1.mean 2.mode 3.median if x 1, x 2, x 3, x n are observations with respective frequencies f 1, f 2,...
TRANSCRIPT
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