Central

Central Tendency

The word “average” denotes a “representative” or “typical value” of a whole set of observations. It is a single figure which describes the whole set of observations. Since a typical value usually occupies central position, so that some observations are lager and others are smaller than it, average are also known as measure of central tendency. There are 3 measures of central tendency-Mean, Median and Mode. Again Mean is of 3 types – Arithmetic Mean (A.M.),Geometric Mean(G.M.), Harmonic Mean(H.M.)

Arithmetic Mean (A.M)

The most popular and widely used measure for representing the entire data by one value is what most laymen call an “average” and what the statisticians call the arithmetic mean. Its value is obtained by adding together all the observations and by dividing this total by the number of observations.

Types of A.M There are two types of A.M 1)Simple A.M. 2)Weighted A.M.

1)Simple A.M. Formula:-

∑ xi x=

n

Where i = 1,2, ………,n.

ILLUSTRATION 1:Q) What is A.M. of 8,1,6? Solution:- Here n = 3 and i = 1,2,3. x = (8+1+6)/3 = 15/3 = 5

ILLUSTRATION 2:Q) Monthly income (in rupees) of 10 employees working in a firm is as follows:4487,4493,4502,4446,4475,4492,4572,4516, 4468 and 4489. Find the average monthly income. Solution:- Let income be denoted by X. ∑X = (4487+4493+4502+4446+4475+4492+4572+4516+ 4468 + 4489) = 44940 X = ∑X /n = 44940/10 = 4494

2) Weighted A.M. Formula:-

∑ fi xi X

=

N Where N = ∑ fi and i = 1,2, ………,n.

ILLUSTRATION 3:- Weighted A.M. in case of simple frequency distribution Q) A.M. of 8,1,6 with frequencies (or weight) 3,2,5 respectively is

Value

( xi )

Frequency ( fi )

fi xi

8

3

24

1

2

2

6

5

30

Total

∑ fi = 10

∑ fi xi = 56

∑ fi xi X

=

∑ fi = 56/10

=

Ans)TheWeighted

5.6 (answer ) A.M. is 5.6

ILLUSTRATION 3:- Weighted A.M. in case of grouped frequency distribution The following are the figures of profits earned by 1400 companies during 2003-04. Profits (Rs. Lakhs)

No of companies

200 ------------- 400

500

400 ------------- 600

300

600 ------------- 800

280

800 ------------- 1000

120

1000 ------------- 1200

100

1200 ------------- 1400

80

1400 ------------- 1600

20

Total

1400

Profits (Rs. Lakhs)

Mid- Points x

No of companies f

fx

200 ------- 400

300

500

1,50,000

400 ------- 600

500

300

1,50,000

600 ------ 800

700

280

1,96,000

800 ------ 1000

900

120

1,08,000

1000 ------ 1200

1100

100

1,10,000

1200 ----- 1400

1300

80

1,04,000

1400 ----- 1600

1500

20

30,000

Total

X

N = 1400

= ∑ fi xi /N = 8,48,000/1400= 605.71

∑ fi xi= 8,48,000

Simplified Calculation for A.M. •

The calculations of A.M can be considerably simplified based on the following theory:We can always reduce the given observations xi by subtracting a convenient number c , preferably near the middle and obtain deviations yi = xi – c.The mean of these deviation yi, which can be calculated easily, is added to the constant c,giving the required mean of x i.e.

If yi = xi – c, then

x=c+y

Illustration:-1 Q) Compute the mean of daily wages (x) of the 65 employees working in a factory from the frequency table given below.

x f

55

65

75

85

95

105

115

Total

8

10

16

14

10

5

2

65

Solving by direct method x (daily wages)

X

f

fx

55

8

440

65

10

650

75

16

1200

85

14

1190

95

10

950

105

5

525

115

2

230

Total

N=∑ fi = 65

∑ fi xi = 5185

= ∑ fi xi /N = 5185/65 =79.77 ( Ans)

Solving by simplified method x (daily wages)

f

y = x – 85

fy

55

8

-30

-240

65

10

-20

-200

75

16

-10

-160

85

14

0

0

95

10

10

100

105

5

20

100

115

2

30

60

Total

N=∑ fi = 65

∑ fiyi =-340

x =c + y = 85 + ∑ fiyi /N = 85 – 5.23 = 79.77(Ans)

