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Sum
The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.
No. of Mangoes | 50 - 100 | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 |
No. of trees | 33 | 30 | 90 | 80 | 17 |
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Solution
Class
(Number of working hours) |
Frequency (Number of workers) fi |
Cumulaive frequency less than the upper limit |
50 - 100 | 33 | 33 |
100 - 150 | 30 | 63 |
150 - 200 (Median Class) |
90 | 153 |
200 - 250 | 80 | 233 |
250 - 300 | 17 | 250 |
N = 250 |
From the above table, we get
L (Lower class limit of the median class) = 150
N (Sum of frequencies) = 250
h (Class interval of the median class) = 50
f (Frequency of the median class) = 90
cf (Cumulative frequency of the class preceding the median class) = 63
Now, Median = \[L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\]
\[= 150 + \left( \frac{\frac{250}{2} - 63}{90} \right) \times 50\]
= 150 + 34.44
= 184.44 mangoes
= 184 mangoes
Hence, the median of data is 184 mangoes.
Concept: Tabulation of Data
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