This presentation covered the following topics : 1. Variance 2. Standard Deviation 3. Meaning and Types of Skewness 4. Related Examples and is useful for B.Sc & M.Sc students.

1. ByBy
Tushar Bhatt
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2. Variance
Standard Deviation
Meaning and Types of Skewness
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( For individual observations , n = total number of observations )( For individual observations , n = total number of observations )( For individual observations , n = total number of observations )( For individual observations , n = total number of observations )

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7. :
; , -
Solu Now the givendata is discrete
f di iX A A Assumed Mean d X Ai if
∴
∑
= + = =
∑
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i if
i∑

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11. 2
: H ere g iven o b servatio n s are co n tin u o u s
T o fin d : (i) S D (ii) M ean (iii) V arian ce (iv) C o efficien t o f variatio n
( )
1 0
7 0 (1 3 8 ) (4 2 )
7 0
0 .1 4 9 6 6 0 1 7 6 4
0 .1 4 7 8 9 6
0 .1 4 8 8 .8 6
1 2 .4 4
x
S o lu
i S D σ =
= −
= −
=
= ×
=
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1 2 .4 4
4 2
( ) 1 5 1 0 2 1
7 0
(
fd
ii M ea n A i
f
ii
=
= + × = + × =
∑
∑
( )
2 2
) (1 2 .4 4 ) 1 5 4 .7 5
1 2 .4 4
( ) C o efficien t o f v ariatio n 1 0 0 1 0 0 5 9 .2 4
2 1
xi V a ria n ce
S D
iv
M ea n
σ= = =
= × = × =

12. Ex – 4 : Find mean, Standard Deviation, Variance and coefficient of variation for
the following data :
(i) 16,26,36,46,56,66,76
(ii)
X 20 21 22 23 24
f 6 4 5 1 2
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(iii)
Class 0-4 4-8 8-12 12-16 16-20
Frequency 4 6 8 5 2

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14. It is clear from the sided diagram that
in a symmetrical skewed distribution
Mean = Median = Mode, the spread of
frequencies is the same on both sides
of the central point of the curve . Mean = Median = Mode
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If the frequency curve has a longer tail
to the right , i.e. The mean is to the
right of mode then the distribution is
said to be have positively skewed. Mode MeanMedian

15. If the curve is more elongated to the
left , then it is said to have negative
skewed distribution.
ModeMean Median
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16. Relative
Measure of
Absolute
Measure of
Skewness
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Measure of
Skewness

17. Absolute Measure of Skewness
- It is calculated by using the formula
- Where = Absolute measure of skewness
- But it is not useful when given data has two series with different
units.
kS M ea n M o d e= −
kS
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18. Relative Measure of Skewness
- It is calculated by using Karl pearson’s coefficient of skewness is
given by
- Where = Karl pearson’s coefficient of skewness.
3( )
O Rkp kp
M ea n M o d e M ea n M ed ia n
S S
S D S D
− −
= =
S
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- Where = Karl pearson’s coefficient of skewness.kpS

19. Note :
( ) 0
( ) 0
( ) 0
k kp
k kp
k kp
i If S S then Measure of skewness is symmetric
ii If S or S then Measure of skewness is positively skewed
iii If S or S then Measure of skewness is negatively skewed
= =
>
<
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20. Ex-1 : Calculate absolute measure of skewness for the data :
20, 25, 30, 35, 40,45,50,55,60,30 .
Solu : Here we want to find : Sk
Now here given data is individual
kS M ea n M o d e= −
20 25 30 35 40 45 50 55 60 30 390
43.33ix
Mean
+ + + + + + + + +
= = = =
∑
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20 25 30 35 40 45 50 55 60 30 390
43.33
9 9
30
43.33 30 13.33 0
i
k
Mean
n
Mode
S Mean Mode then given observations are positively skewed
+ + + + + + + + +
= = = =
=
= − = − = >
∑

21. Ex – 2 : Find absolute measure of skewness for the following data :
(i) 16,26,36,46,56,66,76,26
(ii) X 20 21 22 23 24
f 6 4 5 1 2
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(iii) Class 0-4 4-8 8-12 12-16 16-20
Frequency 4 6 8 5 2

22. Solu : (2)
(i) Here we want to find : Sk
Now here given data is individual
kS M ea n M o d e= −
16 26 36 46 56 66 76 26 348
43.5
8 8
ix
Mean
n
+ + + + + + +
= = = =
∑
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8 8
26
43.5 26 17.5 0k
n
Mode
S Mean Mode then given data is positively skewed
=
= − = − = >

