TOPICS
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 (xi)Number of students (fi)fixi
10110
20120
363108
404160
503150
562112
604240
704280
72172
80180
882176
923276
95195
TotalΣfi = 30Σfixi = 1779
Now, x = Σfixi/Σfi = 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 teachersNo. of states/U.T. (fi)xidi = xi - 50ui =(xi - 50)/10fixifidifiui
15 - 25620-30-3120-180-18
25 - 351130-20-2330-220-22
35 - 45740-10-1280-70-7
45 - 554500020000
55 - 65460101240404
65 - 75270202140404
75 - 8518030380303
TotalΣfi = 35Σfixi = 1390Σfidi = -360Σfiui = -36
Here we take a = 50, h = 10, then di = xi – 50 From the table above, we obtain Σfi = 35, Σfixi = 1390, Σfidi = – 360, Σfiui = –36. Using the direct method, x = Σfixi/Σfi = 1390/35 = 39.71 Using the assumed mean method, x = a + Σfidi/Σfi = 50 + (-360)/35 = 39.71 Using the step-deviation method, x = a + (Σfiui/Σfi) 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 takenNumber of bowlers (fi)xidi = xi - 200ui =(xi - 200)/20fiui
20 - 60740-160-8-56
60 - 100580-120-6-30
100 - 15016125-75-3.75-60
150 - 25012200000
250 - 3502300100510
350 - 45034002001030
TotalΣfi = 45Σfiui = -106
Here we take a = 200, h = 20, then di = xi – 200 From the table above, we obtain Σfi = 45, Σfiui = -106. Therefore, x = a + (Σfiui/Σfi) 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 wickets0123456
No. of matches1132111
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 sizeNo. of families
1 - 37
3 - 58
5 - 72
7 - 92
9 - 111
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 intervalNo. of students
10 - 252
25 - 403
40 - 557
55 - 706
70 - 856
85 - 1006
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 intervalFrequencyCumulative Frequency (cf)
Below 14044
140 - 14574 + 7 = 11
145 - 1501811 + 18 = 29
150 - 1551129 + 11 = 40
155 - 160640 + 6 = 46
160 - 165546 + 5 = 51
Totaln = 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 intervalsFrequencyCumulative Frequency (cf)
0 - 10022
100 - 20057
200 - 300x7 + x
300 - 4001219 + x
400 - 5001736 + x
500 - 6002056 + x
600 - 700y56 + x + y
700 - 800965 + x + y
800 - 900772 + x + y
900 - 1000476 + x + y
Totaln = 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.
  Statistics       
  
ClassesNo. of shopsCumulative Frequency (cf)
5 - 1022
10 - 15122 + 12 = 14
15 - 20214 + 2 = 16
20 - 25416 + 4 = 20
25 - 30320 + 3 = 23
30 - 35423 + 4 = 27
35 - 40327 + 3 = 30
Totaln = 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. Statistics 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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