ---------------- (i)
2! 3!
Which is the general relation between central and raw moments.
In particular, on putting r  1, 2, 3, 4, in (i) we get,

1  1  1  0

 2   2  2 11  1 2   2  1 2

3   3  31 2  31 2 1  1 3

  3  3 2 1  2 1 3

 4   4  4 13  6 1 2  2  4 1 3 1  1 4

  4  4  3 1  6  2 1 2  31 4
Which are the relation between raw and central moments for particular four moments.

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Question-6: The mean, median and the coefficient of variation of 100 observations are found to be 90, 84 and
80 respectively. Find the coefficient of skewness of the above system of 100 observations.
3Mean  Median  3M  M d 
Solution: We know, the skewness, S k   ------------(i)
STANDARD deviation 
C.V  mean 80  90
Here Standard deviation,     72
observations 100
390  84 
 Sk   0.25 Ans.
72
Question-7: Calculate the skewness and kurtosis of the following data
x 1 2 3 4 5 6 7 8 9
y=f(x) 1 6 13 25 30 22 9 5 2

Solution: From the given data we construct the following table:

x u f fu fu 2 fu 3 fu 4
1 -4 1 -4 16 -64 256
2 -3 6 -18 54 -162 486
3 -2 13 -26 52 -104 208
4 -1 25 -25 25 -25 25
5 0 30 0 0 0 0
6 1 22 22 22 22 22
7 2 9 18 36 72 144
8 3 5 15 45 135 405
9 4 2 8 32 128 512
N   f  113  fu  10  fu 2  282  fu 3  2  fu 4  2058

Now, 1  c 
 fu  1  10  0.088496
N 113

 2  c 2 
 fu 2  12  282  2.49558
N 113

3  c 3 
 fu 3  13  2
 0.017699
N 113

 4  c 4 
 fu 4  14  2058  18.2124
N 113

1  0

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 2   2  1 2  2.49558   0.088496 2  2.48775

 3   3  3 2 1  2 1 3
 0.017699  3   0.088496   2.49558  2   0.088496   0.6789
3

 4   4  4 3 1  6  2 1 2  31 4

 18.2124  4  0.017699   0.088496   6  2.49558   0.088496   3 0.088496  18.3357
2 4

 32 0.6789 2
Therefore the Skewness, 1    0.02994 Ans.
 23 2.487753
4 18.3357
and Kurtosis,  2    2.96268 Ans.
 2 2.487752
2