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EMBS


Unit 4


Measures of Dispersion

Question and Answer

Q1) explain moments

Solution

Moments

Moments are a set of statistical parameters to measure a distribution. Four moments are commonly used:

•         1st moment - Mean (describes central value)

•         2nd moment - Variance (describes dispersion)

•         3rd moment - Skewness (describes asymmetry)

•         4th moment - Kurtosis (describes peakedness)

The formula for calculating moments is as follows:

1st moment =

 

2nd moment =

 

3rd moment =

 

4th moment =

 

Raw Moments and Central Moments

The n-th moment about zero of a probability density function f(x) is the expected value of Xn and is called a raw moment or crude moment. The moments about its mean μ are called central moments; these describe the shape of the function, independently of translation.

A moment mu_n of a probability function P(x) taken about 0,

mu_n^'

=

<x^n>

(1)

https://mathworld.wolfram.com/images/equations/RawMoment/Inline6.gif

=

intx^nP(x)dx.

(2)

The raw moments mu_n^' (sometimes also called "crude moments") can be expressed as terms of the central moments mu_n (i.e., those taken about the mean mu) using the inverse binomial transform

mu_n^'=sum_(k=0)^n(n; k)mu_kmu_1^('n-k),

(3)

With mu_0=1 and mu_1=0 (Papoulis 1984, p. 146). The first few values are therefore

mu_2^'

=

mu_2+mu_1^('2)

(4)

mu_3^'

=

mu_3+3mu_2mu_1^'+mu_1^('3)

(5)

mu_4^'

=

mu_4+4mu_3mu_1^'+6mu_2mu_1^('2)+mu_1^('4)

(6)

mu_5^'

=

mu_5+5mu_4mu_1^'+10mu_3mu_1^('2)+10mu_2mu_1^('3)+mu_1^('5).

(7)

The raw moments mu_n^' can also be expressed in terms of the cumulates kappa_n by exponentiating both sides of the series

lnphi=ln(sum_(k=0)^infty((it)^k)/(k!)mu_k^')=sum_(n=0)^inftykappa_n((it)^n)/(n!),

(8)

Where phi is the characteristic function, to obtain

sum_(k=0)^infty((it)^k)/(k!)mu_k^'=exp(sum_(n=0)^inftykappa_n((it)^n)/(n!)).

(9)

The first few terms are then given by

mu_1^'

=

kappa_1

(10)

mu_2^'

=

kappa_1^2+kappa_2

(11)

mu_3^'

=

kappa_1^3+3kappa_1kappa_2+kappa_3

(12)

mu_4^'

=

kappa_1^4+6kappa_1^2kappa_2+3kappa_2^2+4kappa_1kappa_3+kappa_4

(13)

mu_5^'

=

kappa_1^5+10kappa_1^3kappa_2+15kappa_1kappa_2^2+10kappa_1^2kappa_3+10kappa_2kappa_3+5kappa_1kappa_4+kappa_5.

(14)

The raw moment of a multivariate probability function P(x_1,x_2,...) can be similarly defined as

mu_(m,n,...)^'=<x_1^mx_2^n...>.

(15)

Therefore,

mu_(n,0,...,0)^'=mu_n^'.

(16)

The multivariate raw moments can be expressed in terms of the multivariate cumulants. For example,

mu_(1,1)^'

=

kappa_(0,1)kappa_(1,0)+kappa_(1,1)

(17)

mu_(2,1)^'

=

kappa_(0,1)kappa_(1,0)^2+2kappa_(1,0)kappa_(1,1)+kappa_(0,1)kappa_(2,0)+kappa_(2,1).

(18)

 

In probability theory and statistics, a central moment is a moment of a probability distribution of a random variable about the random variable's mean; that is, it is the expected value of a specified integer power of the deviation of the random variable from the mean. The various moments form one set of values by which the properties of a probability distribution can be usefully characterized. Central moments are used in preference to ordinary moments, computed in terms of deviations from the mean instead of from zero, because the higher-order central moments relate only to the spread and shape of the distribution, rather than also to its location.

Sets of central moments can be defined for both univariate and multivariate distributions.

 

Q2) Calculate Bowley’s Coefficient of Skewness from the following test scores:

Sl. N o

Test scores

1

17

2

17

3

26

4

27

5

30

6

30

7

31

8

37

 

Solution:

 

First quartile (Q1)

Qi= [i * (n + 1) /4] th observation

Q1= [1 * (8 + 1) /4] th observation

Q1 = 2.25 th observation

Thus, 2.25 th observation lies between the 2nd and 3rd value in the ordered group, between frequency 17 and 26

 

First quartile (Q1) is calculated as

Q1 = 2nd observation +0.75 * (3rd observation - 2nd observation)

Q1 = 17 + 0.75 * (26 – 17) = 23.75

 

Second quartile()

Q2= [2 * (8 + 1) /4] th observation

Q2 =   4.5th Observation

So, 4.5th observation lies between 4th and 5th value in ordered group, between frequency 27 and 30.

