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Question 6
The following table gives the distribution of students of two sections according to the marks obtained by them:
Section ASection BMarksFrequencyMarksFrequency0−1030−10510−20910−201920−301720−301530−401230−401040−50940−501
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Question 6
The following table gives the distribution of students of two sections according to the marks obtained by them:
Section ASection BMarksFrequencyMarksFrequency0−1030−10510−20910−201920−301720−301530−401230−401040−50940−501
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
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Solution
We can find the class marks of the given class intervals by using the following formula.
Class mark=Upper class limit + Lower class limit2
Section ASection BMarksClass marksFrequencyMarksClass marksFrequency0−10530−105510−2015910−20151920−30251720−30251530−10351230−40351040−5045940−50451
We can find the class marks of the given class intervals by using the following formula.
Class mark=Upper class limit + Lower class limit2
Section ASection BMarksClass marksFrequencyMarksClass marksFrequency0−10530−105510−2015910−20151920−30251720−30251530−10351230−40351040−5045940−50451
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