A TV reporter was given a task to prepare a report on the rainfall of the city Dispur of India in a particular year. After collecting the data, he analyzed the data and prepared a report on the rainfall of the city. Using this report, he drew the following graph for a particular time period of 66 days.

Based on the above graph, answer the following questions:

(i) Identify less than type ogive and more than type ogive from the given graph.

(ii) Find the median rainfall of Dispur

(iii) Obtain the Mode of the data if mean rainfall is 23.4cm

Let us prepare table for cumulative frequencies for curve 1 and 2

(i) For curve 1 first cumulative frequency is least and last one is highest. Hence, it less than type ogive. Whereas for curve 2 first cumulative frequency is heighest and last one is least. Hence, it greater than type ogive.

(ii) Calculation of median :

Here, total frequency = ∑ f_i = N = 66

cumulative frequency just greater than or equal to 33 is 40.

Therefore, corresponding class, 20 -30 is a median class.

Here, L = 20, c.f. = 32, h = 10, f = 8

∴ Median = 20 + 1.25

∴ Median = 21.25

Hence, median rainfall of Dispur is 21.25 cm.

(iii) Mean rainfall = 23.4 cm

We know that,

Mean – Mode = 3 (Mean – Median)

∴ 23.4 – Mode = 3 (23.4 – 21.25)

∴ 23.4 – Mode = 3 (2.15)

∴ 23.4 – Mode = 6.45

∴ Mode = 23.4 – 6.45

∴ Mode = 16.95

Hence, Modal rainfall of Dispur is 16.95 cm.