Further Simplification In some cases, the values of xi are such that after subtracting a constant c, the deviation can further be reduced on division by a constant factor d i.e. xi – c If yi = ------------------ then, d Where d= common width

x=c+dy

Solving illustration:-1 by this method X

f

x – 85 Y=------------------10

fy

55

8

-3

-24

65

10

-2

-20

75

16

-1

-16

85

14

0

0

95

10

1

10

105

5

2

10

115

2

3

6

Total

N = 65

∑fy= -34

Y = ∑ fiyi /N = -34/65 X = 85 + 10(-34/65 ) = 85 - 340/65 = 85 - 5.23 = 79.77(Ans)

ILLUSTRATION 2:Given the following frequency distribution, calcalculate the mean. Monthly wages (Rs)

Number of worker

12.5 ---- 17.5

2

17.5 ---- 22.5

22

22.5 ---- 27.5

10

27.5 ---- 32.5

14

32.5 ---- 37.5

3

37.5 ---- 42.5

4

42.5 ---- 47.5

6

47.5 ---- 52.5

1

52.5 ---- 57.5

1

Total

63

Class Boundaries

Frequency (f)

Mid-value ( x)

X – 35 Y =-----------------5

fy

12.5 ---- 17.5

2

15

-4

-8

17.5 ---- 22.5

22

20

-3

-66

22.5 ---- 27.5

10

25

-2

-20

27.5 ---- 32.5

14

30

-1

-14

32.5 ---- 37.5

3

35

0

0

37.5 ---- 42.5

4

40

1

4

42.5 ---- 47.5

6

45

2

12

47.5 ---- 52.5

1

50

3

3

52.5 ---- 57.5

1

55

4

4

Total

63

∑fy = - 85

Y = ∑ fiyi /N = -85/63 X = 35 + 5 (-85/63) = 30 – 6.75 = 28.25 (Ans)

ILLUSTRATION 3:Find the mean of the following distribution Age in years

Number of petrsons

15 ----- 19

37

20 ----- 24

81

25 ----- 29

43

30 ----- 34

24

35 ----- 44

9

45 ----- 59

6

Total

200

Class Limit Frequency (f)

Mid-value ( x)

X – 27 Y =------------5

fy

15 ----- 19

37

17

-2

-71

20 ----- 24

81

22

-1

-81

25 ----- 29

43

27

0

0

30 ----- 34

24

32

1

24

35 ----- 44

9

39.5

2.5

22.5

45 ----- 59

6

52

5

30

Total

200

∑fy = - 78.5

Y = ∑ fiyi /N = - 78.5/200 X = 27 + 5 (-78.5/200) = 27 – 1.96 = 25.04 (Ans)

Mathematical properties of A.M. The important mathematical property of A.M are:• The algebraic sum of deviation of observations from their A.M is zero i.e.

Σ ( xi – x ) = 0 where x = simple A.M Σ fi( xi – x ) = 0 where x = weighted A.M

For example:Illustration 1:- Consider the observation 8,1 and 6. Here x = (8+1+6)/3 = 5 The algebraic sum of deviation of observations from their A.M is

Σ ( xi – x ) = (8 - 5) + (1 – 5) + (6 – 5) = 0(proved)

Illustration 2:- Consider the observation 8,1,6 are weighted by 3,2,5 respectively. Then x = ( 8x3 +1x2 + 6x5 )/10 = 56/10 = 5.6 And the algebraic sum of deviations is Σ fi( xi – x ) = 3(8 – 5.6)+2(1-5.6)+5(6-5.6) = 3 x 2.4 + 2 x (-4.6) + 5 x 0.4 = 0 (proved)

2. The sum of the squared of deviation of a set of observations has the smallest value, when deviation is taken from their A.M. i.e.

i)Σ ( xi – A)2 is minimum, when A = simple A.M. ii)Σfi( xi – A)2 is minimum, when A = weighted A.M.

3. If two groups contain n1 and n2 observations with mean x1 and x2 respectively, then the mean ( x ) of the composite group of n1+n2 observations is,

n1 x1 + n2 x2 x

= -------------------------n1+n2

Example Q) There are two branches of an establishment employing 100 and 180 persons respectively. If the arithmetic means of the monthly salaries paid by the two branches are Rs. 275 and Rs.225 respectively, find the arithmetic mean of the salaries of the employees of the establishment as a whole.

