Examples of Arithmetic Mean

Example (4):

The following data shows distance covered by 100 persons to perform their routine jobs.

Distance (Km)

0 - 10

10 - 20

20 - 30

30 - 40

Number of Persons

10

20

40

30

Calculate Arithmetic Mean by Step-Deviation Method; also explain why it is better than direct method in this particular case. 

Solution:

The given distribution belongs to a grouped data and the variable involved is ages of “distance covered”. While the “number of persons” Represent frequencies.

Distance Covered in (Km)

Number of Persons
f

Mid Points
x

u = \left( {\frac{{x - 5}}{{10}}} \right)

fu

0 - 10

10

5

 - 1

 - 10

10 - 20

20

15

0

0

20 - 30

40

25

 + 1

40

30 - 40

30

35

 + 2

60

Total

\sum f = 100

 

 

\sum fu = 90

Now we will find the Arithmetic Mean as \overline X = A + \frac{{\sum  fu}}{{\sum f}} \times h
Where
            A = 15,    \sum fu = 90,    \sum f = 100   and   h = 10
            \overline X = 15 +  \frac{{90}}{{100}} \times 10 = 24 Km
Explanation:

Here from the mid points (x) it is very much clear that each mid point is multiple of 5 and there is also a gap of 10 from mid point to midpoint i.e. class size or interval (h). Keeping in view this, we should prefer to take method of Step-Deviation instead of Direct Method.

Example (5):

The following frequency distribution showing the marks obtained by 50 students in statistics at a certain college. Find the arithmetic mean using (1) Direct Method (2) Short-Cut Method (3) Step-Deviation.

Marks

20 - 29

30 - 39

40 - 49

50 - 59

60 - 69

70 - 79

80 - 89

Frequency

1

5

12

15

9

6

2

Solution:

 

Direct Method

Short-Cut
Method

Step-Deviation
Method

Marks

f

x

fx

D = x - A

fD

u = \frac{{x - A}}{h}

fu

20 - 29

1

24.5

24.5

 - 30

 - 30

 - 3

 - 3

30 - 39

5

34.5

172.5

 - 20

 - 100

 - 2

 - 10

40 - 49

12

44.5

534.5

 - 10

 - 120

 - 1

 - 12

50 - 59

15

54.5

817.5

0

0

0

0

60 - 69

9

64.5

580.5

10

90

1

9

70 - 79

6

74.5

447.5

20

120

2

12

80 - 89

2

84.5

169.5

30

60

3

6

Total

50

 

2745

 

20

 

2

(1) Direct Method:

                                    \overline X = \frac{{\sum  fx}}{{\sum f}} = \frac{{2745}}{{50}} = 54.9 or 55 Marks
(2) Short-Cut Method:
                           \overline X = A + \frac{{\sum  fD}}{{\sum f}}       Where A = 54.5
                                = 54.5 +  \frac{{20}}{{50}} = 54.5 + 0.4 = 54.9Marks
(3) Step-Deviation Method:
                           \overline X = A + \frac{{\sum  fu}}{{\sum f}} \times h   Where A = 54.5        h = 10
                                 = 54.5 + \frac{2}{{50}}  \times 10
                                 = 54.5 + 0.4 = 54.9 Marks

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