Class 10 NCERT Math Solution

Unit-14(Examples)

Example-1 :- The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. Statistics

Solution :-

    
      
        
          Marks obtained (x_i) Number of students (f_i) f_ix_i
        
        
          10 1 10
        
        
          20 1 20
            
        
          36 3 108
              
        
          40 4 160
         
        
          50 3 150
          
        
          56 2 112
             
        
          60 4 240
         
        
          70 4 280
        
        
          72 1 72
         
        
          80 1 80
         
        
          88 2 176
          
        
          92 3 276
         
        
          95 1 95
          
        
          Total Σf_i = 30 Σf_ix_i = 1779
                                                                                     
      
     
    Now, x = Σf_ix_i/Σf_i = 1779/30 = 59.3
    Therefore, the mean marks obtained is 59.3.

Example-2 :- : The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section. Statistics

Solution :-

    
      
        
          Percentage of female teachers No. of states/U.T. (f_i) x_i d_i = x_i - 50 u_i =(x_i - 50)/10 f_ix_i f_id_i f_iu_i
        
        
          15 - 25 6 20 -30 -3 120 -180 -18
        
        
          25 - 35 11 30 -20 -2 330 -220 -22
            
        
          35 - 45 7 40 -10 -1 280 -70 -7
              
        
          45 - 55 4 50 0 0 200 0 0
         
        
          55 - 65 4 60 10 1 240 40 4
          
        
          65 - 75 2 70 20 2 140 40 4
             
        
          75 - 85 1 80 30 3 80 30 3
          
        
          Total Σf_i = 35 Σf_ix_i = 1390 Σf_id_i = -360 Σf_iu_i = -36
                                                                                     
      
     
    Here we take a = 50, h = 10, then d_i = x_i – 50
    From the table above, we obtain Σf_i = 35, Σf_ix_i = 1390, Σf_id_i = – 360, Σf_iu_i = –36.

    Using the direct method, 
    x = Σf_ix_i/Σf_i = 1390/35 = 39.71

    Using the assumed mean method,
    x = a + Σf_id_i/Σf_i = 50 + (-360)/35 = 39.71

    Using the step-deviation method,
    x = a + (Σf_iu_i/Σf_i) x h = 50 + (-36/35) x 10 = 39.71

    Therefore, the mean percentage of female teachers in the primary schools of rural areas is 39.71.

Example-3 :- : The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? Statistics

Solution :-

    
      
        
          Number of wickets taken Number of bowlers (f_i) x_i d_i = x_i - 200 u_i =(x_i - 200)/20 f_iu_i
        
        
          20 - 60 7 40 -160 -8 -56
        
        
          60 - 100 5 80 -120 -6 -30
            
        
          100 - 150 16 125 -75 -3.75 -60
              
        
          150 - 250 12 200 0 0 0
         
        
          250 - 350 2 300 100 5 10
          
        
          350 - 450 3 400 200 10 30
              
        
          Total Σf_i = 45 Σf_iu_i = -106
                                                                                     
      
     
    Here we take a = 200, h = 20, then d_i = x_i – 200
    From the table above, we obtain Σf_i = 45, Σf_iu_i = -106.

    Therefore, x = a + (Σf_iu_i/Σf_i) x h = 200 + (-106/45) x 20 = 200 - (2120/45) = 200 - 47.11 = 152.89

    This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89

Example-4 :- The wickets taken by a bowler in 10 cricket matches are as follows:
2, 6, 4, 5, 0, 2, 1, 3, 2, 3
Find the mode of the data.

Solution :-

  Let us form the frequency distribution table of the given data as follows:
   
    
        
          No. of wickets 0 1 2 3 4 5 6
        
        
          No. of matches 1 1 3 2 1 1 1
                                                                                    
      
     
  Therefore, 2 is the number of wickets taken by the bowler in the maximum number (i.e., 3) of matches. 
  So, the mode of this data is 2.

Example-5 :- A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household: Statistics Find the mode of this data.