Characteristic s

Group I

Number of n1 = 100 observations

Group II n2 = 80

Mean salary (Rs)

Composite Groups n1+n2= 180

X X1= 275

X2 = 225

or

x

n1 x1 + n2 x2 100 x 275 + 80 x 225 = -------------------------- = -------------------------------------n1+ n2 275 + 225

or

X

= 45500/180 = 275.78 (Ans)

Frequency distribution with open end class Consider the following salary distribution of the employees. Salary(Rs ‘000)

Number of employee

below 10

4

10 – 20

6

20 – 30

10

30 – 40

20

40 and above

10

In case of open end class it is generally assumed that the open-end class has the same width as the adjacent class i.e. Salary (Rs‘000)

Mid-value Number of employees (fi)

X – 25 yi =------------10

fiyi

0 – 10 (say)

5

4

-2

-8

10 – 20

15

6

-1

-6

20 – 30

25

10

0

0

30 – 40

35

20

1

20

40 – 50(say)

45

10

2

20

Total

50

Σfiyi= 26

Y = ∑ fiyi /N = 26/50 X = 25 + 10 (26/50) = 25 + 10 x 0.52 = 25 + 5.2 30.2 (Ans)

Advantages 1) The computation of A.M is easy and does not involve any laborious numerical calculations. Even if all observations are not known individually,A.M can be found, provided their sum and the number of observations are known. 2) A.M can be treated algebrically.Given the A.M and the number of observations in each of the several groups,A.M of the composite group can be easily determined by using algebraic formula.

3) It is a very stable and reliable average as regards sampling fluctuations. If many samples are drawn from the same population and each time several measures of central tendency calculated, it will be found that A.M. fluctuates less from sample to sample than any other measure.

Disadvantages 1) It can not be calculated unless the exact magnitude of all observations and their number is known accurately.If some of the extreme values are missing, the accuracy of A.M is greatly affected.A.M. cannot be calculated from a grouped frequency distribution with open-end class, unless some assumptions are made regarding the sizes of these classes.

2) The greatest disadvantage of A.M is that it is highly affected by the presence of even a few extremely large or small observations. For example:A.M. of 2,5,8,4,6,9 is 34/6 = 5.67, it represents the set of observation properly, since most of the observation lies closer to this 5.67. But A.M of 2,5,8,4,6,9,71 is 105/7= 15, which is nearly three times of the previous A.M. 5.67 and most of the observations are much smaller than 15.So here the A.M. is not a proper representative of the whole group of observations because of the presence of the extreme values.

3) A.M may not be an actual value of the variable For example:- Consider the following example, Family size ( xi ) Frequency ( fi ) fi xi 2

4

8

3

3

9

4

3

12

Total

∑ fi = 10

∑ fi xi = 29

∑ fi xi X

= -----------------------------

∑ fi =

=

29/10 2.9 (answer )

A.M. of the family size come to 2.9.However although there may be 2 or 3 family member in a family, 2.9 family member is meaningless.

Problem:- 1 Find the missing frequencies in the following distribution, when it is known that A.M.= 11.09

Class frequency

9.3 – 9.7

9.8 -10.2

10.3 – 10.8 10.7 – 11.2

11.3 – 11.7

11.8 – 12.2

12.3 – 12.7

12.8 – Total 13.2

Frequency

2

5

f3

14

6

3

1

f4

60

Class Limits

Frequency

Mid- values

X – 11 yi =------------0.5

fiyi

9.3 – 9.7

2

9.5

-3

-6

9.8 – 10.2

5

10.0

-2

-10

10.3 – 10.7

f3

10.5

-1

- f3

10.8 – 11.2

f4

11.0

0

0

11.3 – 11.7

14

11.5

10

14

11.8 – 12.2

6

12.0

2

12

12.3 – 12.7

3

12.5

3

9

12.8 – 13.2

1

13.0

4

4

Total

∑ fi = 60

-

-

∑ fi xi = 23 – f3

23 – f3 x = 11.9 = 11.0 + 0.5 x ------------60 0.5 Or 11.9 – 11.0 = ------------- x (23 – f3) 60 Or f3= 12.2 Since the frequency f3 must be a whole number, f3= 12. Again total frequency is or 2+5+ f3 + f4 + 14+6+3+1 = 60 Or f3 + f4 = 60 - 31 = 29 Since f3= 12 , we have f4 = 17

Problem - 2 The mean of 200 observations was 50. Later on, it was discovered that two observations were wrongly read as 92 and 8 instead of 192 and 88.Find out the correct mean. Solution:Here x = 50 and N =200 ∑X = 200 x 50 = 10,000 Correct ∑X = 10,000 – (92+8) + (192+88) = 10,000-100 + 280 = 10,180 Correct mean = 10,180/200 = 50.9