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The India Meteorological Department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results.
The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023.
Rainfall (mm) | No. of Sub-divisions
200-400 | 3
400-600 | 4
600-800 | 7
800-1000 | 4
1000-1200 | 3
1200-1400 | 3
Based on the information given above, answer the following questions:
(i) Write the modal class.
(ii) (a) Find the median of the given data.
OR
(b) Find the mean rainfall in the season.
(iii) If a sub-division having at least $800 \operatorname{mm}$ rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall?
The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023.
Rainfall (mm) | No. of Sub-divisions
200-400 | 3
400-600 | 4
600-800 | 7
800-1000 | 4
1000-1200 | 3
1200-1400 | 3
Based on the information given above, answer the following questions:
(i) Write the modal class.
(ii) (a) Find the median of the given data.
OR
(b) Find the mean rainfall in the season.
(iii) If a sub-division having at least $800 \operatorname{mm}$ rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall?
Show SolutionHide Solution↓
Sol. (i) Modal Class $= 600 – 800$
(ii)(a) Rainfall (mm) | No. of Sub-divisions ($f_i$) | $cf$
200-400 | 3 | 3
400-600 | 4 | 7
600-800 | 7 | 14
800-1000 | 4 | 18
1000-1200 | 3 | 21
1200-1400 | 3 | 24
$N = 24$
Median Class $= 600 - 800$
Median $= 600 + \frac{\frac{24}{2} - 7}{7} \times 200$
$= \frac{5200}{7}$ or $742.8 \operatorname{mm}$ (approx.)
OR
(ii)(b) Rainfall (mm) | No. of Sub-divisions ($f_i$) | $x_i$ | $f_i x_i$
200-400 | 3 | 300 | 900
400-600 | 4 | 500 | 2000
600-800 | 7 | 700 | 4900
800-1000 | 4 | 900 | 3600
1000-1200 | 3 | 1100 | 3300
1200-1400 | 3 | 1300 | 3900
$\sum f_i = 24$ | $\sum f_i x_i = 18600$
Mean $= \frac{18600}{24} = 775$
$\therefore$ Mean rainfall $= 775 \operatorname{mm}$
(iii) Required number of sub – divisions $= 4 + 3 + 3 = 10$
(ii)(a) Rainfall (mm) | No. of Sub-divisions ($f_i$) | $cf$
200-400 | 3 | 3
400-600 | 4 | 7
600-800 | 7 | 14
800-1000 | 4 | 18
1000-1200 | 3 | 21
1200-1400 | 3 | 24
$N = 24$
Median Class $= 600 - 800$
Median $= 600 + \frac{\frac{24}{2} - 7}{7} \times 200$
$= \frac{5200}{7}$ or $742.8 \operatorname{mm}$ (approx.)
OR
(ii)(b) Rainfall (mm) | No. of Sub-divisions ($f_i$) | $x_i$ | $f_i x_i$
200-400 | 3 | 300 | 900
400-600 | 4 | 500 | 2000
600-800 | 7 | 700 | 4900
800-1000 | 4 | 900 | 3600
1000-1200 | 3 | 1100 | 3300
1200-1400 | 3 | 1300 | 3900
$\sum f_i = 24$ | $\sum f_i x_i = 18600$
Mean $= \frac{18600}{24} = 775$
$\therefore$ Mean rainfall $= 775 \operatorname{mm}$
(iii) Required number of sub – divisions $= 4 + 3 + 3 = 10$