Question

India Meteorological Department observes seasonal and annual rainfall every year in different subdivisions of our country. It helps them to compare and analyse the results. The table given below shows subdivision wise seasonal (monsoon) rainfall (mm) in 2018. Rainfall (mm) Number of subdivisions 200–400 2 400–600 4 600–800 7 800–1000 4 1000–1200 2 1200–1400 3 1400–1600 1 1600–1800 1

Based on the above information, answer the following questions.

1. Write the modal class.
2. Find the median of the given data.
3. Find the mean rainfall in this season.
4. If a subdivision having at least 1000 mm rainfall during monsoon season is considered as a good rainfall subdivision then how many subdivision had good rainfall?

Answer 1

i. Modal class

The modal class is the class interval corresponding to the highest frequency.

Maximum frequency is 7, which corresponds to the class interval 600–800.

Therefore, the modal class is 600–800.

ii. Median rainfall

Step 1: Construct the cumulative frequency table

Rainfall (mm)	Number of subdivisions \((f_i)\)	Cumulative frequency \((cf)\)
200–400	2	2
400–600	4	6
600–800	7	13
800–1000	4	17
1000–1200	2	19
1200–1400	3	22
1400–1600	1	23
1600–1800	1	24
Total	\[N = \sum f_i = 24\]

Step 2: Identify the median class \[\frac{N}{2} = \frac{24}{2} = 12\]

The cumulative frequency just greater than 12 is 13, which corresponds to the class interval 600–800.

Hence, the median class is 600–800.

Lower limit of median class, \[l = 600\]

Class size, \[h = 200\]

Frequency of median class, \[f = 7\]

Cumulative frequency of preceding class, \[cf = 6\]

Step 3: Calculate the median

\[\text{Median} = l + \left[\frac{\frac{N}{2} - cf}{f}\right] \times h\] 

\[\text{Median} = 600 + \left[\frac{12 - 6}{7}\right] \times 200\] 

\[\text{Median} = 600 + \left(\frac{6}{7} \times 200\right)\] 

\[\text{Median} = 600 + \frac{1200}{7}\] 

\[\text{Median} = 600 + 171.43 = 771.43\text{ mm}\]

iii. Mean rainfall

Class Interval	Class Mark \((x_i)\)	Frequency \((f_i)\)	\(f_i x_i\)
200–400	300	2	600
400–600	500	4	2000
600–800	700	7	4900
800–1000	900	4	3600
1000–1200	1100	2	2200
1200–1400	1300	3	3900
1400–1600	1500	1	1500
1600–1800	1700	1	1700
Total		\[\sum f_i = 24\]	\[\sum f_i x_i = 20400\]

Formula: \[\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}\]

Calculation: \[\bar{x} = \frac{20400}{24} = 850\text{ mm}\]

iv. Number of subdivisions with good rainfall

A subdivision having at least 1000 mm rainfall corresponds to all classes from 1000 mm and above 1000–1200, 1200–1400, 1400–1600 and 1600–1800: \[\text{Number of subdivisions} = 2 + 3 + 1 + 1 = 7\]