23. Solu : (2)
(ii) Here we want to find : Sk
Now here given data is discrete
kS M ea n M o d e= −
Xi fi di = xi-A fidi
20 6 -2 -12
21 4 -1 -4
22 = A22 = A22 = A22 = A 5 0 0
23 1 1 1
24 2 2 4
Total 18 -11
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Total 18 -11

24. Solu :2(ii)
11
22 22 0.61 21.39
18
20 (that observation havingmaximum frequency)
i i
i
f d
Mean A
f
Mode
= + = − = − =
=
∑
∑
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21.39 20 1.39 0kS Mean Mode thengiven datais positively skewed= − = − = >

25. Solu :2(iii)
Here given data is continuous therefore
:
(1)
To find
f d
Mean A i
f
f f
= + ×
 −
∑
∑
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1
1 2
(2)
2
(3)
m
m
k
f f
Mode L i
f f f
S Mean Mode
 −
= + × 
− − 
= −

26. Solu :2(iii) (1)
f d
Mean A i
f
= + ×
∑
∑
Class Xi fi di = (xi-A) /i fidi
0-4 2 4 -2 -8
4-8 6 6 -1 -6
8888----12121212 10 = A10 = A10 = A10 = A 8 0 0
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8888----12121212 10 = A10 = A10 = A10 = A 8 0 0
12-16 14 5 1 5
16-20 18 2 2 4
Total 25252525 -5

27. Solu :2(iii)
5
10 4 10 0.8 9.2
25
f d
Mean A i
f
∴ = + × = − × = − =
∑
∑
1
1 2
(2)
2
m
m
f f
Mode L i
f f f
 −
= + × 
− − 
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1
2
1
1 2
8 1 2
8
8
6
5
8 6
8 4 8 1 .6 9 .6
2 2 (8) 6 5
m
m
m
M o d a l cla ss
L
f
f
f
f f
M o d e L i
f f f
= −
=
=
=
=
   − −
= + × = + × = + =   
− − − −  

28. Solu :2(iii)
(3) kS Mean Mode= −
9.2 9.6 0.4 0kS givendatais negatively skewed∴ = − = − <
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29. Ex-3 : Calculate relative measure of skewness for the data :
21, 22, 31, 35, 41,45,51,55,61,30 by using Karl pearson’s
coefficient of skewness and measure of median.
SoluSoluSoluSolu : Here we want to find : Skp
Now here given data is individual
Now, the Karl pearson’s coefficient of skewness is given by
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Now, the Karl pearson’s coefficient of skewness is given by
3( )
O Rkp kp
M ea n M o d e M ea n M ed ia n
S S
S D S D
− −
= =

30. SoluSoluSoluSolu :
Ascending Order : 21, 22, 30, 31, 35, 41,45, 51, 55, 61
Now here given data is individual
21 22 30 31 35 41 45 51 55 61 392
39.2
10 10
10( )
ix
M ean
n
G iven num ber of observations even
+ + + + + + + + +
= = = =
=
∑
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10( )
1
35 412 2
38
2 2
th th
G iven num ber of observations even
n n
observation observation
M edian
=
   
+ +    +   = = =

31. SoluSoluSoluSolu :
2
( )
1697.6
ix X
SD
n
SD
−
=
⇒ =
∑
XiXiXiXi XiXiXiXi----meanmeanmeanmean (Xi(Xi(Xi(Xi----mean)^2mean)^2mean)^2mean)^2
21 -18.2 331.24
22 -17.2 295.84
30 -9.2 84.64
31 -8.2 67.24
35 -4.2 17.64
41 1.8 3.24
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10
169.76
13.03
SD
SD
SD
⇒ =
⇒ =
⇒ =
41 1.8 3.24
45 5.8 33.64
51 11.8 139.24
55 15.8 249.64
61 21.8 475.24
Total 1697.6

32. SoluSoluSoluSolu :
3( ) 3(3 9 .2 3 8) 3 .6
0 .2 8
1 3 .0 3 1 3 .0 3
0
kp
kp
M ea n M ed ia n
N o w S
S D
H ere S th erefo re g iven d a ta is p o sitively skew ed
− −
= = = =
>
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33. Ex-4 : Calculate relative measure of skewness for the data :
by using Karl pearson’s coefficient of skewness and measure
of mode.
SoluSoluSoluSolu : Here we want to find : Skp
Xi 10 20 30 40 50
fi 1.1 6.1 3.1 4.1 5.1
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SoluSoluSoluSolu : Here we want to find : Skp
Now here given data is discrete
Now, the Karl pearson’s coefficient of skewness is given by
3( )
O Rkp kp
M ea n M o d e M ea n M ed ia n
S S
S D S D
− −
= =

