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.
furye 1
Cumulative Frequency
10
50
60
70
20 30 40
Rainfall in cm
Based on the above graph, answer the following questions :
i) Identify less than type Ogive and more than type Ogive form the given graph.
ii) Find the median rainfall of Dispur.
iii) Obtain the Mode of the data if mean rainfall is 23.4 cm.​

Answer 1

Answer:

hahahahahahahhahaa ha

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

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