Mathematics
The data given below represent the marks obtained by 35 students:
21 26 21 20 23 24 22 19 24
26 25 23 26 29 21 24 19 25
26 25 22 23 23 27 26 24 25
30 25 23 28 28 24 28 28
Taking class intervals 19 - 20, 21 - 22 etc., make a frequency distribution for the above data.
Construct a combined histogram and frequency polygon for the distribution.
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Answer
From above data :
Least mark = 19
Greatest marks = 30
Range = 30 – 19 = 11.
Given,
We need to take class intervals as 19 - 20, 21 - 22 etc.
The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,
Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2
=
= 0.5
Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.
Continuous frequency distribution for given data is :
Classes before adjustment | Classes after adjustment | Class mark | Frequency |
---|---|---|---|
19 - 20 | 18.5 - 20.5 | 19.5 | 3 |
21 - 22 | 20.5 - 22.5 | 21.5 | 5 |
23 - 24 | 22.5 - 24.5 | 23.5 | 10 |
25 - 26 | 24.5 - 26.5 | 25.5 | 10 |
27 - 28 | 26.5 - 28.5 | 27.5 | 5 |
29 - 30 | 28.5 - 30.5 | 29.5 | 2 |
Steps of construction of histogram :
Since, the scale on x-axis starts at 16.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 16.5.
Take 2 cm along x-axis = 2 units.
Take 1 cm along y-axis = 2 units.
Construct rectangles corresponding to the above continuous frequency distribution table.
The required histogram is shown in the adjoining figure.
Steps of construction of frequency polygon :
Mark the mid-points of upper bases of rectangles of the histogram.
Join the consecutive mid-points by line segments.
Join the first end point with the mid-point of class 16.5 - 18.5 with zero frequency, and join the other end point with the mid-point of class 30.5 - 32.5 with zero frequency.
The required frequency polygon is shown by thick line segments in the diagram.
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