Exercise 14.1

Question 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 houses
1
2
1
5
6
2
3
Which method did you use for finding the mean, and why?


To find the class mark (xi) for each interval, the following relation is used.
Class mark (xi)






                                              
xi and fixi can be calculated as follows.
Number of plants
Number of houses
(fi)
xi
fixi
0­ − 2
1
1
1 × 1 = 1
2­ − 4
2
3
2 × 3 = 6
4 − 6
1
5
1 × 5 = 5
6 − 8
5
7
5 × 7 = 35
8 − 10
6
9
6 × 9 = 54
10 − 12
2
11
2 ×11 = 22
12 − 14
3
13
3 × 13 = 39
Total
20
162
From the table, it can be observed that




Mean,


Therefore, mean number of plants per house is 8.1.
Here, direct method has been used as the values of class marks (xi) and fi are small.



Question 2:
Consider the following distribution of daily wages of 50 worker of a factory.
Daily wages (in Rs)
100­ − 120
120­ − 140
140 −1 60
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.
To find the class mark for each interval, the following relation is used.



Class size (h) of this data = 20
Taking 150 as assured mean (a), di, ui, and fiui can be calculated as follows.
Daily wages
(in Rs)
Number of workers (fi)
xi
di = xi − 150
fiui
100­ −120
12
110
− 40
− 2
− 24
120­ − 140
14
130
− 20
− 1
− 14
140 − 160
8
150
0
0
0
160 −180
6
170
20
1
6
180 − 200
10
190
40
2
20
Total
50
− 12
From the table, it can be observed that







Therefore, the mean daily wage of the workers of the factory is Rs 145.20.



Question 3:
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.
Daily pocket allowance (in Rs)
11­ − 13
13­ − 15
15 −17
17 − 19
19 − 21
21 − 23
23 − 25
Number of workers
7
6
9
13
f
5
4



To find the class mark (xi) for each interval, the following relation is used.




Given that, mean pocket allowance,
Taking 18 as assured mean (a), di and fidi are calculated as follows.
Daily pocket allowance
(in Rs)
Number of children
fi
Class mark xi
di = xi − 18
fidi
11­ −13
7
12
− 6
− 42
13 − 15
6
14
− 4
− 24
15 − 17
9
16
− 2
− 18
17 −19
13
18
0
0
19 − 21
f
20
2
2 f
21 − 23
5
22
4
20
23 − 25
4
24
6
24
Total
2f − 40
From the table, we obtain









Hence, the missing frequency, f, is 20.



Question 4:
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine 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 women
2
4
3
8
7
4
2


To find the class mark of each interval (xi), the following relation is used.




Class size, h, of this data = 3
Taking 75.5 as assumed mean (a), di, ui, fiui are calculated as follows.
Number of heart beats per minute
Number of women
fi
xi
di = xi − 75.5
fiui
65 − 68
2
66.5
− 9
− 3
− 6
68 − 71
4
69.5
− 6
− 2
− 8
71 − 74
3
72.5
− 3
− 1
− 3
74 − 77
8
75.5
0
0
0
77 − 80
7
78.5
3
1
7
80 − 83
4
81.5
6
2
8
83 − 86
2
84.5
9
3
6
Total
30
4
From the table, we obtain






Therefore, mean hear beats per minute for these women are 75.9 beats per minute.



Question 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 the mean did you choose?


Number of mangoes
Number of boxes fi
50 − 52
15
53 − 55
110
56 − 58
135
59 − 61
115
62 − 64
25
It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore 1/2  has to be subtracted from the lower class limit
of each interval.
, has to be added to the upper class limit and
Class mark (xi) can be obtained by using the following relation.




Class size (h) of this data = 3
Taking 57 as assumed mean (a), di, ui, fiui are calculated as follows.
Class interval
fi
xi
di = xi − 57
fiui
49.5 − 52.5
15
51
− 6
− 2
− 30
52.5 − 55.5
110
54
− 3
− 1
− 110
55.5 − 58.5
135
57
0
0
0
58.5 − 61.5
115
60
3
1
115
61.5 − 64.5
25
63
6
2
50
Total
400
25
It can be observed that





Mean number of mangoes kept in a packing box is 57.19.
Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di.


Question 6:
The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure (in Rs)
100 − 150
150 − 200
200 − 250
250 − 300
300 − 350
Number of households
4
5
12
2
2
Find the mean daily expenditure on food by a suitable method.
To find the class mark (xi) for each interval, the following relation is used.




Class size = 50
Taking 225 as assumed mean (a), di, ui, fiui are calculated as follows.
Daily expenditure (in Rs)
fi
xi
di = xi − 225
fiui
100 − 150
4
125
− 100
− 2
− 8
150 − 200
5
175
− 50
− 1
− 5
200 − 250
12
225
0
0
0
250 − 300
2
275
50
1
2
300 − 350
2
325
100
2
4
Total
25
− 7
From the table, we obtain





Therefore, mean daily expenditure on food is Rs 211.



Question 7:
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
concentration of SO2 (in ppm)
Frequency
0.00 − 0.04
4
0.04 − 0.08
9
0.08 − 0.12
9
0.12 − 0.16
2
0.16 − 0.20
4
0.20 − 0.24
2
Find the mean concentration of SO2 in the air.
To find the class marks for each interval, the following relation is used.




Class size of this data = 0.04
Taking 0.14 as assumed mean (a), di, ui, fiui are calculated as follows.
Concentration of SO2 (in ppm)
Frequency
fi
Class mark
xi
di = xi − 0.14
fiui
0.00 − 0.04
4
0.02
− 0.12
− 3
− 12
0.04 − 0.08
9
0.06
− 0.08
− 2
− 18
0.08 − 0.12
9
0.10
− 0.04
− 1
− 9
0.12 − 0.16
2
0.14
0
0
0
0.16 − 0.20
4
0.18
0.04
1
4
0.20 − 0.24
2
0.22
0.08
2
4
Total
30
− 31
From the table, we obtain






Therefore, mean concentration of SO2 in the air is 0.099 ppm.