Question

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 ogive and more than ogive from the given graph.
(ii)Find the median rainfall of Dispur.
(iii)Obtain the mode of the data if the mean rainfall is 23.4 cm. ​

Answer 1

Step-by-step explanation:

• Less than ogive is curve 1
• More than ogive is curve 2

Answer 2

median = 21 cm , mode = 16.2 cm

Step-by-step explanation:

here ,(i) Curve 1 is the less than ogive (because , in less than ogive , the cumulative frequency increases)

curve 2 is the more than ogive

(ii) Median is the point where less than and more than ogive meet

drawing the perpendicular line from that point ,

we see that

median rainfall = 21 cm

(iii) now , we know that ,

3 median = mode + 2 mean

putting median = 21 , mean = 23.4

3 × 21 = mode + 2×23.4

63-46.8 = mode

mode = 16.2 cm

hence , median = 21 cm , mode = 16.2 cm

#Learn more :

What is the value of the median of the data using the graph in figure of less than ogive and more than ogive?​

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