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Exercise - 14.4

Question-1 :-  The following distribution gives the daily income of 50 workers of a factory. Statistics Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution :-
  
Daily income (in Rs.)(upper class limits)Number of workersCumulative Frequency (cf)
Less than 1201212
Less than 1401412 + 14 = 26
Less than 160826 + 8 = 34
Less than 180634 + 6 = 40
Less than 2001040 + 10 = 50
Totaln = 50
Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows. Statistics

Question-2 :-  During the medical check-up of 35 students of a class, their weights were recorded as follows: Statistics Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution :-
  
Weight (in kg)(upper class limits)Number of students (cf)
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235
Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows. Statistics Here, n = 35. So, n/2 = 35/2 = 17.5 Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore, median of this data is 46.5. Statistics It can be observed that the difference between two consecutive upper class limits is 2.
    
Weight (in kg)(upper class limits)Number of studentsCumulative Frequency (cf)
Less than 3800
38 - 403 - 0 = 33
40 - 425 - 3 = 25
42 - 449 - 5 = 49
44 - 4614 - 9 = 514
46 - 4828 - 14 = 1428
48 - 5032 - 28 = 432
50 - 5235 - 32 = 335
Totaln = 35
Now, n = 35 n/2 = 35/2 = 17.5 is 28. So, this observation lies in the class 46 - 48. Then, l (the lower limit) = 46, cf (the cumulative frequency of the class preceding 46 - 48) = 14, f (the frequency of the median class 46 - 48) = 14, h (the class size) = 2. Therefore, Median = l + [(n/2 - cf)/f] x h = 46 + [(35/2 - 14)/14] x 2 = 46 + [(17.5 - 14)/14] x 2 = 46 + (3.5/14 x 2) = 46 + 3.5/7 = 46 + 0.5 = 46.5 Therefore, the median of data is 46.5. Hence, the value of median is verified.

Question-3 :-  The following table gives production yield per hectare of wheat of 100 farms of a village. Statistics Change the distribution to a more than type distribution, and draw its ogive.

Solution :-
  
Production yeild (lower class limits)Number of farmsCumulative Frequency (cf)
Less than 1202100
Less than 1408100 - 2 = 98
Less than 1601298 - 8 = 90
Less than 1802490 - 12 = 78
Less than 2003878 - 24 = 54
Less than 2001654 - 38 = 16
Totaln = 100
Taking lower class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows. Statistics
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