CHAPTER 14. STASTICS

MEANS OF CENTRAL TENDENCY:

1.MEAN

2.MODE

3.MEDIAN

MEAN OF UNGROUPED DATA

If x1, x2, x3, xn are observations with respective frequencies f1, f2, f3, …, fn, then mean is given as

X =

Mean =

MEAN OF GROUPED DATAMethods to find mean:

1. Direct method :

Class mark =

For a class interval 10 – 20, class mark is 15.For a class interval 15- 35, class interval is 25.

1. Direct method :

MEAN OF GROUPED DATA

X =

MEAN OF GROUPED DATAEXERCISE 14. 1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants

0-2 2-4 4-6 6-8 8-10 10-12

12-14

Number of

houses1 2 1 5 6 2 3

EXERCISE 14. 1

Class interval

Class mark

x

Frequencyf fx

0-2 1

2-4 2

4-6 1

6-8 5

8-10 6

10-12 2

12-14 3

EXERCISE 14. 1

Class interval

Class mark

x

Frequencyf fx

0-2 1 1 1

2-4 3 2 6

4-6 5 1 5

6-8 7 5 35

8-10 9 6 54

10-12 11 2 22

12-14 13 3 39

TOTAL 20

EXERCISE 14. 1

Here, 20

162

So, Mean =

=

ANS. The mean number of plants is 8.1 per house.

= 8.1

EXERCISE 14. 1

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages( in Rs)

100-120

120-140

140-160

160-180 180-200

Number of workers 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Class interval

Class mark

x

Frequencyf fx

100-120 12

120-140 14

140-160 8

160-180 6

180-200 10

TOTAL 50

EXERCISE 14. 1

Class interval

Class mark

x

Frequencyf

di = x- 150 fi x di

100-120 110 12 -40 -480

120-140 130 14 -20 -280

140-160 150 8 0 0

160-180 170 6 20 120

180-200 190 10 40 400

TOTAL = 50 = ?

EXERCISE 14. 1

Let assumed mean A = 150

So, mean = A + ( assumed mean method)

= 150 + ( )

= 150- 4.8

= 145.2

So, the mean daily wage of the workers = Rs. 145.20

EXERCISE 14. 1

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women choosing a suitable method.

Number of heart beats per minute

65-68

68-71

71-74

74-77

77-80

80-83

83-86

Number of

women2 4 3 8 7 4 2

EXERCISE 14. 1

Class interval

Class mark

x

Frequencyf

di = x- 75.5 fi x di

65-68 66.5 2

68-71 69.5 4

71-74 72.5 3

74-77 75.5 8

77-80 78.5 7

80-83 81.5 4

83-86 84.5 2

TOTAL =30 = ?

EXERCISE 14. 1

Class interval

Class mark

x

Frequency

f

d = x- 75.5

f x d

65-68 66.5 2 -9 -18

68-71 69.5 4 -6 -24

71-74 72.5 3 -3 -9

74-77 75.5 8 0 0

77-80 78.5 7 3 21

80-83 81.5 4 6 24

83-86 84.5 2 9 18

TOTAL =30 = ?

So, mean = A + ( assumed mean method)

= 75.5 + ( )

= 75.5 + 0.4

= 75.9

Ans. The mean heart beats per minute = 75.9

EXERCISE 14. 1

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes 50-52 53-55 56-58 59-61 62-64

Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding mean did you choose?

EXERCISE 14. 1

Class interva

l

New CI.

Class

mark x

Frequency

f

di = x- 75.5 fi x di

65-68 66.5 2 -9 -18

68-71 69.5 4 -6 -24

71-74 72.5 3 -3 -9

74-77 75.5 8 0 0

77-80 78.5 7 3 21

80-83 81.5 4 6 24

83-86 84.5 2 9 18

TOTAL =30

= ?

EXERCISE 14. 1

EXERCISE 14. 1