Solution :-

   
    
        
          Family size No. of families
        
        
          1 - 3 7
          
        
          3 - 5 8
        
        
          5 - 7 2
        
        
          7 - 9 2
        
        
          9 - 11 1
                        
      
           
  Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5. 
  So, the modal class is 3 – 5.
  Now, Modal class = 3 - 5, 
  lower limit(l) of modal class = 3,
  class size(h) = 2
  frequency(f₁) of modal class = 8,
  frequency(f₀) of modal class = 7,
  frequency(f₂) of modal class = 2,

  Therefore, Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] x h
  = 3 + [(8 - 7)/{(2 x 8) - 7 - 2)}] x 2
  = 3 + [1/(16 - 9)] x 2
  = 3 + (1/7) x 2
  = 3 + 2/7
  = 3 + 0.286
  = 3.286

  Hence, the mode of the data above is 3.286.

Example-6 :- : The marks distribution of 30 students in a mathematics examination are given in Table 14.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean.

Solution :-

  Refer to Table 14.3 of Example 1. 
    
      
        
          Class interval No. of students
        
        
          10 - 25 2
        
        
          25 - 40 3
            
        
          40 - 55 7
              
        
          55 - 70 6
         
        
          70 - 85 6
          
        
          85 - 100 6
                                                                                    
      
     
  Since the maximum number of students (i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. 
  Therefore, the lower limit(l) of the modal class = 40,
  the class size(h) = 15,
  the frequency(f₁) of modal class = 7,
  the frequency(f₀) of the class preceding the modal class = 3,
  the frequency(f₂) of the class succeeding the modal class = 6.

  Therefore, Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] x h
  = 40 + [(7 - 3)/(14 - 6 - 3)] x 15
  = 40 + (4/5) x 15
  = 40 + 12
  = 52 

  So, the mode marks is 52.
  Now, from Example 1, you know that the mean marks is 62.
  So, the maximum number of students obtained 52 marks, while on an average a student obtained 62 marks.

Example-7 :- A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained: Statistics Find the median height.

Solution :-

    
      
        
          Class interval Frequency Cumulative Frequency (cf)
        
        
          Below 140 4 4
        
        
          140 - 145 7 4 + 7 = 11
            
        
          145 - 150 18 11  + 18 = 29
              
        
          150 - 155 11 29 + 11 = 40
         
        
          155 - 160 6 40 + 6 = 46
          
        
          160 - 165 5 46 + 5 = 51
         
        
          Total n = 51 
                                                                                            
      
    
  Now, n = 51
  n/2 = 51/2 = 25.5 is 29. So, this observation lies in the class 145 - 150. 
  Then, l (the lower limit) = 145,
  cf (the cumulative frequency of the class preceding 145 - 150) = 11,
  f (the frequency of the median class 145 - 150) = 18,
  h (the class size) = 5. 

  Therefore, Median = l + [(n/2 - cf)/f] x h
  = 145 + [(51/2 - 11)/18] x 5
  = 145 + [(25.5 - 11)/18] x 5
  = 145 + (14.5/18) x 5
  = 145 + 72.5/18
  = 145 + 4.027
  = 149.028
  = 149.03 (approx.)    

  So, the median height of the girls is 149.03 cm.
  This means that the height of about 50% of the girls is less than this height, and 50% are taller than this height.

Example-8 :- The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Statistics

Solution :-

  
      
        
          Class intervals Frequency Cumulative Frequency (cf)
        
        
          0 - 100 2 2
        
        
          100 - 200 5 7
            
        
          200 - 300 x 7 + x
              
        
          300 - 400 12 19 + x
         
        
          400 - 500 17 36 + x
          
        
          500 - 600 20 56 + x
          
        
          600 - 700 y 56 + x + y
         
        
          700 - 800 9 65 + x + y
         
        
          800 - 900 7 72 + x + y
          
        
          900 - 1000 4 76 + x + y
         
        
          Total n = 76 + x + y 
                                        
      
    
  It is given that n = 100
  So, 76 + x + y = 100, i.e., x + y = 24 ............(i)
  The median is 525, which lies in the class 500 – 600
  So, l = 500, f = 20, cf = 36 + x, h = 100  

  Therefore, Median = l + [(n/2 - cf)/f] x h
  525 = 500 + [{(100/2) - (36 + x)}/20] x 100
  525 = 500 + [(50 - 36 - x)/20] x 100
  525 - 500 = (14 - x) x 5
  25/5 = 14 - x
  5 = 14 - x
  x = 14 - 5
  x = 9

  Therefore, from (i), we get
  9 + y = 24
  y = 24 - 9
  y = 15 

  Hence, the values of x and y are 9 and 15 respectively.