34. Solu :4
(1)
60
30
i i
i
f d
Mean A
f
Mean
= +
⇒ = +
∑
∑
Xi fi di = (xi-A) fidi
10 1.1 -20 -22
20 6.1 -10 -61
30= A30= A30= A30= A 3.1 0 0
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30
19.5
30 3.08
33.08
Mean
Mean
Mean
⇒ = +
⇒ = +
⇒ =
30= A30= A30= A30= A 3.1 0 0
40 4.1 10 41
50 5.1 20 102
Total 19.5 60

35. Solu :4
(2) Mode = An observationwith maximum frequency =20
Xi fi
10 1.1
20 6.1
30= A 3.1
40 4.1
50 5.1
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50 5.1
Total 19.5

36. SoluSoluSoluSolu :
XiXiXiXi fifififi XiXiXiXi----meanmeanmeanmean (Xi(Xi(Xi(Xi----mean)^2mean)^2mean)^2mean)^2 fifififi(Xi(Xi(Xi(Xi----mean)^2mean)^2mean)^2mean)^2
10 1.1 -23.08 532.69 585.96
20 6.1 -13.08 171.09 1043.63
30 3.1 -3.08 9.49 29.41
40 4.1 6.92 47.89 196.33
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40 4.1 6.92 47.89 196.33
50 5.1 16.92 286.27 1460.06
Total 19.5 3315.38

37. SoluSoluSoluSolu :
2
f ( )
3315.38
19.5
i ix X
SD
N
SD
−
=
⇒ =
∑
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19.5
170.019
13.04
SD
SD
⇒ =
⇒ =

38. SoluSoluSoluSolu :
33.08 20 13.08
1.003
13.04 13.04
kp
kp
Mean Mode
Now weknowthat S
SD
S
−
=
−
⇒ = = =
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13.04 13.04
1.003 0 ,kpNow S givendata is positively skewed= >

39. Ex-5 : Calculate relative measure of skewness for the data :
by using Karl pearson’s coefficient of skewness and measure
of median.
SoluSoluSoluSolu : Here we want to find : Skp
Class 10-20 20-30 30-40 40-50 50-60
fi 2 3 5 1 4
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SoluSoluSoluSolu : Here we want to find : Skp
Now here given data is Continuous
Now, the Karl pearson’s coefficient of skewness is given by
3( )
O Rkp kp
M ea n M o d e M ea n M ed ia n
S S
S D S D
− −
= =

40. (1)
f d
Mean A i
f
= + × =
∑
∑
2
(2)
n
c
Median L i
f
  
−  
  = + × =
 
  
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 

41. 2
(1) 35 10 36.33
15
f d
Mean A i
f
= + × = + × =
∑
∑
ClassClassClassClass fifififi c.f.c.f.c.f.c.f. xixixixi didididi ==== yiyiyiyi fidifidifidifidi====fiyifiyifiyifiyi yi^2yi^2yi^2yi^2 fifififi (yi^2)(yi^2)(yi^2)(yi^2)
10-20 2 2 15 -2 -4 4 8
20-30 3 5 25 -1 -3 1 3
30-40 5 10 35=A35=A35=A35=A 0 0 0 0
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30-40 5 10 35=A35=A35=A35=A 0 0 0 0
40-50 1 11 45 1 1 1 1
50-60 4 15 55 2 8 4 16
Total 15 2 28

42. 7.5 52
(2) 30 10 35
5
,
n
c
Median L i
f
Where
  
−   −   = + × = + × =   
  
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,
15
7.5 30 40
2 2
30 ; 5 ; 5 ; 10
Where
n
lies in the class
L c f i
 
= = − 
 
= = = =

43. 210
15(28) (2) 13.59
15
− =
3( ) 3(3 6 .3 3 3 5)
(4 ) 0 .2 9
1 3 .5 9
kp
M ea n M ed ia n
S
S D
− −
= = =
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(4 ) 0 .2 9
1 3 .5 9
, 0 .2 9 0 ,
kp
kp
S
S D
N o w S g iven d a ta is p o sitively skew ed
= = =
= >

44. Thank youThank you
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