Hence Q2 = 4th observation + 0.50 * (5th observation – 6th observation)

Q2 = 27 + 0.50 * (30 – 27) = 28.5

 

Third quartile (Q3)

Qi= [i * (n + 1) /4] th observation

Q3= [3 * (8 + 1) /4] th observation

Q3 = 6.75 th observation

So, 6.75 th observation lies between the 6th and 7th value in the ordered group, between frequency 30 and 31

 

Third quartile (Q3) is calculated as

Q3 = 6th observation +0.25 * (7th observation – 6th observation)

Q3 = 30 + 0.25 * (31 – 30) = 30.25

 

Therefore, Bowley’s Coefficient of Skewness is calculated as under:

 

 


=

= = - 0.461

 

Q3) Below are the data of hours spent watching television by the 220 students. Calculate Karl Pearsons Co-efficient of Skewness.

 

Hours

No. Of students

10 – 14

2

15 – 19

12

20 – 24

23

25 – 29

60

30 – 34

77

35 – 39

38

40 - 44

8

 

Solution:

 

Hours

No. Of students

x

Fx

X – x

(x – x ) 2

F (x – x ) 2

10 – 14

2

12

24

-17.82

317.49

634.98

15 – 19

12

17

204

-12.82

164.31

1971.67

20 – 24

23

22

506

-7.82

61.12

1405.85

25 – 29

60

27

1620

-2.82

7.94

476.53

30 – 34

77

32

2464

2.18

4.76

366.55

35 – 39

38

37

1406

7.18

51.58

1959.98

40 - 44

8

42

336

12.18

148.40

1187.17

 

220

 

6560

 

 

8002.73

 

Mean = 6560/220 = 29.82

SD = √8002.73/220 = 6.03

 

Mode = L1 + (L2 – L1)   d1

                          d1 +d2

Here modal class is 30 – 34 (Since the frequency is highest)

L1 = 30, L2 = 34, d1 = 17, d2 = 39

 

Mode = 30 + (34 – 30)   17

                          17 + 39

Mode = 30 + x 17

          = 30 + 1.21

          = 31.21

Therefore, Co-efficient of Skewness

 

=

       =

          = - 0.23

 

Q4) the frequency table shows the number of goals the lakers scored in their last twenty matches. What was the range

No. Of goals

Frequency

0

2

1

3

2

3

3

6

4

3

5

1

6

1

7

1

 

Solution

The range is the difference between the lowest and highest values.

The highest value was 7 (They scored 7 goals on 1 occasion)

The lowest value was 0 (They scored 0 goals on 2 occasions)

Therefore the range = 7 - 0 = 7

 

Q5) calculate quartile deviation from the following test scores

Sl. N o

Test scores

1

17

2

17

3

26

4

27

5

30

6

30

7

31

8

37

 

Solution

First quartile (Q1)

Qi= [i * (n + 1) /4] th observation

Q1= [1 * (8 + 1) /4] th observation

Q1 = 2.25 th observation

Thus, 2.25 th observation lies between the 2nd and 3rd value in the ordered group, between frequency 17 and 26

First quartile (Q1) is calculated as

Q1 = 2nd observation +0.75 * (3rd observation - 2nd observation)

Q1 = 17 + 0.75 * (26 – 17) = 23.75

Third quartile (Q3)

Qi= [i * (n + 1) /4] th observation

Q3= [3 * (8 + 1) /4] th observation

Q3 = 6.75 th observation

So, 6.75 th observation lies between the 6th and 7th value in the ordered group, between frequency 30 and 31

Third quartile (Q3) is calculated as

Q3 = 6th observation +0.25 * (7th observation – 6th observation)

Q3 = 30 + 0.25 * (31 – 30) = 30.25

Now using the quartiles values Q1 and Q3, we will calculate the quartile deviation.