Example-9 :- The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution: Statistics Draw both ogives for the data above. Hence obtain the median profit.

Solution :-

  We first draw the coordinate axes, with lower limits of the profit along the horizontal axis, and the cumulative frequency along the vertical axes. 
  Then, we plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3). 
  We join these points with a smooth curve to get the ‘more than’ ogive, as shown in Fig. Now, let us obtain the classes, their frequencies and the cumulative frequency from the table above.
         
  
    
        
          Classes No. of shops Cumulative Frequency (cf)
        
        
          5 - 10 2 2
        
        
          10 - 15 12 2 + 12 = 14
            
        
          15 - 20 2 14 + 2 = 16
              
        
          20 - 25 4 16 + 4 = 20
         
        
          25 - 30 3 20 + 3 = 23
          
        
          30 - 35 4 23 + 4 = 27
            
        
          35 - 40 3 27 + 3 = 30
                
        
          Total n = 30 
                                                                                      
      
    
  Using these values, we plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27), (40, 30) on the same axes as in Fig. to get the ‘less than’ ogive, as shown in Fig.
   
  The abcissa of their point of intersection is nearly 17.5, which is the median. 
  This can also be verified by using the formula. 
  Hence, the median profit (in lakhs) is Rs 17.5.

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Marks obtained (x_i)	Number of students (f_i)	f_ix_i
10	1	10
20	1	20
36	3	108
40	4	160
50	3	150
56	2	112
60	4	240
70	4	280
72	1	72
80	1	80
88	2	176
92	3	276
95	1	95
Total	Σf_i = 30	Σf_ix_i = 1779

Percentage of female teachers	No. of states/U.T. (f_i)	x_i	d_i = x_i - 50	u_i =(x_i - 50)/10	f_ix_i	f_id_i	f_iu_i
15 - 25	6	20	-30	-3	120	-180	-18
25 - 35	11	30	-20	-2	330	-220	-22
35 - 45	7	40	-10	-1	280	-70	-7
45 - 55	4	50	0	0	200	0	0
55 - 65	4	60	10	1	240	40	4
65 - 75	2	70	20	2	140	40	4
75 - 85	1	80	30	3	80	30	3
Total	Σf_i = 35				Σf_ix_i = 1390	Σf_id_i = -360	Σf_iu_i = -36

Number of wickets taken	Number of bowlers (f_i)	x_i	d_i = x_i - 200	u_i =(x_i - 200)/20	f_iu_i
20 - 60	7	40	-160	-8	-56
60 - 100	5	80	-120	-6	-30
100 - 150	16	125	-75	-3.75	-60
150 - 250	12	200	0	0	0
250 - 350	2	300	100	5	10
350 - 450	3	400	200	10	30
Total	Σf_i = 45				Σf_iu_i = -106

Family size	No. of families
1 - 3	7
3 - 5	8
5 - 7	2
7 - 9	2
9 - 11	1

Class interval	No. of students
10 - 25	2
25 - 40	3
40 - 55	7
55 - 70	6
70 - 85	6
85 - 100	6

Class interval	Frequency	Cumulative Frequency (cf)
Below 140	4	4
140 - 145	7	4 + 7 = 11
145 - 150	18	11 + 18 = 29
150 - 155	11	29 + 11 = 40
155 - 160	6	40 + 6 = 46
160 - 165	5	46 + 5 = 51
Total	n = 51

Class intervals	Frequency	Cumulative Frequency (cf)
0 - 100	2	2
100 - 200	5	7
200 - 300	x	7 + x
300 - 400	12	19 + x
400 - 500	17	36 + x
500 - 600	20	56 + x
600 - 700	y	56 + x + y
700 - 800	9	65 + x + y
800 - 900	7	72 + x + y
900 - 1000	4	76 + x + y
Total	n = 76 + x + y

Classes	No. of shops	Cumulative Frequency (cf)
5 - 10	2	2
10 - 15	12	2 + 12 = 14
15 - 20	2	14 + 2 = 16
20 - 25	4	16 + 4 = 20
25 - 30	3	20 + 3 = 23
30 - 35	4	23 + 4 = 27
35 - 40	3	27 + 3 = 30
Total	n = 30

No. of wickets	0	1	2	3	4	5	6
No. of matches	1	1	3	2	1	1	1