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 Which method did you use for finding the mean, and why?

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
    

      

        

          
Number of plants
Number of houses (fi)
xi
fixi
        
        

          
0 - 2
1
1
1
        
        

          
2 - 4
2
3
6
        
    
        

          
4 - 6
1
5
5
        
      
        

          
6 - 8
5
7
35
        
 
        

          
8 - 10
6
9
54
        
  
        

          
10 - 12
2
11
22
        
     
        

          
12 - 14
3
13
39
        
  
        

          
Total
Σfi = 20
Σfixi = 162
        
                                                                             
      
    
 
  Now, x = Σfixi/Σfi = 162/20 = 8.1
  Therefore, the 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 workers of a factory Find the mean daily wages of the workers of the factory by using an appropriate method

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 20
  Taking assumed mean (a) = 150
    

      

        

          
Daily wages (in Rs.)
Number of workers (fi)
xi
di = xi - 150
ui = di/20
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
Σfi = 50
Σfiui = -12
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 150 + (-12/50) x 20
  = 150 - 24/5
  = 150 - 4.8
  = 145.2

  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.

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2

  Given that the mean pocket allowance x = Rs. 18
  Taking assumed mean (a) = 18
    

      

        

          
Daily pocket allowance (in Rs.)
Number of children (fi)
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
2f
        
   
        

          
21 - 23
5
22
4
20
        
        

          
23 - 25
4
24
6
24
        
                 
        

          
Total
Σfi = 44 + f
Σfidi = 2f - 40
        
                                                                             
      
    
 
  Now, x = a + (Σfidi/Σfi) 
  18 = 18 + (2f - 40)/(44 + f)
  18 - 18 = (2f - 40)/(44 + f)
  0 = (2f - 40)/(44 + f)
  0 = 2f - 40
  40 = 2f
  f = 40/2
  f = 20

  Therefore, 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 summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 3
  Taking assumed mean (a) = 75.5
    

      

        

          
Number of heart beats (per min.)
Number of women (fi)
xi
di = xi - 75.5
ui = 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
Σfi = 30
Σfiui = 4
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 75.5 + (4/30) x 3
  = 75.5 + 4/10
  = 75.5 + 0.4
  = 75.9

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

Solution :-

  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 upper class limit and 1/2 has to be substracted from the lower class limit of each interval.
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 3
  Taking assumed mean (a) = 57
    

      

        

          
Number of mangoes
Number of boxes (fi)
xi
di = xi - 57
ui = di/3
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
Σfi = 400
Σfiui = 25
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 57 + (25/400) x 3
  = 57 + 3/16
  = 57 + 0.1875
  = 57.1875
  = 57.19 (approx.)

  Therefore, the mean number of mangoes kept in a packing box is 57.19.
    

Question-6 :- The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 50
  Taking assumed mean (a) = 225
    

      

        

          
Daily expenditure (in Rs.)
Number of households (fi)
xi
di = xi - 225
ui = di/50
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
Σfi = 25
Σfiui = -7
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 225 + (-7/25) x 50
  = 225 - 14
  = 211

  Therefore, the 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: Find the mean concentration of SO₂ in the air.

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 0.04
  Taking assumed mean (a) = 0.14
    

      

        

          
Concentration of SO2 (in ppm)
Frequency (fi)
xi
di = xi - 0.14
ui = di/0.04
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
0.4
1
4
        
 
        

          
0.20 - 0.24
2
0.22
0.8
2
4
        
                          
        

          
Total
Σfi = 30
Σfiui = -31
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 0.14 + (-31/30) x 0.04
  = 0.14 - 0.04133
  = 0.09867
  = 0.099 ppm (approx.)

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

Question-8 :- 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.

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Taking assumed mean (a) = 17
    

      

        

          
Number of days
Number of students (fi)
xi
di = xi - 17
fidi
        
        

          
0 - 6
11
3
-14
-154
        
        

          
6 - 10
10
8
-9
-90
        
    
        

          
10 - 14
7
12
-5
35
        
      
        

          
14 - 20
4
17
0
0
        
 
        

          
20 - 28
4
24
7
28
        
  
        

          
28 - 38
3
33
16
48
        
 
        

          
38 - 40
1
39
22
22
        
                                  
        

          
Total
Σfi = 40
Σfidi = -181
        
                                                                             
      
    
 
  Now, x = a + (Σfidi/Σfi) = 17 + (-181/40)
  = 17 - 4.525
  = 12.475
  = 12.48 (approx.)

  Therefore, the mean number of days is 12.48 days for which a student was absent. 
    

Question-9 :- The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution :-

  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 10
  Taking assumed mean (a) = 70
    

      

        

          
Literacy rate (in %)
Number of cities (fi)
xi
di = xi - 70
ui = di/10
fiui
        
        

          
45 - 55
3
50
-20
-2
-6
        
        

          
55 - 65
10
60
-10
-1
-10
        
    
        

          
65 - 75
11
70
0
0
0
        
      
        

          
75 - 85
8
80
10
1
8
        
 
        

          
85 - 95
3
90
20
2
6
        
                                   
        

          
Total
Σfi = 35
Σfiui = -2
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 70 + (-2/35) x 10
  = 70 - 4/7
  = 70 - 0.57
  = 69.43

  Therefore, the mean literacy rate is 69.43%.