(i) Construct a histogram and frequency polygon of the following distribution:

Marks 0−10 10−20 20−30 30−40 40−50

Number of Students 8 18 35 25 14

(ii) Show that the area under frequency polygon is equal to the area under histogram.

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Solution

(a)

(b)
The area under a histogram and under a frequency polygon is the same because of the fact that we extend the first class interval to the left by half the size of class interval as the starting point of the frequency polygon. Similarly, the last class interval is extended to the right by the same amount as the end point of the frequency polygon. This ensures that the area that was excluded while joining the mid-points is included in the frequency polygon such that the area under the frequency polygon and the area of histogram is the same.

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Class	5-10	10-15	15-20	20-25	25-30
Frequency	20	30	50	40	10

Class interval	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	30	24	52	28	46	10

Marks	0−10	10−20	20−30	30−40	40−50
Number of Students	8	18	35	25	14

Marks Obtained	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	10	16	20	20	22	15	8	5

Class-interval	0−20	20−30	30−40	40−60	60−100
Frequency	10	4	6	14	16