Question-1 :-  The following table shows the ages of the patients admitted in a hospital during a year: Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

  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) = 30
    

      

        

          
Age (in years)
Number of patients (fi)
xi
di = xi - 30
fidi
        
        

          
5 - 15
6
10
-20
-120
        
        

          
15 - 25
11
20
-10
-110
        
    
        

          
25 - 35
21
30
0
0
        
      
        

          
35 - 45
23
40
10
230
        
 
        

          
45 - 55
14
50
20
280
        
  
        

          
55 - 65
5
60
30
150
        
                                         
        

          
Total
Σfi = 80
Σfidi = 430
        
                                                                             
      
    
 
  Now, x = a + (Σfidi/Σfi) = 30 + (430/80)
  = 30 + 43/8
  = 30 + 5.375
  = 35.375
  = 35.38 (approx.)

  Therefore, the mean of this data is 35.38 years.
  It represents that on an average, the age of patient admitted to hospital was 35.38 years.

  Here the maximum class frequency is 23, and the class corresponding to this frequency is 35 – 45. 
  So, the modal class is 35 – 45.
  Now, Modal class = 35 - 45, 
  lower limit(l) of modal class = 23,
  class size(h) = 10
  frequency(f1) of modal class = 23,
  frequency(f0) of modal class = 21,
  frequency(f2) of modal class = 14,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 35 + [(23-21)/{(2x23)-21-14}] x 10
  = 35 + [2/(46-35)] x 10
  = 35 + (2/11 x 10)
  = 35 + 20/11
  = 35 + 1.81
  = 36.81

  Therefore, Mode is 36.8. 
  It represents that the age of maximum number of patients admitted in hospital was 36.81 years. 
    

Question-2 :-  The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Determine the modal lifetimes of the components.

  Here the maximum class frequency is 61, and the class corresponding to this frequency is 60 – 80. 
  So, the modal class is 60 – 80.
  Now, Modal class = 60 - 80, 
  lower limit(l) of modal class = 60,
  class size(h) = 20
  frequency(f1) of modal class = 61,
  frequency(f0) of modal class = 52,
  frequency(f2) of modal class = 38,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 60 + [(61-52)/{(2x61)-52-38}] x 20
  = 60 + [9/(122-90)] x 20
  = 60 + (9/32 x 20)
  = 60 + 45/8
  = 60 + 5.625
  = 65.625
  = 65.63 (approx.)

  Therefore, modal lifetime of electrical coponents is 65.63 hours.  
    

Question-3 :-  The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

  Here the maximum class frequency is 40, and the class corresponding to this frequency is 1500 – 2000. 
  So, the modal class is 1500 – 2000.
  Now, Modal class = 1500 - 2000, 
  lower limit(l) of modal class = 1500,
  class size(h) = 500
  frequency(f1) of modal class = 40,
  frequency(f0) of modal class = 24,
  frequency(f2) of modal class = 33,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 1500 + [(40-24)/{(2x40)-24-33}] x 500
  = 1500 + [16/(80-57)] x 500
  = 1500 + (16/23 x 500)
  = 1500 + 8000/23
  = 1500 + 347.826
  = 1847.826
  = 1847.83 (approx.)

  Therefore, modal monthly expenditure was Rs. 1847.83.  
    

  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) = 500
  Taking assumed mean (a) = 2750
    

      

        

          
Expenditure (in Rs.)
Number of families (fi)
xi
di = xi - 2750
ui = di/500
fiui
        
        

          
1000 - 1500
24
1250
-1500
-3
-72
        
        

          
1500 - 2000
40
1750
-1000
-2
-80
        
    
        

          
2000 - 2500
33
2250
-500
-1
-33
        
      
        

          
2500 - 3000
28
2750
0
0
0
        
 
        

          
3000 - 3500
30
3250
500
1
30
        
  
        

          
3500 - 4000
22
3750
1000
2
44
        
 
        

          
4000 - 4500
16
4250
1500
3
48
        
 
        

          
4500 - 5000
7
4750
2000
4
28
        
                                                         
        

          
Total
Σfi = 200
Σfiui = -35
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 2750 + (-35/200) x 500
  = 2750 - 175/2
  = 2750 - 87.5
  = 2662.5

  Therefore, the mean monthly expenditure was Rs. 2662.5.

Question-4 :-  The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

  Here the maximum class frequency is 10, and the class corresponding to this frequency is 30 – 35. 
  So, the modal class is 30 – 35.
  Now, Modal class = 30 - 35, 
  lower limit(l) of modal class = 30,
  class size(h) = 5
  frequency(f1) of modal class = 10,
  frequency(f0) of modal class = 9,
  frequency(f2) of modal class = 3,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 30 + [(10-9)/{(2x10)-9-3}] x 5
  = 30 + [1/(20-12)] x 5
  = 30 + (1/8 x 5)
  = 30 + 5/8
  = 30 + 0.625
  = 30.625
  = 30.63 (approx.)

  Therefore, the mode of this data is 30.63.  
  It represents that most of the states/U.T have a teacher-student ratio as 30.63.
    

  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) = 5
  Taking assumed mean (a) = 32.5
    

      

        

          
Expenditure (in Rs.)
Number of families (fi)
xi
di = xi - 32.5
ui = di/5
fiui
        
        

          
15 - 20
3
17.5
-15
-3
-9
        
        

          
20 - 25
8
22.5
-10
-2
-16
        
    
        

          
25 - 30
9
27.5
-5
-1
-9
        
      
        

          
30 - 35
10
32.5
0
0
0
        
 
        

          
35 - 40
3
37.5
5
1
3
        
  
        

          
40 - 45
0
42.5
10
2
0
        
 
        

          
45 - 50
0
47.5
15
3
0
        
 
        

          
50 - 55
2
52.5
20
4
8
        
                                                         
        

          
Total
Σfi = 35
Σfiui = -23
        
                                                                             
      
    
 
  Now, x = a + (Σfiui/Σfi) x h = 32.5 + (-23/35) x 5
  = 32.5 - 23/7
  = 32.5 - 3.28
  = 29.22

  Therefore, the mean of data is 29.22.
  It represents that on an average, teacher-student ratio was 29.22.

Question-5 :-  The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

  Here the maximum class frequency is 18, and the class corresponding to this frequency is 4000 – 5000. 
  So, the modal class is 4000 – 5000.
  Now, Modal class = 4000 - 5000, 
  lower limit(l) of modal class = 4000,
  class size(h) = 1000
  frequency(f1) of modal class = 18,
  frequency(f0) of modal class = 4,
  frequency(f2) of modal class = 9,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 4000 + [(18-4)/{(2x18)-4-9}] x 1000
  = 4000 + [14/(36-13)] x 1000
  = 4000 + (14/23 x 1000)
  = 4000 + 14000/23
  = 4000 + 608.695
  = 4608.695
  = 4608.7 (approx.)

  Therefore, the mode of this data is 4608.7 runs.  
    

Question-6 :- A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

  Here the maximum class frequency is 20, and the class corresponding to this frequency is 40 – 50. 
  So, the modal class is 40 – 50.
  Now, Modal class = 40 - 50, 
  lower limit(l) of modal class = 40,
  class size(h) = 10
  frequency(f1) of modal class = 20,
  frequency(f0) of modal class = 12,
  frequency(f2) of modal class = 11,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 40 + [(20-12)/{(2x20)-12-11}] x 10
  = 40 + [8/(40-23)] x 10
  = 40 + (8/17 x 10)
  = 40 + 80/17
  = 40 + 4.7
  = 44.7

  Therefore, the mode of this data is 44.7 cars.