Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way.
Class intervalFrequency(km/h)30−40340−50650−602560−706570−805080−902890−10014
Draw a histogram and frequency polygon representing the data above.
Class intervalFrequency(km/h)30−40340−50650−602560−706570−805080−902890−10014
Draw a histogram and frequency polygon representing the data above.
Clearly, the given frequency distribution is exclusive form.
Along the horizontal axis, we represent the class intervals of length on some suitable scale.
The corresponding frequencies are represented along the vertical axis on a suitable scale.
We construct rectangles with class intervals as the bases and the respective frequencies as the heights.
Let us draw a histogram for this data and mark the midpoints of the top of the rectangles as B, C, D, E, F, G and H respectively. Here, the first class is 30 - 40 and the last class is 90 - 100.
Also, consider the imagined classes 20 - 30 and 100 - 110 each with frequency 0. The class marks of these classes are 25 and 105 at the points A and l, respectively.
Join all these points using a dotted line.
Then, the curve ABCDEFGHI is the required frequency polygon.
Clearly, the given frequency distribution is exclusive form.
Along the horizontal axis, we represent the class intervals of length on some suitable scale.
The corresponding frequencies are represented along the vertical axis on a suitable scale.
We construct rectangles with class intervals as the bases and the respective frequencies as the heights.
Let us draw a histogram for this data and mark the midpoints of the top of the rectangles as B, C, D, E, F, G and H respectively. Here, the first class is 30 - 40 and the last class is 90 - 100.
Also, consider the imagined classes 20 - 30 and 100 - 110 each with frequency 0. The class marks of these classes are 25 and 105 at the points A and l, respectively.
Join all these points using a dotted line.
Then, the curve ABCDEFGHI is the required frequency polygon.
