TOPICS
Exercise - 14.2

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

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) = 30
    
Age (in years)Number of patients (fi)xidi = xi - 30fidi
5 - 15610-20-120
15 - 251120-10-110
25 - 35213000
35 - 45234010230
45 - 55145020280
55 - 6556030150
TotalΣfi = 80Σfidi = 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 : Statistics Determine the modal lifetimes of the components.

Solution :-
  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 : Statistics

Solution :-
  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)xidi = xi - 2750ui = di/500fiui
1000 - 1500241250-1500-3-72
1500 - 2000401750-1000-2-80
2000 - 2500332250-500-1-33
2500 - 3000282750000
3000 - 3500303250500130
3500 - 40002237501000244
4000 - 45001642501500348
4500 - 5000747502000428
TotalΣfi = 200Σfiui = -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. Statistics

Solution :-
  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)xidi = xi - 32.5ui = di/5fiui
15 - 20317.5-15-3-9
20 - 25822.5-10-2-16
25 - 30927.5-5-1-9
30 - 351032.5000
35 - 40337.5513
40 - 45042.51020
45 - 50047.51530
50 - 55252.52048
TotalΣfi = 35Σfiui = -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. Statistics Find the mode of the data.

Solution :-
  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 : Statistics

Solution :-
  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. 
    
CLASSES

Connect with us:

Copyright © 2015-20 by a1classes. All Rights Reserved.