The following frequency distribution shows the marks obtained by the class of 100 students.
Draw the less than ogive for the data in the attached document. Find mean graphically and verify the result by actual method.
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number of students = 10+15+ 30+32+8+5 = 100
total marks obtained = 15 * 10 + 25 * 15 + 35 * 30 + 45 * 32 + 55 * 8 + 65* 5 ] = 3780
mean of the marks = 3780 / 100 = 37.80
Ogive curve is for plotting the cumulative frequency distribution of students against the less than or more than or mid-points of the marks intervals.
We calculate the Median or 50% percentile or 25 percentile or some other. We do not find the mean of the values by using Ogive curve. It is median we can find.
======================================
Marks < 10 < 20 < 30 < 40 < 50 < 60 < 70
number of
Students 0 10 15 30 32 8 5
Cumulative 0 10 25 55 87 95 100
frequency
=================================================
From the cumulative frequency row, 50th student is in the interval 30-40. We say the median is 40 marks. If we take the midpoint of this interval, then we say that median is 35 marks.
From the graph:
Plotting these data values and drawing the graph , gives us the less than ogive curve. Then at 50% of students size, ie., at 100/2 = 50, draw a line from Y axis horizontally. It cuts the ogive curve. Then draw vertical line from there. Read the marks at that point. That is the median.
For the data using less than ogive curve we get : median = 38. marks.
If we had drawn the normal Ogive using the midpoints of the intervals, then median would be 33 marks.
total marks obtained = 15 * 10 + 25 * 15 + 35 * 30 + 45 * 32 + 55 * 8 + 65* 5 ] = 3780
mean of the marks = 3780 / 100 = 37.80
Ogive curve is for plotting the cumulative frequency distribution of students against the less than or more than or mid-points of the marks intervals.
We calculate the Median or 50% percentile or 25 percentile or some other. We do not find the mean of the values by using Ogive curve. It is median we can find.
======================================
Marks < 10 < 20 < 30 < 40 < 50 < 60 < 70
number of
Students 0 10 15 30 32 8 5
Cumulative 0 10 25 55 87 95 100
frequency
=================================================
From the cumulative frequency row, 50th student is in the interval 30-40. We say the median is 40 marks. If we take the midpoint of this interval, then we say that median is 35 marks.
From the graph:
Plotting these data values and drawing the graph , gives us the less than ogive curve. Then at 50% of students size, ie., at 100/2 = 50, draw a line from Y axis horizontally. It cuts the ogive curve. Then draw vertical line from there. Read the marks at that point. That is the median.
For the data using less than ogive curve we get : median = 38. marks.
If we had drawn the normal Ogive using the midpoints of the intervals, then median would be 33 marks.
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