 A survey was conducted by a group of students as a part of their environment awareness program, 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  01  24  46  610  810  1012  1214 
number of houses  1  2  1  5  6  2  3 
Which method did you use for finding the mean, and why?
Solution:
To find the class mark (xi) for each interval, the following relation is used.
Class mark 𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕+}^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
xi and fixi can be calculated as follows.
Number of plants  number of houses  x_{i}  f_{i}x_{i} 
0 – 2  1  1  1 
2 – 4  2  3  2 x 3 = 6 
4 – 6  1  5  1 x 5 = 5 
6 – 8  5  7  5 x 7 = 35 
8 – 10  6  9  6 x 9 = 54 
10 – 12  2  11  2 x 11 = 22 
12 – 14  3  13  3 x13 = 39 
Total  20  162 
From the table, it can be observed that
∑f_{i} = 20
∑f_{i}x_{i} = 162
_{= }^{162}/_{20 = 8.1}
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.
 Consider the following distribution of daily wages of 50 worker of a factory.
Daily wages(in Rs)  100120  120140  140160  160180  180200 
number of workers  12  14  8  6  10 
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
To find the class mark for each interval, the following relation is used.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
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(f_{i})  x_{i}  d_{i} = x_{i} – 150  u_{i}=^{di}/_{20}  f_{i}u_{i} 
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
∑f_{i} = 50
∑f_{i}x_{i} = 12
_{=150+(}^{12}/_{50)20}
_{ = 150 – }^{24}/_{5}
= 150 – 4.8 = 145.2
Therefore, mean number of plants per house is 145.20
 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 
Solution:
To find the class mark (xi) for each interval, the following relation is used.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
Given that, mean pocket allowance,
Taking 18 as assured mean (a), d_{i} and f_{i}d_{i} are calculated as follows.
Daily pocket allowance(in Rs)  Number of children (f_{i})  class mark x_{i}  d_{i} = x_{i} – 18  f_{i}d_{i}  f_{i}u_{i} 
11 – 13  7  12  6  – 42  24 
13 15  6  14  4  24  14 
15 – 17  9  16  2  18  0 
17 – 19  13  18  0  0  6 
19 – 21  f  20  2  2f  20 
21 – 23  5  22  4  20  12 
23 – 25  4  24  6  24  24 
total  ∑f_{i} = 44 + f

2f – 40 
From above,
∑f_{i} = 44 + f
∑f_{i}u_{i} = 2f – 40
18 = 18 +^{(2f40)}/_{(44+f)}
2f – 40 = 0
2f = 40
f = 20
Hence, the missing frequency, f, is 20.
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 
Solution:
To find the class mark of each interval (xi), the following relation is used.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
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  𝒖_{𝒊} = ^{𝒅𝒊}/_{𝟑}  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
∑f_{i} = 30
∑f_{i}u_{i} = 4
= 75.5+(^{4}/_{30})x3
= 75.5 + ^{4}/_{30} x 3
= 75.5 – 0.4
= 75.9
Therefore, mean hear beats per minute for these women are 75.9 beats per minute.
 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?
Solution:
Number of mangoes  Number of boxes f_{i} 
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 added to the upper class limit and 1/2 has to be subtracted from the lower class limit of each interval.
Class mark (xi) can be obtained by using the following relation.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
Class size (h) of this data = 3
Here we are using Step deviation method
Taking 57 as assumed mean (a), di, ui, fiui are calculated as follows.
class interval  f_{i}  x_{i}  d_{i} = x_{i}57  u_{i} = ^{di}/_{3}  f_{i}u_{i} 
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 
∑f_{i} = a + (^{∑fiui}/_{∑fi})xh
= 57 + (^{25}/_{400})x3
= 57 + ^{3}/_{16} = 57 + 0.1875
= 57.1875 = 57.19
Mean number of mangoes kept in a packing box is 57.19.
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.
Solution:
To find the class mark (xi) for each interval, the following relation is used.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
Class size = 50
Taking 225 as assumed mean (a), di, ui, fiui are calculated as follows.
Daily expenditure(in Rs)  f_{i}  x_{i}  d_{i} = x_{i} – 225  u_{i} == ^{di}/_{50}  f_{i}u_{i} 
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 
∑f_{i} = 25
∑f_{i}u_{i} = a + (^{∑f}_{i}^{u}_{i}/ _{∑fi})xh
= 225 + (^{7}/_{25})x50
= 225 – 14
= 211
Therefore, mean daily expenditure on food is Rs 211.
 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 SO_{2}  Frequency 
00.0 – 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.
Solution:
To find the class marks for each interval, the following relation is used.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
Class size of this data = 0.04
Taking 0.14 as assumed mean (a), di, ui, fiui are calculated as follows.
Concentration of SO_{2}  frequency fi  class mark xi  di = xi – 0.14  ui = ^{di}/_{0.04}  fiui 
0.000.04  4  0.02  0.12  3  12 
0.040.08  9  0.06  0.08  2  18 
0.080.12  9  0.10  0.04  1  9 
0.120.16  2  0.14  0  0  0 
0.160.20  4  0.18  0.04  1  4 
0.200.24  2  0.22  0.08  2  4 
total  30  31 
∑f_{i} = 30
∑fiui = 31
=0.14 +(^{31}/_{30})(0.04)
=0.14 – 0.04133
=0.09867
=0.099ppm
Therefore, mean concentration of SO_{2} in the air is 0.099 ppm.
 A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days  06  610  1014  1420  2028  2838  3849 
number of students  11  10  7  4  4  3  1 
Solution:
To find the class mark of each interval, the following relation is used.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
Taking 17 as assumed mean (a), di and fidi are calculated as follows.
Number of days  number of students fi  xi  di = xi – 17  fidi 
06  11  3  14  154 
610  10  8  9  90 
1014  7  12  5  35 
1420  4  17  0  0 
2028  4  24  7  28 
2838  3  33  16  48 
3840  1  39  22  22 
total  40  181 
∑fi = 40
∑fidi = 181
= 17 + (^{181}/_{40})
= 17 – 4.525
= 12.475
= 12.48
Therefore, the mean number of days is 12.48 days for which a student was absent.
9: The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate(in%)  4555  5565  6575  7585  8595 
number of cities  3  10  11  8  3 
Solution:
To find the class marks, the following relation is used.
𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}
Class size (h) for this data = 10 Taking 70 as assumed mean (a), di, ui, and fiui are calculated as follows.
Literacy rate(in%)  number of cities fi  xi  di = xi – 17  ui = ^{di}/_{10}  fidi 
4555  3  50  20  2  6 
5565  10  60  10  1  10 
6575  11  70  0  0  0 
7585  8  80  10  1  8 
8595  3  90  20  2  6 
total  35  2 
∑fi = 35
∑fidi = 2
= 70 + (^{2}/_{35})x10
= 17 – ^{20}/_{35 }= 17 – ^{4}/_{7}
= 70 – 0.57
= 69.43
Therefore, mean literacy rate is 69.43%
1 thought on “Statistics Exercise 14.1 – Class 10”
Comments are closed.