QD = (Q3 - Q1) / 2

QD = (30.25 – 23.75) / 2 = 3.25

 

Q6) computation of quartile deviation for grouped test scores

Class

Frequency

9.3-9.7

22

9.8-10.2

55

10.3-10.7

12

10.8-11.2

17

11.3-11.7

14

11.8-12.2

66

12.3-12.7

33

12.8-13.2

11

 

Solution

Class

Frequency

Class boundaries

CF

9.3-9.7

2

9.25-9.75

2

9.8-10.2

5

9.75-10.25

2 + 5 = 7

10.3-10.7

12

10.25-10.75

7 + 12 = 19

10.8-11.2

17

10.75-11.25

19 + 17 = 36

11.3-11.7

14

11.25-11.75

36 + 14 = 50

11.8-12.2

6

11.75-12.25

50 + 6 = 56

12.3-12.7

3

12.25-12.75

56 + 3 = 59

12.8-13.2

1

12.75-13.25

59 + 1 = 60

 

First quartile (Q1)

Qi= [i * (n ) /4] th observation

Q1 = [1*(60)/4]th observation

Q1 = 15th observation

So, 15th value is in the interval 10.25-10.75

Group of Q1 = 10.25-10.75

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q1 = (10.25 + ( 0.5/ 12)* (1* (60/4) – 7)

Q1 = 10.58

 

Third quartile (Q3)

Qi= [i * (n) /4] th observation

Q3= [3 * (60) /4] th observation

Q3 = 45th observation

So, 45th value is in the interval 11.25-11.75

Group of Q3 = 11.25-11.75

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q3 = (11.25 + ( 0.5/ 14)* (3* (60/4) – 36)

Q3 = 11.57

QD = (Q3 - Q1) / 2

QD = (11.57 – 10.58) / 2 =  0.495

 

Q7) Computation of Mean deviation in grouped data

Class interval

15 – 19

20 – 24

25 – 29

30 – 34

35 – 39

40 – 44

45 - 49

Frequency

1

4

6

9

5

3

2

Solution

 

Class Interval

F

X

FX

D

FD

15 – 19

1

17

17

15

15

20 – 24

4

22

88

10

40

25 – 29

6

27

162

5

30

30 - 34

9

32

288

0

0

35 - 39

5

37

185

5

25

40 - 44

3

42

126

10

30

45 - 49

2

47

94

15

30

 

N = 30

 

∑fx = 960

 

= 170

 

Mean = 960/30 = 32

MD = 170 / 30 = 5.667

Coefficient of mean deviation

Coefficient of mean deviation = (5.67/32)*100 = 17.71

 

Q8) calculate mean deviation from the median

Class

5  -15

15 - 25

25 - 35

35 - 45

45 – 55

Frequency

5

9

7

3

8

 

Solution

x

f

Cf

Midpoint x

x –median

F(x-m)

5 -15

5

5

10

17.42

87.1

15 -25

9

14

20

7.42

66.78

25 -35

7

21

30

2.58

18.06

35 -45

3

24

40

12.58

37.74

45- 55

8

32

50

22.58

180.64

 

32

 

 

 

390.32

 

Since n/2 = 32/2 = 16, therefore the class is 25 – 35 is the median.

Median =

 

Median = 25+16-14 *10 = 27.42

   7

MD from median is  390.32/32 = 12.91

 

Q9) calculate the standard deviation using the direct method

Class interval

Frequency

30 – 39

3

40 – 49

1

50 – 59

8

60 – 69

10

70 – 79

7

80 – 89

7

90 – 99

4

 

Solution

Class interval

Frequency

Midpoint x

Fx

X – x

(x – x ) 2

F (x – x ) 2

30 – 39

3

34.5

103.5

-33.5

1122.25

3366.75

40 – 49

1

44.5

44.5

-23.5

552.25

552.25

50 – 59

8

54.5

436.0

-13.5

182.25

1458

60 – 69

10

64.5

645.0

-3.5

12.25

122.5

70 – 79

7

74.5

521.5

6.5

42.25

295.75

80 – 89

7

84.5

591.5

16.5

272.25

1905.75

90 – 99

4

94.5

378.0

26.5

702.25

2809

 

40

 

2720

 

 

10510

 

Mean = 2720/40 = 68

SD = √10510/40 = 16.20

.

 

Q10) calculate the mean and standard deviation of hours spent watching television by the 220 students.

Hours

No. Of students

10 – 14

2

15 – 19

12

20 – 24

23

25 – 29

60

30 – 34

77

35 – 39

38

40 - 44

8

 

Solution

Hours

No. Of students

x

Fx

X – x

(x – x ) 2

F (x – x ) 2

10 – 14

2

12

24

-17.82

317.49

634.98

15 – 19

12

17

204

-12.82

164.31

1971.67

20 – 24

23

22

506

-7.82

61.12

1405.85

25 – 29

60

27

1620

-2.82

7.94

476.53

30 – 34

77

32

2464

2.18

4.76

366.55

35 – 39

38

37

1406

7.18

51.58

1959.98

40 - 44

8

42

336

12.18

148.40

1187.17

 

220

 

6560

 

 

8002.73

 

Mean = 6560/220 = 29.82

SD = √8002.73/220 = 6.03

 


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